AP Physics Final exam

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For which of the cases shown above is the acceleration of the mass (FA+mg sinθ)/m?

Ramp1 only Feedback

A crane lowers a 450. kg steel beam as shown above at a constant velocity of 2.50 m/s. Find the tension in the vertical cable.

4410 N

Convert 0.092 kg to standard units

92 g

A quantity that measures the resistance of a body to a change in its motion is

mass

The prefix for one one-thousandth (10-3) is

milli

A 15 kg object is pulled to the left by a tension of 40. N applied at 30o above the horizontal. A second force R is applied such that the object moves at a constant velocity. What is the work done by the weight, W, as the object moves 5.0 m to the left?

0 J W=Fx were F is the component of the force in the axis of the displacement x. As the weight is perpendicular to the motion the work done by the weight is 0 J.

Convert 0.092 kg to standard units

0.092 kg

Assume an astronaut, including his space suit, has a mass of 150kg. In space, if he experiences an acceleration of 0.45 m/s2 as he pushes on a 500. kg satellite, what is the magnitude of the acceleration of the satellite?

0.14 m/s2 F = ma = 150 kg * 0.45 m/s2 = 67.5 N By the third law, the reaction force has the same magnitude: a = F/m = 67.5N / 500 kg = 0.135 m/s2

Convert 814 mm to standard units.

0.814 m

A 40.0 kg box is being pulled up a 25.00 ramp by a force of 320. N which is directed parallel to the ramp. The kinetic coefficient of friction between the box and the ramp is 0.300. What is the acceleration of the box?

1.18 m/s2

A man is walking along a flatbed car on a train in the same direction the train is traveling. If the train's speed is 11 m/s east and the man is walking 3.0 m/s relative to the train, how fast is the man traveling relative to a stationary observe standing next to the train?

14 m/s east vMT + vTG = vMG 3 east + 11 east = 14 east

A 2000 kg truck rounds an unbanked curve with a 50m radius. The static coefficient friction between the tires and the road is 0.450. Find the maximum safe speed of the truck so that it does not slip on the curve.

14.8 m/s

A box is lifted by two ropes. The first rope exerts a force of 600 N at an angle of 35° to the right of vertical. The second cable exerts a force of 1300 N. If the box is accelerating directly upward, at what angle to the left of the vertical is the second rope pulling?

15° For the box to accelerate directly upward, the forces in the x direction must sum to zero. Since the angle is measured from vertical, the x component are the force times the sin of the angle. 600 sin 35°-1300 sinθ=0. 344=1300 sinθ. θ = sin-1(344/1300) = 15°

How far must a farmer push his 250 kg to do 1200 J worth of work on it if he pushes with a constant 75 N of force?

16 m W=Fx x=W/F=1200/75=16m

A man is walking along a flatbed car on a train in the same direction the train is traveling. If the train's speed is 15 m/s south and the man is walking 2.0 m/s relative to the train, how fast is the man traveling relative to a stationary observe standing next to the train?

17 m/s south vMT + vTG = vMG 2 south + 15 south = 17 south

A car is observed from above traveling in the positive x direction with an initial speed of 15 m/s and then begins accelerating at a constant rate. The car accelerates at 5.0 m/s2 in the y axis for 2.0 s. What is the magnitude of the car's velocity after 2.0 s?

18 m/s x velocity remains 15 y velocity v=v0+at v=0+(5)(2)=10 d=sqrt(152+102)=18 m/s

A man goes out for a walk. He first walks 120 m and then rests. He then continues on his walk for another 64.8 m. What is the total length of his walk?

180 m 120+64.8=184.8 Significant figures are limited by 120 as it does not have a significant digit in the 1's column. So the answer rounded to the nearest 10's column is 180 m.rounds up to 520 km.

A force is applied to a 0.25kg puck on an air hockey table resulting in a 75 m/s2 acceleration, find the net force on the puck.

19 N F=ma F=(0.25)(75)=19 N

Mass, m2, accelerates to the right on a horizontal tabletop with a kinetic coefficient of friction of 0.150. It is connected over a pulley to m1 on the left and over another pulley to m3 on the right as shown above. Assume frictionless, massless pulleys. If m1 = 2.0 kg, m2 = 8.0 kg, and m3 = 7.0 kg, find the acceleration of m2.

2.19 m/s2

A 0.17 kg ball, traveling in the +x direction at 3.0 m/s, strikes a stationary second ball of the same mass. If the second ball is traveling 1.5 m/s at 60° below the +x axis after the collision, while the first ball deflects 30° above the +x axis after the collision, what is the final speed of the first ball?

2.6 m/s

What is the displacement of a particle that has a position versus time of x(t) = t3-3t from t= 1.0 s to the particle's final position at t=3.0s?

A 40.0 kg crate is pulled along a floor with a kinetic coefficient between the crate and the floor of 0.300. What horizontal force is required to accelerate the crate to 2.20 m/s2?

207 N The correct answer is 207 N. From an FBD, summing the forces in the y direction leads to N=mg. In the x direction, FA - f = ma. FA = ma+f = ma + μkN = ma + μk mg = 40*2.2 + 0.3*40*9.8 = 207N

The graph above shows the component of force in the direction of displacement as a function of displacement. Determine the work done by the force acting on an object as the object is displaced from 0m to 24 m.

-6 J From 0m to 12m, the area of the under the graph is the triangle ½·12·(-4)= -24J. From 12m to 18m, the area of the under the graph the triangle ½·6·4=12J. From 18m to 21m, the area of the under the graph the triangle ½·3·4=6J. From 21m to 24m, the area is zero The total work done on the object from 0m to 24m is (-24) +12 + 6 = -6 J.

A man is running 12 m/s at 120o. What is the x component of his velocity?

-6.0 m/s vx=vcos() vx=12cos(120)=-6.0 m/s If your calculator shows 9.8 you are in radians and need to be in degrees

68 N Fa = ma = 150 kg * 0.45 m/s2 = 67.5 N By the third law, the reaction force has the same magnitude: Fs=68 N (Sig Figs)

A constant net force of 30. N is applied to a 15 kg box initially at rest. What is the velocity of this box after 4.0 s?

8.0 m/s F=ma a=F/m=30/15=2 v=v0+at v=0+(2)(4)=8.0 m/s

A constant net force of 30. N is applied to a 15 kg box initially at rest. What is the velocity of this box after 4.0 s?

8.0 m/s F=ma a=F/m=30/15=2 v=v0+at v=0+(2)(4)=8.0 m/s

The energy required to run a 60W light bulb for one hour would be enough energy to lift a 25 kg child through a vertical distance of

880 m P = W/t = mgh / t. h = P·t / (mg) = 60·3600 / (25·9.8) = 880 m.

If the momentum of a mass M is increased by a factor of 3, its kinetic energy will be multiplied by a factor of

9 For a given mass, M, if momentum, Mv, is multiplied by 3, v must increase to 3v. In K= ½ Mv2, if v increase to 3v K2= ½ M(3v)2 K2= ½ M 9v2 K2=9K

A 0.3 kg ball is thrown against a wall with the force on the ball represented by the graph above. Determine the magnitude of the impulse imparted to the ball.

90 N·s

Neglecting air resistance, at what angle should a garden hose be aimed in order for the water to land with the greatest flight time?

90°

A box resting on a table experiences the force of gravity pulling downward and the normal force from the table pushing upward. These two forces are opposite in direction and

According to Newton's first law, if the net force on an object is zero (gravity pulling down on the box, while the table is pushing up on the box with a force of the same magnitude), the object is in equilibrium, stationary in this case.

Using the a-t graph above determine the change in velocity from t=6.0s to t=10.0s.

Correct answer: -28 m/s For an a-t graph the change in velocity between two points is the area bound by that graph v=area=1/2BxH v=1/2(4)(-14)=-28 m/s

A ball mass M is traveling to the right with a velocity 2v when it collides with a ball mass M traveling to the left at v. The first ball bounces back with a velocity -v. What is the velocity of the second ball?

Correct answer: 2v M(2v)+(M)(-v)=M(-v)+Mvf Mv=-Mv+Mvf 2Mv=Mvf vf=2v

Accelerating a 2.0 kg object from 4.0 m/s to 8.0m/s requires what magnitude of applied force, if the force is applied for 0.25s?

Correct answer: 32 N Δt=Δp=mvf - mvf . F= m(vf - v0) / Δt = 2·(8-4) / 0.25 = 32N.

Force is measure in Newtons (kgm/s2), acceleration in m/s2, and mass in kilograms. Which of the following equations would be dimensionally correct?

F=ma

A mass m is pulled up a ramp with a kinetic coefficient of friction μk. Which of the following vector orientations is most appropriate for a free body diagram?

only south one is erected

In perfectly elastic collison

Momentum is conserved Kinetic Energy is conserved

If the vector sum of all of the forces acting on an object is NOT zero,

the object's velocity will change.

The nucleus of an atom contains which of the following particles?

Neutrons and protons

A 3050kg car travels around a test track at 10.0m/s. The radius of the track, R, is 45.0m. A. The coefficient of friction between the road and the track is 0.400. What is the maximum force of friction between the road and the car? B. What is the maximum velocity that the car can travel as it circles the track? C. What is the actual centripetal acceleration of the car as it travels around the track?

22B. This equation is for orbital motion, G is the universal gravitational constant C. you are given the velocity and a=v^2/r

What is the kinetic energy of a 2 kg mass whose linear momentum is 6 kg·m/s?

9 J v=p/m=6/2=3m/s. K = ½ mv2= ½ · 2 · 32 = 9J.

Which of the following statements is true about momentum?

Momentum and impulse have the same units. The correct answer is "Momentum and impulse are measured in the same units." Kg m/s is the same as Ns=(kg· m/s2)· s

Apparent weight is equal to your weight when the surface you are standing on

Moves at a constant velocity or is at rest. From the FBD, N-mg=ma, so N=mg+ma = mg + 0 for an object at rest or travelling at a constant velocity.

Newtonian mechanics is not concerned with

The make up of protons and neutrons

Two masses m1 and m2 are connected in a "modified Atwood's" machine as shown above. Assume that the table is frictionless and the pulley is frictionless and massless. m2 accelerates downward. Compare the tension in the string to the weight of m2. As the system accelerates in the direction shown,

The tension is smaller than m2g

The "net force" applied to an object is what causes:

an object's change in velocity.

Objects that are described by Newton's third law must:

have forces applied that always exist in pairs.

For two objects moving as system

internal forces can not cause an acceleration of the system

An onion traveling northward at 4 m/s collides with a stationary watermelon. The onion bounces back to the south, while the watermelon moves very slowly to the north. During the collision, the greater magnitude of impulse is experienced by

neither; the magnitude of their impulses is the same.

A ball mass M is traveling to the right with a velocity 2v when it collides with a ball mass M traveling to the left at v. The two stick together upon impact. What is the velocity of the system after the collision?

v/2 M(2v)+(M)(-v)=(M+M)vf Mv=2Mvf v/2=vf

Starting from rest, a 4.0 kg box slides for a distance of 5.0 m down a ramp making an angle of 60° with horizontal. Find the work done on the box by friction if the kinetic coefficient of friction between the ramp and the box is 0.20.

- 19.6 J Wfriction. = μkmg cosθ·s · cos180° = 0.2 · 4 · 9.8 · 0.5 · 5 (-1) = - 19.6

A car traveling in a straight line at 14 m/s when the driver puts on the brakes and comes to a top in 35 s. What is the average acceleration of the car while it is braking?

-0.40 m/s2 v0=14, v=0 (stops), t=35, a=? v=v0+at a=(v-v0)/t a=(0-14)/35=-0.400 m/s2

A 42.5 kg girl is traveling across a lake in a 16 kg ultralight canoe at 0.15 m/s toward the shore. She stops rowing and someone on the shore throws a 3.0 kg ball at 5.0 m/s to the girl. Find the velocity of the girl after she catches the ball.

0.10 m/s backward Treating forward as positive, the girl and the canoe have a combined mass of 58.5 kg. m1v10+m2v20 = (m1+m2)vf 58.5 · 0.15 + 3 · (-5) = (58.5 + 3)·v1f -6.23 = 61.5· v1f v1f = -0.1 m/s , negative answer indicates backward.

A 3.0 kg toy truck is moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0 kg toy car cart that is initially moving to the left with a speed of 2.0 m/s. After the collision, the 3.0 kg car is moving to the left with a speed of 1 m/s. What is the final velocity of the 5.0 kg car?

0.8 m/s to the left m1v10+m2v20 = m1v1f+m2v2f . 3·1 + 5·(-2) = 3·(-1)+5v2f . -7=-3+5v2f -4 = 5v2f v2f = -4/5 = 0.8 m/s to the left

An elevator is accelerating with a 57.0 kg girl standing on a scale in an elevator. If the scale reads 450. N, what is the magnitude of the acceleration of the elevator?

1.91 m/s2 From the FBD, N-mg = ma, a=(N-mg)/m = (450-57*9.8)/57 = 1.91 m/s2

Convert 19.2 hours to minutes.

1150 min

150 J The area under the graph is 30 squares, each square representing 1m·5N = 5J. 30·5J = 150J.

The force on a 1.50 kg stone dropped onto a surface is represented by the graph above. Determine the magnitude of the impulse imparted to the surface.

24 N·s The area under the graph is 4 N·s + 12 N·s + 8 N·s = 24 N·s

Sarah is traveling west toward an intersection with a constant velocity of 21.2 m/s. John is traveling north toward the same intersection at a constant velocity of 12.8 m/s. How fast is Sarah travelling relative to John?

24.8 m/s, 31.1 degrees south of west. vJG=12.8 m/s north, vSG=21.2m/s west vSJ = vSG + vGJ = vSG + (-vJG) vSJ = 21.2m/s west + 12.8 m/s south (opposite of John's north) west and south point to the third quadrant, south of west θSJ = 31.1 degrees |vSJ|=√(21.22+12.82) = 24.8 m/s

If you are lowering a book weighing 5.0 N by applying an upward force of 3.0N on the book, the magnitude of the normal force on your hand due to the book is:

3.0 N

If you are applying an upward force of 3.0 N to a book weighing 2.0N, the normal force on your hand due to the book is:

3.0 N downward By Newton's third law, the force of the hand on the book (3.0 N upward) is equal and opposite to the force of the book on the hand (3.0 N downward).

A boy pushes his 15 kg toy car across a level floor for 6.0 m. How much force did he apply if he did 20. J of work on the car?

3.3 N W=Fx F=W/x=20/6=3.3 N

An astronaut weighing 85.0 N on the Moon, where the acceleration of gravity is 1.62 m/s2, has what weight on Earth?

510 N W=mg m=W/g=85/1.62= 52.5 on Earth W=(52.5)(9.8)=510 N

An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.

59 N in the negative x direction

Find the mass of a truck that accelerates from rest to 35.0 m/s in a distance of 150. m when a net force of 2500 N is applied to it.

610 kg v2=v02+2ax. a=(v2-v02)/(2x)=(352)/(300)=4.083 m/s2. m=F/a=2500/4.08=610 kg

What is the area of a circle with a 4.578 cm radius? (carefully consider the number of significant figures)

65.8418 cm2

A cannon sits on the edge of a castle wall that is 5.0 m above the ground. A cannon ball leaves the cannon with an initial velocity of 25 m/s at 30o to the horizontal. At its highest point how far has the cannon ball traveled vertically?

8.0 m at max height vy=0 v0y=vsin() v0y=25sin(30)=12.5 vy2=v0y2+2ay 0=(12.5)2+2(-9.8)y y=8.0 m

An airplane flying South at 85 m/s encounters a wind that blows on the plane at 35 m/s West. What is the magnitude of the planes velocity relative to an observer on the ground?

92 m/s The two velocity values are at right angles to each other and can be treated as components of the resultant velocity. v=sqrt(vx2+vy2) v=sqrt(852+352)=92 m/s

A stone is in free fall from the top of a tall hill. While in free fall, the stone will gain an equal amount of

Which of the following results in zero net work being done?

A box is pulled across a rough floor at constant velocity.

A toy car travels 218 mm in 1.50 hours. A. Convert each of the units into standard units. B. What is the speed of the car if speed=distance/time? C. Convert your answer to Engineering Notation.

A. 218 mm (1 m/1000 mm) = 0.218 m 2 Points 1.50 hr (60min/hr)(60s/min) = 5400 s 2 Points B. 0.218/5400 = 0.00004037 2 Points 0.0000404 or 4.04 x 10-5 m/s 2 Points C. 40.4 x 10-6 m/s 2 Points

A small boy stands on the edge of a cliff overlooking the ocean 15.0m below. He throws a rock at 10.0 m/s, 35.0o above the horizontal x axis. A. What is the highest point reached by the rock? B. What is the final velocity of the rock?

A. v0y=v0sinθ v0y=10sin(35)=5.74 m/s 1 Point for recognizing that vy=0 at the 1 Point max height vy2=v0y2+2ay y=(vy2-v0y2)/2a y=((0)2-(5.74)2)/2(-9.8) 1 Point y=1.68m 1 Point B. v0x=v0cosθ v0x=10cos(35)=8.19 m/s 1 Point vy2=v0y2+2ay 1 Point vy2=(5.74)2+2(-9.8)(-15)=327 vy=18.1 m/s 1 Point for recognizing that the velocity must be downward vy=-18.1 1 Point v=√(vx2+vy2) v=√((8.19)2+(-18.1)2)=19.9 m/s 1 Point θ=tan-1(vy/vx) θ=tan-1(-18.1/8.19)=-65.7o 1 Point or 65.7o below the x axis

For the velocity to remain constant the x component of the tension, T must be equal to the force R. R points opposite to the motion so either the motion or the force must be used as a negative value. R=Tcos() W=Tcos()x W=(40)cos(30)(-5)=-170 J

A ball attached to a string of length R is twirled in a horizontal circle at a constant speed v. If the radius of the circle is changed to ¼ R and the same centripetal force is applied by the string, the new speed of the ball is

One-half the original speed

One-half the original speed a = v12/r. v1 = √(ar). v2= √[a*(1/4)r]=1/2 √(ar)= 1/2 v1 .

A flying squirrel can spread its body out, forming kind of a parachute to take advantage of wind resistance. In doing so, they can easily reach terminal velocity, meaning they can fall with a constant speed. This also means

The force of air resistance on the flying squirrel is equal and opposite to its weight.

Two objects are dropped from the same height. One object is twice the mass of the other. Compare the kinetic energies of the objects as they reach the ground.

The heavier object has twice the kinetic energy as the other. -Both objects start from the same height. Set the ground as UG=0 J. Then the light object starts with UG = mgh and the heavy object starts with UG = (2m)gh = 2 mgh, twice as much. All of this energy will be converted to kinetic energy, twice as much for the heavy object.

A ball is thrown directly upward. Neglecting air resistance, which one of the following statements describes the energy transformation of the ball as it rises?

The kinetic energy decreases and the potential energy increases.

A ball is thrown directly upward. Neglecting air resistance, which one of the following statements is true about the net force acting on the ball at the top of the path?

The net force is always directed downward throughout the entire freefall problem, on the way up, on the way down AND at the top; and the magnitude of the net force is mg, the weight of the ball.

Object A is twice the mass of object B, As mass A collides with mass B, compare the magnitudes of the impulse that A imparts to B with the impulse that B imparts to A.

The same impulse as B.

Which one of the following is characteristic of an elastic collision?

Total kinetic energy is conserved.

An inelastic collision of two objects is characterized by which of the following?

Total momentum of the system is conserved.

A passenger in a car is throwing a ball into the air while the car travelling at a consent speed making a making a hard right turn. Which of the following is INCORRECT?

Turning is a change in direction, therefore a change to the velocity vector. So the car, being the reference frame, is accelerating and is NOT an inertial reference frame.

A woman takes a stroll to the local store. She covers 2.72km in 37.5min. What is her average speed in meters per second?

d=2.72km(1000m/1km)=2720m 1 Point t=37.5min(60s/min)=2250s 1 Point s=d/t=2720/2250=1.21m/s 1 Point for correctly showing work and having a between 2 and 4 digits in the answer and having the correct units 1 Point

Which of the following variables can be different in the x and y axis for a moving object?

displacement (x or y) acceleration (a) velocity (v)

In the trajectory shown below, the velocity vector at point B is best represented by which of the following vectors?

east

In the trajectory shown below, the velocity vector at point C is best represented by which of the following vectors?

east/south

Two masses m2 and m1 = 3m2 are connected in a "modified Atwood's" machine as shown above. Assume that the table is frictionless and the pulley is frictionless and massless. As m2 accelerates downward, the acceleration of the masses is

g/4 3m2 x a = Tm2a = m2g -T4m2 x a = m2 x ga = g/4----the correct answer

An object in motion has a constant force applied at a 1200 angle to the displacement of an object in motion. The work done by the force

is negative. Whatever the magnitudes of force and displacement, cos 1200 = -0.5, so the work will be negative.

The prefix for one thousand (103) is

kilo

A net force is exerted on an object toward the north. The object

may be moving in any direction.

In the trajectory shown below, the acceleration vector at point B is best represented by which of the following vectors?

south

In the trajectory shown below, the y component of the velocity vector at point C is best represented by which of the following vectors?

south

An object moves with an initial velocity vx and accelerates in the y axis with an acceleration ay.

the equation y=1/2ayt2 is valid the equation x=vxt is valid

A plane is flying horizontally with its engines applying a force. A strong wind blows upward at an angle pushing the plane to go faster. Which of the following could be the FBD for this situation?

west,north/west/south

Given two forces equal in magnitude, F1 pointing upward and F2 pointing to the left, which of the following vectors best represents the net force?

west/ north

Newton's first law does NOT apply to objects:

when the reference frame is accelerating.

A man is running 12 m/s at an angle of -75o. What is the y component of his velocity?

-12 m/s vy=vsin() vy=12sin(-75)=-12 m/s (with two sig figs) If your calculator shows 11 you are in radians and need to be in degrees

A cannon sits on the edge of a castle wall that is 5.0 m above the ground. A cannon ball leaves the cannon with an initial velocity of 25 m/s at 30o to the horizontal. What is the vertical component of the velocity when the object strikes the ground?

-16 m/s vy=25sin(30)=12.5 vy2=v0y2+2ay vy2=12.52+2(-9.8)(-5) vy2=254 vy=16, but it is down so vy=-16 m/s

A spring balance has a spring constant of 8000. N/m. If m=30.0 kg are placed on the spring balance, the displacement due to the weight of the mass is

0.0368 m

A spring balance has a spring constant of 8000. N/m. If m=30.0 kg are placed on the spring balance, the displacement due to the weight of the mass is

0.0368 m From F=-kx, mg = 294N, so the spring is pulling back with an equal and opposite restoring force. -294 = -8000 x, x = 294/8000 = 0.0368 m

A 0.10 kg toy car is rolled at 10m/s toward a stack of wooden blocks. If the car bounced off of the blocks with a speed of 8m/s, knocking the blocks over, what was the maximum contact time between the car and the blocks if 45N is the minimum force required to knock over the blocks.

0.040 s F·Δt=Δp=mΔv. Δt=( mΔv)/F = [0.1·(8-(-10))] / (-45) = (0.1·18)/45=0.040s.

1.18 m/s2 Summing the forces in the y direction find that N=mg cosθ = 40*9.8*cos250=355N. Summing the forces in the x direction, FA-mgsinθ-f = ma, FA-mgsinθ-μkN = ma, a=( FA-mgsinθ-μkN)/m = (320-166-0.3*355)=47.8/40=1.18m/s2

How much power is required to lift a 49 kg crate a vertical distance of 5.0 m in 20.0 s?

120 W

Two boxes m1 and m2 are resting on a frictionless surface and connected by a rope. A horizontal force, F, pulls box m1 to the right. Box m2 is also pulled to the right since it is connected to box m1. If the acceleration of both boxes is 2.43 m/s2, the mass of m1 is 43.0 kg, and the mass of m2 is 20.0 kg, find the force, F.

153 N The acceleration of the system is a = F/( m1 + m2). ( m1 + m2)a = F. F=2.43 (20 +43) = 153 N

A soccer player kicks a 0.410 kg soccer ball from rest. If the player's foot was in contact with the ball for 0.0510 s and the ball's initial speed after the collision was 21.0 m/s, what was the magnitude of the average force on the ball?

169 N F·Δt=mv-0, F = mv / Δt = 0.41·21 / 0.051 = 169N.

A constant force of 45 N is applied at an angle θ with the horizontal to slide a box along a horizontal surface for a distance of 4.0 meters. If the vertical component of the force is 12N, find the work done by the force when moving the box.

173 J The force in the direction of displacement needs to be found, which requires knowing the angle. This is found from the vertical component of the force, even though the vertical component does no work. sinθ = (12/45) = 15.50 . W=Fs cosθ = 45·4·cos 15.50 = 173 J.

A 0.1 kg stone is dropped onto the floor and bounces straight up with a 10 m/s speed just before and just after the collision. Find the magnitude of impulse on the stone as a result of the collision.

2 N·s J=Δp=mvf - mvf = m(vf - v0) = 0.1·(10-(-10)] = 2N·s.

What is the displacement of a particle that has a position versus time of x(t) = t3-3t from t= 1.0 s to the particle's final position at t=3.0s?

20 displacement = final displacement - initial displacement = x(3s) - x(1s) = 18 - (-2) = 20 m

A 30.0 kg box slides down a frictionless ramp inclined 27.00 with horizontal. Find the normal force on the box.

262 N

Two forces, equal in magnitude, are pulling an object forward 15.0 meters across a surface. The magnitude of each force is 12.0N, but one is pulling at 35.0° to the left of the forward direction and the second force is pulling at 35.0° to the right of forward. Find the total work done by both forces.

295 J The total work is the scalar sum of the work done by each force. W=12·15·cos35° + 12·15·cos35° = 295 J.

A 42.5 kg boy is traveling across a lake in a 16 kg ultralight canoe at 0.15 m/s. If he stops rowing and wishes to stop the canoe throwing his 2.5 kg oar, how fast and in what direction would he need to throw the oar relative to the lake?

3.7 m/s, forward (m1+m2)v0 = m1v1f+m2v2f . (58.5 +2.5)· 0.15 = 0 + 2.5 v1f v1f = 9.15 / 2.5 = 3.66 m/s, positive answer indicates forward.

A crane lifts a 250. kg steel beam as shown above at a constant acceleration of 2.50 m/s2. Find the tension in the vertical cable.

3080 N The correct answer is 3080 N. From an FBD which includes as a single object T1, T2 and the steel beam, the force vectors are T3 up and mg down. Summing the forces in the y direction results in T3 -mg=ma. T3 =ma+mg=m(a+g)=250*(2.5+9.8)=3075N.

What is the mass of a child with a weight of 350 N at the surface of the Earth?

36 kg W=mg m=W/g=(350)/(9.8)= 36 N

A 10. kg rock is dropped from rest from a 79m tower. How long does it take for the rock to reach the ground?

4.0 s y=v0t+1/2at2 v0=0 -79=1/2(-9.8)t2 -158=-9.8t2 16.1=t2 t=4.0s

Given two vectors: A=4i+ 2j and B=5i-3j Find the magnitude of: A-B

5.1

What is the momentum of a child weighing 437. N walking in a straight line at 1.22 m/s?

54.5 kg·m/s

A ball mass 2M is traveling to the right with a velocity v when it collides with a ball mass M traveling to the left at 2v. The first ball bounces back with a velocity -v. What is the velocity of the second ball?

Correct answer: 2v Using the equation for a collision where the two objects do not stick together 2Mv+(M)(-2v)=2M(-v)+Mvf 0=-2Mv+Mvf 2Mv=Mvf vf=2v

An object has a net force, F, applied to it resulting in acceleration, a. If the same object accelerates at another time with an acceleration 4a, what must be the net force on the object?

Correct answer: 4F F=ma m=F/a F/a=F2/4a F2=(F)(4a)/a=4F

A roller coaster car is on a vertical circular loop with radius, r. What is the minimum speed the car must travel if the car is to NOT fall off the track?

Correct answer: sqrt(rg)

Which of the following is a fundamental unit of the International System of Units?

Second

A piece of clay traveling at 20. m/s collides and sticks to a door that is stationary. Which of the following statements could be true. I. The total momentum of the clay and door is equal to the original momentum of the clay. II. The total kinetic energy of the clay and door is equal to the original kinetic energy of the clay.

Statement I only.

An object moves initially in the x axis with a velocity vx and accelerates in the y axis with an acceleration ay.

The acceleration must cause an increase in the magnitude of the velocity.

Which of the following are NOT vectors? (Please choose all that are correct)

The age of the Earth A persons height

A mass, m, is at rest on a ramp at an angle of θ with horizontal. If the block is not in motions, the magnitude of the frictional force on the block is equal to

mg sin θ

A golf ball and a lump of clay have equal mass and are thrown with equal speed against a wall. The golf ball bounces back with nearly its initial speed while the clay sticks to the wall. Which objects experiences the greater momentum change?

the golf ball Golf ball: pf-pi=-mv-mv=-2mv. The clay: 0-mv = -mv.

53.3 N The correct answer is 53.3 N. Assume +y is up for m1, to the right for m2 and down for m3. Friction acts only on m2 and opposes the rightward motion. Treating the tension as an internal force, the acceleration of the system is a=( m3g - m1g - μkm2g) / (m1+m2+m3) = 9.8( 7-2-0.15*8) / (2+8+7) = 2.19 m/s2. From m3 FBD, m3g-T = m3a, T = m3g- m1a = m1(g-a) = 7*(9.8-2.19) = 53.3 N

-12 J From 0m to 12m, the area of the under the graph is the triangle ½·12·(-4) = -24J. From 12m to 18m, the area of the under the graph the triangle ½·6·4 = 12J. The total work done on the object from 0m to 24m is (-24) +12= -12 J.

A cannon sits on the edge of a castle wall that is 15 m above the ground. A cannon ball leaves the cannon with an initial velocity of 25 m/s at 30o to the horizontal. After 3.0 s What is the y component of the velocity?

-17 m/s v0y=vsin() v0y=25sin(30)=12.5 vy=v0y+at vy=12.5-(9.8)(3)=-17 m/s

A woman swims directly across a river at 3.0 m/s but the water is running downstream at 5.0 m/s. Which of the following could be considered the angle of the woman's velocity relative to the shore?

-31o The shore must be the x axis for the angle to be relative to the shore. No indication of any direction being positive so ()=tan-1(y/x) ()=tan-1(3/5)=31o other possible angles would be -31o if 3 was negative 149o if 5 was negative 211o if both values were negative

A 2.0 kg ball has a velocity of 6.0 m/s to the right. If the momentum change of the ball is -20. kgm/s, what is the final velocity of the ball?

-4.0 m/s Δp=pfinal - pinitial Δp= mvfinal - mvinitial Δp= m (vfinal - vinitial) -20=2(v-6 m/s) -20=2v-12 -8=2v v=-4.0 m/s

A cannon sits on the edge of a castle wall that is 5.0 m above the ground. A cannon ball leaves the cannon with an initial velocity of 25 m/s at 30o to the horizontal. What is the vertical displacement of the cannon ball when it strikes the ground?

-5.0 m The cannon ball starts at 5.0 m above the ground so when it lands it has fallen 5.0 my y=-5.0 m

A ball mass M is traveling to the right with a velocity v when it collides with a ball mass M traveling to the left at 2v. The two stick together upon impact. What is the velocity of the system after the collision?

-v/2 Mv+(M)(-2v)=(M+M)vf -Mv=2Mvf -v/2=vf

A 40.0 kg box is being pushed, with a 15.0N force 25.0° below horizontal, along a horizontal, frictionless surface. What is the acceleration of the box?

0.340 m/s2 The force in the horizontal direction is 15 cos 25° = 13.6 N. a = F / m = 13.6 N / 40 kg = 0.340 m/s2

If a force F is required to extend a spring a distance 3X, how far will it be extended by force 4F?

12x

If a force F is required to extend a spring a distance 3X, how far will it be extended by force 4F?

12x The correct answer is 12x. From hooks law, F = -k (3x). Then with the same spring, the same k, -k(x2) = 4F. Solve for k = -F/(3x) = -4F / (x2). Reciprocate both sides, 3x / F = x2 / 4F. x2 = 12x.

132 J From 0 to 6m, the area of the triangle is ½·6·8=24J. From 6m to 12m the area of the rectangle is 6·8=48J. From 12m to 21m the area is 9·4 + ½·9·4=54J. From 21m to 24m, the area is ½·3·4=6J The total work done on the object is 24+48+54+6=132 J.

A rocket is traveling in the y axis with a velocity of 30.0 m/s. A wind blows horizontally causing to the rocket to accelerate in the +x axis. After 5.00 s the velocity is now 40.0 m/s at 75.0o. What was the magnitude of the rocket's acceleration?

2.00 m/s2 v0y=0 (initial velocity was in the y axis) v=40cos(75)=10 v=v0+at a=(v-v0)/t=(10-0)/5=2.0 m/s2

Sarah is traveling west toward an intersection with a constant velocity of 14.8 m/s. John is traveling north toward the same intersection at a constant velocity of 13.9 m/s. How fast is Sarah travelling relative to John?

20.3 m/s, 43.2 degrees south of west. vJG=13.9 m/s north, vSG=14.8m/s west vSJ = vSG + vGJ = vSG + (-vJG) vSJ = 14.8m/s west + 13.9 m/s south (opposite of John's north) west and south point to the third quadrant, south of west θSJ = 43.2 degrees |vSJ|=√(14.82+13.92) = 20.3 m/s

A sprinter running in a straight line passes through two gates at a constant speed of 6 m/s. If the first gate is a displacement of 10.0 meters from the starting line, and the second gate is a displacement of 30.0 meters from the starting, how much time elapsed between when sprinter passed the first and second gates?

3.33 s vavg = Δx/Δt = (30-10) / (t-0) = 6 m/s, t = 20/6 = 3.33 s

A car travels at 12.0 m/s over a bump that has a 36.0 m circular radius of curvature. Find the apparent weight of a 55.0 kg passenger when the car is at the top of the bump.

319 N The correct answer is 319 N, Apparent weight is the normal force. From an FBD a the top of the bump, mg points down, toward the center of the circular path that forms the roadway (positive centripetal), the normal force points upward, away from the center of the circle (negative centripetal). Newton's second law in the centripetal direction is mg-N=mv2/r. N= mg-mv2/r = m(g-v2/r) = 55(9.8-144/36) = 55*5.8=319N.

A car accelerates at 3.00 m/s2 so that its displacement is 15.0 m after 2.0s. What was the initial velocity of the car?

4.5 m/s v0=?, x=15, t=2, a=3 x=v0t+1/2at2 15=v(2)+1/2(3)(2)2 15=2v+6 15-6=9=2v v=4.5 m/s

Two boxes m1 and m2 are resting on a frictionless surface and connected by a rope. A horizontal force of 243 N pulls box m1 to the right. Box m2 is also pulled to the right since it is connected to box m1. If the acceleration of both boxes is 3.47 m/s2 and the mass of m1 is 27.0 kg, find the mass of m2.

43.0 kg

A ball mass 2M is traveling to the right with a velocity v when it collides with a ball mass M traveling to the left at 2v. The first ball bounces back with a velocity -v. If the initial kinetic energy of the second ball is K, what is the kinetic energy of ball after the collision?

4K 2Mv+(M)(-2v)=2M(-v)+Mvf 0=-2Mv+Mvf 2Mv=Mvf vf=2v Energy K=1/2Mv2 both M and 1/2 are constant so K proportional to v2 K2/K=vf2/v2 K2=K(2v)2/v2 K2=4Kv2/v2 K2=4K

An object with a mass m has a net force, F, applied to it resulting in acceleration. Another object has a force 4F applied to it resulting in the same acceleration. What is the mass of the second object?

4m F=ma a=F/m F/m=4F/m2 m2=4Fm/F=4m

For a car accelerating from rest, assuming all other variables remain unchanged, if you increased the time from t to 2t you would expect the displacement to change from x to

4x x=v0t+1/2at2 v0 is zero so x=1/2at2 x/t2=a/2 and a/2 is a constant so x/t2=x'/(2t)2 x'=(2t)2x/t2=4t2x/t2=4x

Starting from rest, on a straight road, a car accelerates at 3.5 m/s2. What is the cars velocity after it has gone 6.0 m?

6.5 m/s v0=0 (from rest), v=?, x=6.0, a=3.5 v2=v02+2ax v2=2(3.5)(6) v2=42 v=sqrt(42)=6.5 m/s

Add the following vectors. A 3.75 m at 34.7o West of North B 8.31m East

A=3.75 θA=124.7 1 Point B=8.31 θB=0 1 Point Ax=Acosθ=(3.75)cos(124.7)=-2.13 1 Point Ay=Asinθ=(3.75)sin(124.7)=3.08 1 Point Bx=Bcosθ=(8.31)cos(0)=8.31 1 Point By=Bsinθ=(8.31)sin(0)=0 1 Point Rx=Ax+Bx=(-2.13)+8.31=6.18 1 Point Ry=Ay+By=3.08+0=3.08 1 Point R=√(Rx2+Ry2)=√(6.18)2+(3.08)2)=6.90m 1 Point θ=tan-1(Ry/Rx)=tan-1(3.08/6.18)=26.5o θ=26.5o (implied from +x) 1 Point or 26.5o North of East

A 50.0 kg box, initially at rest, is pushed with an applied horizontal 250.N force across a floor with a static coefficient of friction of 0.500 and kinetic coefficient of friction of 0.400. The box will

Accelerate

A 5.0 kg ball headed east strikes a 4.0 kg ball headed west. The balls rebound away from each other.

Kinetic energy could be conserved in this collision

An acceleration is applied to an object

There must be a change in either vx or vy

The reference point for gravitational potential energy

can be chosen arbitrarily.

A net force is exerted on an object toward the south. The object

is accelerating toward the south.

If the net work done on an object is zero, then the object's kinetic energy

is not changing. WNET = ΔK = Kfinal - Kinitial = 0 so the kinetic energy is not changing.

A ball of mass m is attached to a thin string and whirled in a vertical circle of radius r. The tension in the string at point C where the ball is moving with speed v is

mv2/r + mg

A ball of mass m is attached to a thin string and whirled in a vertical circle of radius r. The tension in the string at point C where the ball is moving with speed v is

mv2/r + mg The correct answer is mv2/r + mg. At the bottom of the circle, mg is down (negative centripetal) and T is pointing upward (positive centripetal) T-mg = mac. T = mac + mg = mv2/r + mg

During a serve (assume a stationary ball), a tennis racquet strikes a tennis ball, of mass m, imparting a speed of v to the ball. Which if the following best represents the magnitude of the average force experienced by the ball during the collision?

m·v / Δt

In the trajectory shown below, the acceleration vector at point A is best represented by which of the following vectors?

south

A 10.0 kg pendulum is pulled back to an angle of 5.0o and released. Which of the following Free Body Diagrams could possible represent the forces acting on the pendulum at max displacement?

west/north/ south

A 2.00 kg cannon ball is shot at an angle of 60.0o with an initial velocity of 25.0m/s. Using energy considerations determine A. Kinetic energy of the ball when it is 15.0m above the ground? B. What is the velocity of the cannon ball at a height of 15.0m?

A. For recognizing that the angle does not matter 1 Point K0+Ug0=K+Ug 1 Point K0=K+Ug 1 Point K=K0-Ug K=½mv02-mgh 1 Point K=½(2)(25)2-(2)(9.8)(15) 1 Point K=331J 1 Point B. K=½mv2 1 Point v=√(2K/m) 1 Point v=√(2(331)/(2)) 1 Point v=18.2m/ 1 Point

A student carries a physics book with a weight of 20. N for a horizontal distance of 30. m at constant velocity of 4.0 m/s. How much work did the student do on the book?

0 J mg exerts a downward force, the student exerts an upward normal force, resulting in no motion in the vertical direction. Both of these forces are perpendicular to the horizontal displacement, W=F·s·cos900 = F·s·0 = 0J.

Express 12700 meters in engineering notation.

0.0123Mm , Not Selected It is already in engineering notation. , Not Selected 12.7km

An initial velocity of 1.5 m/s is imparted to a 60. kg object at the bottom of a frictionless ramp inclined at 250 above horizontal. As the object slides up the ramp, find the distance along the ramp the object travels to reach its maximum height before turning around and sliding down the ramp.

0.27 m

Starting from rest, a 5.00 kg box slides down a ramp making an angle of 60.0° with horizontal. The kinetic coefficient of friction between the ramp and the box is 0.300. Find the speed of the box after it has traveled a distance of 7.00 m.

10.3 m/s WNET = Wgravity + Wfriction. = mg sinθ · s ·cos0° + (μkmg cosθ)·s · cos180° = 5 · 9.8 · 0.866 · 7 - 0.2 · 5 · 9.8 · 0.5 · 7 = 297 - 34 = 263 J. WNET=ΔK. 263 = ½ mv2-0 V = √(2·263/5) = 10.3 m/s

Calculate the impulse applied to a 15. kg mass if a force of 20. N is applied to the object for 5.0 s.

100 kgm/s (2 sig figs) J=(20)(5)=100kgm/s (2 sig figs)

A rocket lifts off upward with an acceleration of 55 m/s2 but a sudden gust of wind causes a horizontal acceleration of 15 m/s2 to the left. What is the angle of the acceleration of the rocket at this instant relative to an observer on the ground?

105o Standard convention is that up is positive and right is positive. ()=tan-1(y/x) ()=tan-1(55/-15)=-75o as the x component is negative we add 180 -75+180=105o

Using the a-t graph above determine the change in velocity from t=0 to t=6.0 s.

12.0 m/s For an a-t graph the change in velocity between two points is the area bound by that graph v=area=1/2BxH v=1/2(6)(4)=12 m/s

A person with a mass of 45 kg weighs how much on Mars, where g = 3.78 m/2 ?

170 N W=mg=(45)(3.78)= 170 N

18 m/s x velocity remains 15 y velocity v=v0+at v=0+(5)(2)=10 d=sqrt(152+102)=18 m/s

In the diagram above the slope has an angle of 30.0o and the coefficient of friction between box B and the surface is 0.200. Box B has a mass of 15.0kg and box A has a mass of 20.0kg. A. What is the force of friction acting on box B? B. What is the acceleration of the system? C. What is the tension in the rope connecting box A to box B?

21A. the angle was 30, not 20 and you have to multiply by u - this is just an attempt to solve for N B. where is this equation coming from? it is not correct as this equals (M+m)a, not a C. this equation is not correct either - which box is it for?

A 50. kg girl on ice skates, initially at rest, pushes against a wall, extending her arms 0.25m for 0.10s causing her to accelerate. Find the magnitude of the force applied by the girl to the wall.

2500 N Fg=ma x= (1/2) a t2 + v0t a = 2 x / t2 a = 2*0.25 / 0.12= 50 m/s2 Fg=ma = 50 kg * 50 m/s2 = 2500N Third Law Fg=Fw=2500N

A dog walks 25.0 m West and then 7.50 m North. What is the magnitue of the sum of these two vectors?

26.1 m R=sqrt(Ax2+Ay2) R=sqrt(252+7.52)=26.1 m

Two objects, one twice the mass of the other, are connected by a string as shown above. A force, F, is the only force acting on the system and is pulling the smaller mass to the right. Find the tension in the string between the two masses.

2F/3 The acceleration of the system is a = F / (3m). From an FBD of the 2m object, T=(2m)a=2m*F/(3m)=2F/3

A roller coaster car travels at a constant speed around a vertical circular loop. At this constant speed what is s the value of the apparent weight at the bottom of the loop if the car is just about to fall off the track at the top?

2mg

2v 2Mv+(M)(-2v)=2M(-v)+Mvf 0=-2Mv+Mvf 2Mv=Mvf vf=2v

Three sticks of length 6.0 cm, 8.0 cm and 10.0 cm are assembled into a right triangle. What is the angle of the where the 6.0 cm stick connects to the 10.0 cm stick?

53º T-6.0 cm becomes the adjacent side so cosθ=(6/10) θ=cos-1(86/10)=53o or 3-4-5 triangle so Pythagorean triple, with angles of 53º and 37º.

Add 46.73 + 11 and determine the answer in significant figures.

An impulse of 30.0 km/s is applied to a 15. kg mass. What was the value of the force if it was applied for 5.0 s?

6.0 N J=Ft F=J/t F=30/5=6.0 N

A car is traveling at 7.0 m/s when the driver applies a constant braking force. The car travels an additional 1.5 m, while braking, before coming to a complete stop. If the car had been moving at 14 m/s, how far would it have continued to move after the brakes were applied?

6.0 m First we need to find the braking force from the work energy theorem, W=ΔK: F·s·cos180°=0 - ½ mv2. F·1.5·(-1) = -½ m 49. F = 16.3m. Next apply the work energy theorem at the higher speed: F·s·cos180°=0 - ½ mv2. 16.3m·s·(-1) = - ½ m·196. Mass and -1 cancel, so s= 196 / (2·16.3) = 6m.

A boat is observed from above traveling in the positive y direction with an initial speed of 25 m/s and then begins accelerating at a constant rate. The car accelerates at 6.0 m/s2 in the x axis for 3.0 s. What is the magnitude of the boat's displacement after 3.0 s?

80. m y displacement y=vt=(25)(3)=75 x displacement x =voyt+½at2 the initial velocity is 0 in the y axis so x=1/2(6)(3)2 y=27 displacement = √(272+752)=80. m

84 J From 6m to 12m, the area of the under the graph is the rectangle 6·4=24J. From 12m to 18m, the area of the under the graph is the rectangle + triangle is 6·4 + ½·6·4=36J. From 18m to 24m, the area of the under the graph is the triangle is ½·6·8=24J. The total work done on the object from 6 to 24m is 24+36+24=84 J.

A ball is observed to accelerate down a hill from 2.0 m/s to 10. m/s. The acceleration is measured as 5.0 m/s2. How far does the ball roll?

9.6 m v0=2, v=10, x=?, a=5 v2=v02+2ax x=(v2-v02)/2a x=(102-22)/2(5)=9.6 m

A car is observed from above traveling in the positive x direction with an initial speed of 15 m/s and then begins accelerating at a constant rate in the y axis. The car reaches a final velocity of 25 m/s at 40.o after 2.0 s. What is the magnitude of the car's acceleration during this time period?

9.6 m/s2 v0y=0 vy=25cos(40)=19.1 v=v0+at a=(v-v0)/t a=(19.1-0)/2=9.6 m/s2

which one of the following situations will there be an increase in kinetic energy?

A man pushes a child on a swing causing the swing to go faster.

A boys is being pulled forward by two cows tied to ropes. The first cow pulls with a force of 75.0N at an angle of 30.0o. the second pulls with a force of 50.0N at an angle of 125o. A. What is the net force on the boy? B. The boy has a mass of 60.0kg. What is the magnitude and direction of his acceleration?

A. 75cos(30)=65.0 75sin(30)=37.5 Both 1st Force Components 1 Point 50cos125=-28.7 50sin125=41 Both 2nd Force Components 1 Point Fx=65+(-28.7)=36.3 1 Point Fy=37.5+41=78.5 1 Point F=sqrt(36.32+78.52)=86.5 N 1 Point ()=tan-1(78.5/36.3)=65.2o 1 Point B. a=F/m 1 Point a=86.5/60=1.44 m/s2 1 Point at 65.2o 1 Point Correct Units and Reasonable (2-4) Sig Figs 1 Point

In the diagram above the slope has an angle of 40.0o and the coefficient of friction between both boxes and the surface is 0.350. Box A has a mass of 10.0kg and box B has a mass of 5.00kg. A. What is the force of friction acting on each box? B. What is the acceleration of the system? C. What is the tension in the rope connecting box A to box B?

A. A=cos40degree*9.8*0.350*10=26.275N or 26.78NB=cos40degree*9.8*0.35*5=13.13766N or 13.138N B. A=mAgsinx-friction-T=MaA 5*9.8*sin40degree-26.275-T=10aB=mBgbsinx-friction+T=mbA 5*9.8*sin40degree-13.137+T=15a15*9.8*sin40degree-39.413=15a a=55.07677/15=3.6717=3.672m/s^2 C. mBgsinx-friction-mBa=TT=5*9.8*sin40degree-13.138-3.672*5T=31.4965928746-13.138-18.36T=-0.0014

A 48 kg ice skater is applying an average force of 76.0N to maintain a constant speed of 8.9 m/s. Assume the applied force is constant. At a later time, the skater stops applying the force and coasts to a complete stop. A. What is the force being applied by the ice on the skater? B. Using the work energy theorem, determine the amount of work needed to bring the skater to a stop. C. Calculate the distance between the point where the skater quit applying a force and the point at which the skater came to rest.

A. As there is a constant velocity the net force must be 0 while the skater is moving so Fi=-76.0 N 1 Point negative sign 1 Point B. W=1/2mv2-1/2mv02 1 Point W=1/2(48)(0)2-1/2(48)(8.9)2 1 Point W=-1900 J 1 Point C. W=Fx 1 Point x=W/F x=-1900/-76 1 Point x=25 m 1 Point Reasonable sig figs (1-3) 1 Point Correct Units on all 3 1 Point

A physics teacher is performing a demonstration of centripetal force by swinging a 4.50 kg bucket of water at the end of a 1.00 m rope in a vertical circle. Assume that the bucket of water can be treated as a point mass and that its speed is constant around the circle. a) describe the FBD at the top of the loop, and state whether the vectors are positive or negative along the centripetal axis. b) describe the FBD at the bottom of the loop, and state whether the vectors are positive or negative along the centripetal axis. c) find the speed of the speed at which the teacher will get wet (that is, the water does not stay in the bucket). d) At the speed found in part a, if the rope breaks when the tension exceeds 70N, state whether the rope will break during this demonstration. Support your answer.

A. At the top: mg points down (positive centripetal). T points down (positive centripetal). 1 point for both vectors B. At the bottom: mg points down (negative centripetal). T points up (positive centripetal). 1 point for both vectors C. Tension will be least at the top of the path, 1 point ΣFc=mac T+mg=mv2/r 1 point If T=0, tension in the string is lost and the bucket and water fall. 1 point mg=mv2/r 1 point g= v2/r v=√(gr) =√(9.8*1.00) =3.13 m/s 1 point D. At the bottom of the loop ΣFc=mac T-mg= mv2/r 1 point T = mv2/r+mg=m(v2/r+g) T=4.5(3.132 / 1 + 9.8) = 88.2N 1 point Yes, a 88.2 N force will break a 70N rope 1 point

A toy car track is set up as shown in the diagram above, where: hA = 5.20m hB = 3.80 m hC = 4.60 m The radius of the loop is 0.700 m. If the toy car has a mass of 0.720 kg and is released from rest at point A, then calculate: a. The total energy of the car. b. The gravitational potential energy of the car at point B. c. The kinetic energy of the car at point B. d. The speed of the car at the top of the loop (point C). e. The normal force at the top of the loop (point C).

A. E = K + U 1 point = 0 + mghA = 0.72 · 9.8 · 5.2 = 36.7 J 1 point B. UB = mghB = = 0.72 · 9.8 · 3.8 = 26.8 J 1 point C. KA + UA = KB + UB 0 + 36.7 = KB + 26.8 1 point KB = 36.7 - 26.8 = 9.9 J 1 point D. KA + UA = KC + UC 1 point 0 + 36.7 = KC + mghC 1 point 36.7 = KC + 0.72 · 9.8 · 4.6 = KC + 32.5 KC = 4.2 = ½ m vC2 vC2 = 2 · 4.2 / 0.72 = 11.7 vC = 3.42 m/s 1 point E. N + mg = mvC2/r 1 point N= mvC2/r - mg = m(vC2/r - g) = 0.72 · (3.422/0.7- 9.8) = 4.97 N 1 point

A 35.0 N force pushes a 5.00 kg block up a ramp for a distance of 3.00 meters as shown above. The ramp makes a 30° angle with horizontal and the applied force is parallel to the surface of the ramp. The coefficient of friction between the block and the ramp is 0.250. a) What is the work done by the force, F=35N, to push the block up the ramp a distance of 3.0m? b) What is the work done by gravity? c) What is the work done by the normal force? d) What is the work done by friction?

A. Force and displacement are in the same direction. W=F·s·cosθ 1 point = 35·3·cos0°= 105J 1 point B. The angle between displacement and gravitation force is seen in an FBD to be: θ=120° 1 point W||=mg·cos 120° 1 point W||=5·9.8· (-0.5) = -24.5 J W=Fx=(-24.5)(3)=-73.5J 1 point C. Normal force is perpendicular to displacement. W = N · s · cos 90° = 0J 1 point D. The frictional force is f=μkN 1 point = μk mg cosθ 1 point = 0.25·5·9.8·cos30° = 10.6N 1 point Work done by friction : W=f·s·cos180° = 10.6 · 3 · (-1) = -31.8J 1 point

A box with a mass of 17.0 kg is free to slide down a ramp. The ramp makes an angle of 25.00 with horizontal and is 63 m long. The kinetic coefficient of friction between the ramp and the box is 0.230. a) Describe the FBD for the box. Be sure to include both the name of the force and the direction. b) Calculate the force of friction acting on the box. c) How long does it take to get to the bottom of the ramp?

A. Friction Up the slope 1 point mg down 1 point normal force perpendicular to the ramp 1 point B. In the perpendicular axis N=mgcos() f=uN=umgcos() 1 point f=(0.23)(17)(9.8)cos(25) f=34.7 N 1 point C. in the parallel axis mgsinθ-f=ma. 1 point (17)(9.8)sin(25)-34.7=17a a=2.1 1 point v2=vo2+2ax 1 point the initial velocity is zero so v=√(2ax) v= √(2)(2.1)(63) 1 point v=16.3 m/s 1 point

A projectile is fired with an initial speed of 45 m/s at an angle of 48 degrees above the horizontal on a long, flat firing range. Determine: (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the horizontal displacement (range), (d) the velocity (magnitude and direction) of the projectile 1.5 s after firing.

A. Given vox=45 cos48 = 30.1=vx, ay = -9.8, voy =45sin48=33.4 vy=0 at the top y=? v2=vo2 + 2ay 1 point y=(vy2-voy2)/(2ay)=(02-33.42)/2(-9.8) y=56.9 m 1 point B. t=? y=v0yt+1/2at2 1 point 0=33.4t+1/2(-4.9)t2 1 point -33.4t=-4.9t2 t=6.82 s 1 point C. x/t=vx, x = vxt = 205 m 1 point D. vx= 30.1 m/s t=1.5 vy=ayt+voy=(-9.8)(1.5)+33.4 1 point vy=18.7 1 point v=√((30.1)2+(18.7)2)= 35.4 m/s 1 point θ=tan-1(18.7/30.1)=31.9o 1 point

A projectile is fired with an initial speed of 55 m/s at an angle of 33 degrees above the horizontal on a long, flat firing range. Determine: (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the horizontal displacement (range), (d) the velocity (magnitude and direction) of the projectile 1.5 s after firing.

A. Given vox=55cos33 =46.1=vx, ay = -9.8, voy =55sin33=30.0 vy=0 at the top y=? v2=vo2 + 2ay 1 point y=(vy2-voy2)/(2ay)=(02-302)/2(-9.8) y=45.9 m 1 point B. t=? y=v0yt+1/2at2 1 point 0=30t+1/2(-4.9)t2 1 point -30t=-4.9t2 t=6.12 s 1 point C. x/t=vx, x = vxt = 282 m 1 point D. vx= 46.1 m/s t=1.5 vy=ayt+voy=(-9.8)(1.5)+30 1 point vy=15.3 1 point v=√((46.1)2+(15.3)2)= 48.6 m/s 1 point θ=tan-1(15.3/46.1)=18.4o 1 point

A baseball bat imparts an impulse to a 0.175 kg soft ball as shown in the graph above (assume all values on the graph have 3 significant digits). If the soft ball approaches the bat at 20.0 m/s a) Find the magnitude of the impulse imparted to the ball. b) How fast is the ball traveling as it leaves the bat.

A. Impulse is the area under the graph. 1 point Between 0 and 2 s: 2 N·s 1 point Between 2 and 8 s: 6 N·s 1 point Between 0 and 8 s: 2+6=8 N·s 1 point Award 3 previous points for ½ 8·2 B. J=Δp 1 point J=mvf - mv0 = m(vf - v0) 1 point Initial velocity and impulse are in opposite directions: 1 point -8= 0.175·(vf -20) vf = (-8/0.175)+20 = -25.7 m/s 1 point Correct Units on both answers 1 point Reasonable Sig Figs (2-4) 1 point

A child imparts a force to a 4.0 kg toy truck as shown in the graph above. (assume all values have 3 significant digits) a) Find the magnitude of the impulse imparted to the truck. b) The truck starts at rest. How fast is the truck traveling as it leaves the child's hand.

A. Impulse is the area under the graph. 1 point Between 0 and 6 s: 12 N·s 1 point Between 6 and 8 s: 4 N·s 1 point Between 0 and 8 s: 12+4=16 N·s 1 point Award three previous points for ½ 4·8 B. J=Δp 1 point J=mvf - mv0 1 point J= mvf - 0 1 point Vf=J/m = 16/4 = 4 m/s 1 point Correct Units 1 point Reasonable Sig Figs (2-4) 1 point

In the graph below, find a. the acceleration for each of the 3 segments i. t=0s to 3s ii. t=3s to 5s iii.t=5s to 7s b. the displacement covered from t=5 to t=7.

A. Indication that student was considering slopes to determine acceleration. 2 points i. a1 = 1.0 m/s2. 1 point ii. a2 = 0 m/s2 1 point iii. a1 = -3 m/s2 1 point B. Indication that student was considering area to find displacement. 2 points Correct positive areas from t=5 to t=6: 1.5m 1 point Correct negative area from t=6 to t=7: -1.5 m 1 point Correct total displacement of 0 m 1 point

Three forces are applied to a 2.00 kg object on a frictionless ramp as shown above. The ramp makes a 30° angle with horizontal. F1 is a 20.0N force applied to the object, parallel to the ramp. F2 is a 40.0 N force applied in the horizontal direction. F3 is a 10.0N force applied in a direction perpendicular to the ramp. If the object is moved 1.20 meters up the incline: a) Find the work done by F1. b) Find the work done by F2. c) Find the work done by F3. d) Calculate the net work done on the object.

A. W1= Fscosθ 1 point θ=0° 1 point W1=(20)(1.2)cos0°=24.0 J 1 point B. θ=30° 1 point W2=Fscosθ W2=(40)(1.2)cos30° W2=41.6 J 1 point C. θ=90° 1 point W3=Fscosθ W3=(10)(1.2)cos90° W3= 0 J 1 point D. Work done by the weight W4=mgsinθx W4=(2)(9.8)sin(30)(-1.2)=-11.8 J 1 point WNET=W1+W2+W3+W4 WNET=41.6+24+0-11.7=53.9 J 1 point Correct units on all final answers. 1 point

A 10.0 kg box is pulled up a slope by a rope applying a force of 100.N parallel to the surface. The slope is angled at 30.0o to the horizontal. The coefficient of friction between the box and the surface is 0.200. Answer all questions using Work and Energy concepts. A. How much work is done by the rope while pulling the box 7.00m? B. How much work is done by friction? C. The initial velocity of the box was 2.00 m/s. What was the final velocity?

A. W=Fx 1 Point W=(100)(7) 1 Point W=700.J 1 Point B. W=-fx 1 Point W=-μNx ∑Fperp=N+Wperp=0 N=Wperp 1 Point N=mgcosθ N=(10)(9.8)cos(30)=84.9 W=-(0.2)(84.9)(7) 1 Point W=-119J 1 Point C. W||=-mgsinθx 1 Point W||=-(10)(9.8)sin(30)(7)=-343 W=700-119-343= W=ΔK=½mv2-½mv02 238=½(10)v2-½(10)(2)2 1 Point v=7.18m/s 1 Point

A 5.00 kg box is pushed down a slope by applying a force of 25.0N parallel to the surface. The slope is angled at 40.0o to the horizontal. The coefficient of friction between the box and the surface is 0.250. Answer all questions using Work and Energy concepts. A. How much work is done by the 25.0N force while pulling the box 6.00m? B. How much work is done by friction? C. The initial velocity of the box was 4.00 m/s. What was the final velocity?

A. W=Fx 1 Point W=(25)(6) 1 Point W=150J 1 Point B. W=-fx 1 Point W=-μNx ∑Fperp=N+Wperp=0 N=Wperp 1 Point N=mgcosθ N=(5)(9.8)cos(40)=37.5 W=-(0.25)(37.5)(6) 1 Point W=-56.2J 1 Point C. W||=mgsinθx 1 Point W||=(5)(9.8)sin(40)(6)=189 WTOT=150+(-56.2)+189=283 W=ΔK=½mv2-½mv02 283=½(5)v2-½(5)(4)2 1 Point v=11.4m/s 1 Point

A roller coaster car with a mass of 275kg travels in a vertical loop with a radius of 20.0m. A. The normal force at the top is measured to be 250N. What is the velocity of the car? B. The car maintains a constant velocity as it circles the loop. What is the normal force acting on the car when it is pointed horizontally (it is at the furthest position on the right or left of the circular part)? C. What is the value of the normal when the car is at the bottom of the loop?

A. ac=v^2/r9.8=v^2/20=root(9.8*20)V=14m/sFc=mv^2/rV=root(Fc*r/m)=root(250*20/275)=4.264m/s? B. Fc=mv^2/rFc=275*(14)^2/20=2695N C. F=Fc+mgF=2695+275*10=5445N? 250+2695=2945N 22A. you are not including the normal force N+mg=mv^2/r B. wy are you using 14, you calculated the velocity as 4.264 (I see you have v=14 in your solution to A, but how did you get two values for v? C. I'm not sure what you are doing here - Fc is not an actual force but the sum of the forces in the centripetal axis N-mg=mv^2/r is the correct relationship 250 is the normal at the top, not at the bottom

A ball is rolling across the floor so that it is traveling along the positive x axis at 5.00 m/s. Due to the bend of the carpet fibers it experiences an acceleration of 1.25 m/s2 at an angle of -125.0o from the x axis. A. What is the x component of the acceleration? B. What is the y component of the acceleration? C. What is the x component of the balls displacement after 3.00s? D. What is the y component of the balls displacement after 3.00s? E. What is the displacement of the ball over 3.00s?

A. ax=acosθ ax=(1.25)cos(-125)=-0.717 m/s2 1 Point B. ay=asinθ ay=(1.25)sin(-125)=-1.02 m/s2 1 Point C. for recognizing that v0x=5.00 m/s 1 Point and v0y=0 x=v0xt+½axt2 x=(5)(3)+½(-0.717)(3)2=11.8 m 1 Point D. y=v0yt+½ayt2 1 Point y=(0)(3)+½(-1.02)(3)2=-4.59 m 1 Point E. v=√(x2+y2) v=√((11.8)2+(-4.59)2) 1 Point v=12.7 m 1 Point θ=tan-1(y/x) θ=tan-1(-4.59/11.8) 1 Point θ=-21.3o 1 Point or 21.9o South of East

A truck is carrying a 250. kg refrigerator, which is free to slide in the bed of the truck. The static coefficient of friction between the bed of the truck and the refrigerator is 0.550. The truck accelerates up a hill inclined at 17o above the horizontal. a) Describe the FBD of the refrigerator. Please indicate both the name of the force and the direction. b) Calculate the force of friction between the truck bed and the refrigerator. c) Find the maximum acceleration the truck (and refrigerator) may have without the box sliding as it travels up the 17.00 incline.

A. f up the slope 1 point mg down 1 point normal force perpendicular to the bed 1 point Loss of 1 Point for each additional force B. f=μsN 1 point in the perpendicular axis N=mgcosθ 1 point f=μsmgcosθ f=(0.55)(250)(9.8)cos(17) 1 point f=1290N 1 point C. In the parallel axis f-mgsinθ=ma. 1 point (1290)-(250)(9.8)sin(17)=(250)a 1 point a=2.29 m/s2 1 point

A 1200. kg truck traveling south at 15.0 m/s collides with a 750. kg car traveling 25.0 m/s at 30.0o. The bumpers of the two vehicles stick together. A. Calculate the magnitude and direction of the two after the collision. B. Is the collision elastic or inelastic. Justify your answer.

A. for applying m1v10+m2m20=(m1+m2)v in the x axis 1 Point (1200)(0)+(750)(25)cos(30)=(1200+750)vx 1 Point vx=8.33 1 Point for applying m1v10+m2m20=(m1+m2)v in the y axis 1 Point (1200)(-15)+(750)(25)sin(30)=(1200+750)vy 1 Point vy=-4.42 1 Point v=sqrt(8.332+4.422) v=9.43 m/s 1 Point ()=tan-1(-4.42/8.33) ()=-28.0o 1 Point B. Inelastic 1 Point The two objects stick together 1 Point

A 10.0kg lab cart is pushed at 5.00m/s directly toward a 7.50kg lab cart moving the opposite direction at 3.00m/s. The two carts collide and stick together. A. What is the velocity of the carts after the collision? B. What is the impulse on the 10.0kg cart? C. What is the force applied to the 10.0kg cart if the collision lasted for 0.500s? D. Is the collision elastic or inelastic. A statement is not enough, you must show proof.

A. m1v01+m2v02=(m1+m2)v 1 Point v=[m1v01+m2v02]/(m1+m2) 1 Point v=[(10)(5)+(7.5)(-3)]/(10+7.5) 1 Point v=1.57m/s 1 Point for any indication that the direction is the 1 Point same as the direction of the 10.0kg cart B. J=Ft=mΔv=m(v-v0) J=(10)(1.57-5) 1 Point J=-34.3kgm/s C. J=Ft F=J/t 1 Point F=-34.3/0.5=-68.6N D. For correctly calculating K0 K0=½m1v012+½m2v022 K0=½(10)(5)2+½(7.5)(3)2 1 Point K0=156J For correctly calculating K K=½(m1+m2)v2 K=½(10+7.5)(1.57)2 1 Point K=21.6J the collision is inelastic 1 Point

A 15.0kg lab cart, A, is pushed at 6.00m/s directly toward a 20.0kg lab cart, B, moving the opposite direction at 10.0m/s. The two carts collide and and cart B rebounds at 3.64m/s. A. What is the velocity of cart A after the collision? B. What is the impulse on cart A? C. What is the force applied to cart A if the collision lasted for 0.250s? D. Is the collision elastic or inelastic. A statement is not enough, you must show proof.

A. m1v01+m2v02=m1v1+m2v2 1 Point v2=[(m1v01+m2v02)-(m1v1)/(m2) 1 Point v=[((15)(6)+(20)(-10))-((20)(3.64))]/(15) 1 Point v=-12.2m/s 1 Point for any indication that the direction is the 1 Point opposite to the original direction of cart A B. J=Ft=mΔv=m(v-v0) J=(15)(6-(-12.2)) 1 Point J=273kgm/s C. J=Ft F=J/t 1 Point F=273/0.25=1090N D. For correctly calculating K0 K0=½m1v012+½m2v022 K0=½(15)(6)2+½(20)(10)2 1 Point K0=1270J For correctly calculating K K=½m1v12+½m2v22 K=½(15)(12.2)2+½(20)(3.64)2 1 Point K=1250J the collision is elastic 1 Point

A man whose weight is 450. N stands on an scale in an elevator traveling downward at constant acceleration of -0.500 m/s2. a) Describe the FBD for the man. Give the name and direction for the forces. b) What is the mass of the man as he accelerates? c) Find the normal force acting on the man. d) What force must be equal to the normal force acting on the man? e) Which of Newton's three laws accounts for your answer.

A. normal up 1 Point weight down 1 Point no other force 1 Point B. m=W/g 1 Point m=450/9.8= m=45.9 kg 1 Point C. sumF=ma N-W=ma 1 Point N=ma+W=ma+mg N=(45.9)(-0.5)+(45.9)(9.8) 1 Point N=427 N 1 Point D. Normal is the surface pushing up on the man reaction is Man pushing down on the surface 1 Point E. Third Law 1 Point

Two dogs pull a 25.0kg sled forward with an acceleration of 3.00m/s2. The first dog applies a force of 50.0N at an angle of 35.0o below the x axis (the positive x axis is in the direction of acceleration, or forward). A. What is the magnitude and direction of the net force acting on the sled? B. What is the magnitude and direction of the force applied by the second dog?

A. sumF=ma 1 Point sumF=(25)(3)=75.0N 1 Point 75.0 N Forward 1 Point B. As the acceleration occurs in the x axis, the y components must cancel out Fy2=F1sin() 1 Point Fy2=50sin(-35)=-28.7 1 Point for recognizing that Fy2 must be in the opposite direction from Fy1 so positive Fy2=+28.7 1 Point The sum of Fx1 and Fx2 cause acceleration 50cos35+Fx2=(25)(3) 1 Point Fx2=34.0 1 Point F2=sqrt(28.72+342)=44.5 N 1 Point ()=tan-1(28.7/34)=40.2o 1 Point

The Atwood machine above has two masses attached by a rope that is hung over a pulley. Assume that the pulley is massless and does not introduce any friction into the system. M has a value of 5.00 kg. A. Calculate the acceleration of the system. B. Calculate the tension on the rope attaching the two masses. C. The 3M mass starts in a position 1.00 m above the ground. How long does it take to hit the ground after the system is released from rest?

A. using down as the positive direction for the 3m mass equation for the 3m mass 3mg-T=3ma 1 Point for the 2m mass T-2mg=2ma 1 Point combine to get 3mg-T+t-2mg=(3m+2m)a mg=5ma 1 Point a=(5)(9.8)/5(5) a=1.96 m/s2 1 Point B. Using either equation solve for T 3mg-T=3ma 1 Point T=3mg-3ma T=2(5)(9.8)-3(5)(1.96) 1 Point T=127 N 1 Point C. x=v0t+1/2at2 1 Point v0=0 x=1/2at2 t=sqrt(2x/a) t=sqrt(2(1)/1.96) 1 Point t=1.01 s 1 Point

An airplane flying at 50.0m/s accelerates at 3.00 m/s2 while covering a distance of 750.m. A. What is the final velocity of the airplane? B. How long does it take to cover the 750.m distance?

A. v0=50.0, a=3.00, x=750, v= v2=v02+2ax 1 Point v2=(50)2+2(3)(750) 1 Point v2=7000 1 Point v=83.7m/s 1 Point B. v0=50.0, a=3.00, x=750, v= v=v0+at 1 Point t=(v-v0)/a 1 Point t=(83.7-50)/3 1 Point t=11.2s 1 Point for the correct unit on both answers 1 Point for a reasonable number of significant digits in both answers (2-4) 1 Point

A car is traveling with a velocity of 25.0 m/s at an angle of 125o from the x axis. It accelerates in the x axis at 1.75 m/s2, and in the y axis at 3.25 m/s2. The acceleration lasts for 5.00s. A. What is the x component of the initial velocity? B. What is the y component of the initial velocity? C. What is the x component of the final velocity after 5.00s? D. What is the y component of the final velocity after 5.00s? E. What is the final velocity of the car?

A. v0x=v0cosθ v0x=(25.0)cos(125)=-14.3 m/s 1 Point B. v0y=v0sinθ v0y=(25.0)sin(125)=20.5 m/s 1 Point C. vx=v0x+axt 1 Point vx=(-14.3)+(1.75)(5)=-5.55 m/s 1 Point D. vy=v0y+ayt 1 Point vy=(20.5)+(3.25)(5)=36.8 m/s 1 Point E. v=√(vx2+vy2) v=√((-5.55)2+(36.8)2) 1 Point v=37.2 m/s 1 Point θ=tan-1(vy/vx) θ=tan-1(36.8/-5.55) 1 Point θ=-81.4o θ=81.4o North of West 1 Point or 98.6o North of East or 98.6o

A small boy stands on the edge of a cliff overlooking the ocean 15.0m below. He throws a rock at 10.0 m/s, 35.0o below the horizontal x axis. A. How long does the rock to hit the water? B. How far from the base of the cliff does the rock land?

A. v0y=v0sinθ v0y=10sin(-35)=-5.73 m/s 1 Point y=v0yt+½at2 1 Point -15=-5.73t+½(-9.8)t2 1 Point 0=-4.9t2-5.73t+15 Quadratic Equation 1 Point 5.73±√((-5.73)2-4(-4.9)(15))/2(-4.9) (5.73±18.1)/-9.8 t=1.26s 1 Point B. vx=v0cosθ vx=10cos(-35)=8.19 m/s 1 Point vx=x/t 1 Point x=vxt=(8.19)(1.26)=10.3 m 1 Point For reasonable number of digits in 1 Point each answer (2-4) For correct units on both answers 1 Point

A 2525kg rocket accelerates from rest by expelling gases out of the back of the engines. The rocket engines push out 125kg which is accelerated from 0 m/s to 50.0 m/s over the 4.00 m length of the exhaust tube. A. What is the acceleration of the gas? B. What is the force applied to the gas? C. If the engines are run for 20.0 s what would be the displacement of the rocket if it starts from rest?

A. v2=v02+2ax 1 Point a=(v2-v02)/2x a=(502-02)/2(4) 1 Point a=312 m/s2 1 Point B. F=ma 1 Point F=(125)(312) F=39,000 N 1 Point C. a=F/m 1 Point a=39,000/2525 a=15.4 1 Point x=v0t+1/2at2 1 Point x=1/2(15.4)(20)2 1 Point x=3080 m 1 Point

A 1500. kg truck is traveling along a straight, level road at a constant speed of 15.0 m/s when the driver removes his foot from the accelerator. After 18.0 s, the truck's speed is 10.0 m/s. A. Calculate the acceleration of the truck. B. Assuming a constant net force was applied opposite the direction of motion, what was the magnitude of the net force acting on the truck during the 18.0 s? C. If the same force had been applied to a truck with a mass of 2500. kg, also traveling at 15.0 m/s, how far it would it travel before coming to a complete stop?

A. v=at+v0 1 points a=(v-v0)/t a=(10-15)/18 1 points a=0.278 m/s2 1 points B. F=ma 1 points F=(1500)(0.278) F=417 N 1 points C. a=F/m 1 points a=417/2500 a=0.167 1 points v2=v02+2ax 1 points x=(v2-v02)/2a x=(02-152)/2(0.167) 1 points x=1350 m 1 points

A 10.0kg cart traveling East at 5.00 m/s collides with a 7.50kg cart traveling 3.00m/s at an angle of 55.0o. The two carts collide and stick together. What is the velocity of the cars after they stick together?

A. vxA=5.00 1 Point vyA=0 1 Point vxB=3cos55=1.72 1 Point vyB=3sin55=2.46 1 Point In the x axis mAv0Ax+mBv0Bx=(mA+mB)vx 1 Point vx=[mAv0Ax+mBv0Bx]/(mA+mB) vx=[(10)(5)+(7.5)(1.72)]/(10+7.5)=3.59 1 Point in the y axis mAv0Ay+mBv0By=(mA+mB)vy 1 Point vy=[mAv0Ay+mBv0By]/(mA+mB) vy=[(10)(0)+(7.5)(2.46)]/(10+7.5)=1.05 1 Point For determining the magnitude v=√(vx2+vy2) v=√((3.59)2+(1.05)2)=3.74m/s 1 Point For determining the direction θ=tan-1(vy/vx) θ=tan-1(1.05/3.59)=16.3o 1 Point

A spaceship, far from any planet, with an initial velocity in the x direction of 34.0 m/s, an acceleration in the x direction of 12.0 m/s2, a velocity in the y direction of 26.0 m/s and a y directed acceleration of 16.0 m/s2. After 8.00 seconds, find (a) x (b) vx (c) y (d) vy (e) the magnitude and direction of velocity at t= 8 s

A. x=1/2axt2+voxt 1 point = 1/2*(12)*82+34*8 = 656 m 1 point B. vx=axt+vox 1 point = 12*8+34 = 130 m/s 1 point C. y=1/2ayt2+voyt 1 point = 1/2*(16)*82+26*8 = 720m 1 point D. vx=axt+vox 1 point = 16*8+26 = 154 m/s 1 point E. sqrt(vx2+vy2) = 202 m/s 1 point tan-1(vy/vx) = 50 degrees above +x 1 point

Two skaters are standing on their skates so they are essentially on a frictionless surface. The first girl, named Beth, has a mass of 35.0 kg and pushes the second girl Minji. Minji accelerates from rest to cover 10.0m in 7.50s. A. What is Minji's acceleration? B. Beth accelerates to 8.00m/s in the same time but in the opposite direction. What is the force applied to Beth? C. What is Minji's mass?

A. x=v0t+1/2at2 1 Point from rest so v0=0 a=2x/t2 a=2(10)/7.52 1 Point a=0.356 m/s2 1 Point B. v=v0+at 1 Point a=(v-v0)/t a=(8-0)/7.5 1 Point a=1.07 1 Point F=ma 1 Point F=(35)(1.07) F=37.4 N 1 Point C. m=F/a 1 Point m=37.4/0.356 m=105 kg 1 Point

Using the following graph A. Determine the acceleration during the time periods i. t=0.0s to t=2.0s ii. t=8.0s to t=11.0s B. Determine the Displacement during the time periods i. t=2.0s to t=5.0s ii. t= 6.0s to t=11.0s

Ai. for any indication of an attempt to calculate the slope of the graph 1 Point v=slope=rise/run=(v-v0)/(t-t0) 1 Point v=(6.0-2.0)/(2-0)=2.0 m/s2 1 Point Aii. v=(-2-(-7))/(11-8)=1.7 m/s2 1 Point Bi. For any indication that the student determined the area under the line 1 Point x=1/2(base x height) = 1/2(5-2)(6-0)=9.0 m 1 Point Bii. For correctly calculating the displacement from 6s to 8s 1 Point x1=(8-6)(-7-0)=-14m For correctly calculating the displacement from 8s to 11s rectangular area x2=(11-8)(-2-0)=-6.0m 1 Point triangular area x3=1/2(11-8)(-7-(-2))=-7.5m 1 Point x=(-14)+(-6)+(-7.5)=-28m For correct units on all answers

Object A has twice the mass of object B. If the two objects bounce off a floor with equal speeds, both just before and just after the collision, and object A is in contact with the floor for twice the time, the force the floor applies to object A is

Compare FA·2Δt=(2m)v-(-2mv) with FB·Δt=mv-(-mv), dividing both sides of the first equation by 2 shows FA= FB.

A 2.0 kg ball has a velocity of 6.0 m/s to the right. If the momentum change of the ball is 20. kgm/s, what is the final velocity of the ball?

Correct answer: 16 m/s Δp=pfinal - pinitial Δp= mvfinal - mvinitial Δp= m (vfinal - vinitial) 20=2(v-6 m/s) 20=2v-12 32=2v v=16 m/s

Correct answer: 2.6 m/s

A passenger in a car is throwing a ball into the air while the car travelling at a consent speed making a making a hard right turn. Which of the following is INCORRECT?

Correct answer: The car is an inertial reference frame

A pickup truck, carrying a box on a level road, accelerates forward from a stoplight. Assuming that the box does not slide on the bed of the truck, as the truck is accelerating forward, the frictional force exerted on the box by the bed of the truck is in which direction?

Forward The correct answer is forward. If there was no frictional force on the box, the inertia of the box would leave it in place, thus slide out of the back of the truck as the truck accelerates forward. So there must be a force pulling the box forward, the frictional force.

Given two boxes, one twice the mass of the other, resting on a frictionless surface as shown above, which situation leads to a greater contact force between the two boxes. Assume the force in the left picture is the same magnitude as the force in the right picture.

Pushing to the right , Not Selected

When an object travels in a circular path at a constant speed, which of the following statements is true about the direction of the object's velocity and acceleration vectors.

The velocity vector is perpendicular to the acceleration vector.

A car and cement truck are traveling down a level road at the same speed and slam on their brakes and skid to a stop. Assuming they both produce the maxium amount of friction possible and have the same coefficients of friction between their tires and the road, which one will stop in a shorter distance?

Their stopping distances are identical.

Which of the following is true of the net work done on an object whose speed is decreasing?

WNET < 0, and ΔK < 0

A box resting on a table experiences the normal force from the table pushing upward, while the table experiences the normal force of the box pushing downward. These two forces are opposite in direction and

have equal magnitudes because they are action / reaction forces.

A moving railroad car collides with and sticks to an identical railroad car that is initially at rest. After the collision, the kinetic energy of the two cars is

is one fourth as much as before. p0 = pf . mv0 + 0 = (m+m)vf . m·v0 = 2·m·vf . v0 = 2vf . K0 = ½ mv02 Kf = ½ (2m)(v0/2)2 = 2/4 ( ½ mv02) = ½ K0

Two identical spring, each with same N coils and the same spring constant, k. If the spring are connected end to end to make a longer spring with 2N coils, what is the spring constant of the new spring? Correct answer:

k/2

Which of the following would be considered derived units in the Internal System?

kgm/s2

Momentum will be conserved only when two object collide, if

the net external force acting on the objects is zero.

The direction of the acceleration when an object is in uniform circular motion is

toward the center of the circular path.

If x is measured in meters, v and vo in m/s and a in m/s2, which of the following is NOT dimensionally consistent?

v2=vo2+2a

A 10.0kg cart, A, traveling East at 5.00 m/s collides with a 7.50kg cart, B, traveling 3.00m/s at an angle of 55.0o. The two carts collide and immediately after the collision Cart A is traveling 4.00 m/s at an angle of 105o. What is the velocity of cart B immediately after the collision?

vx0A=5.00 1 Point (both components) vy0A=0 vx0B=3cos55=1.72 1 Point (both components) vy0B=3sin55=2.46 vxA=4.00cos105=-1.04 1 Point (both components) vyA=4.00sin105=3.86 In the x axis mAv0Ax+mBv0Bx=mAvAx+mBvBx 1 Point (10)(5)+(7.5)(1.72)=(10)(-1.04)+(7.5)vBx vBx=9.77 1 Point in the y axis mAv0Ay+mBv0By=mAvAy+mBvBy 1 Point (10)(0)+(7.5)(2.46)=(10)(3.86)+(7.5)vBy vBy=-2.69 1 Point For determining the magnitude v=√(vx2+vy2) v=√((9.77)2+(2.69)2)=10.1m/s 1 Point For determining the direction θ=tan-1(vy/vx) θ=tan-1(-2.69/9.77)=-15.4o 1 Point For correct sig figs (2-4) and correct units on both the magnitude and direction 1 Point

Three masses are hung from the ceiling by two ropes. The masses and the lengths of the ropes are all identical. Which situation will have the greatest tension in the ropes.

widest inverse triangle

A mass, m, is at rest on a ramp at an angle of θ with horizontal. The magnitude of the frictional force on the block is always less than or equal to

μs mg cosθ

If a 10 kg ball and a 40 kg box are dropped from the top of a building at the same time, then the acceleration would be approximately:

-10 m/s2 for both objects.

A man is running 12 m/s at 210o. What is the x component of his velocity?

-10. m/s vx=vcos() vx=12cos(210)=-10. m/s If your calculator shows -11 you are in radians and need to be in degrees

A rock is thrown vertically downward at an initial velocity of -50. m/s. What is the velocity of the rock after 2.0 s?

-70. m/s v=v0+at v=(-50)+(-9.8)(2) v=-70. m/s

Given a particle with the displacement versus time graph shown above, what is the velocity of the particle at t = 4.5 s?

2.0 m/s Δx/Δt = 2/1 m/s = 2.0 m/s

Convert 211.3 s to minutes.

3.522 min

A ball is observed to accelerate down a hill from 2.0 m/s to 10. m/s. The acceleration lasts for 5.0s. What is the displacement of the ball?

30. m v0=2, v=10, t=5, x=? x=1/2(v+v0)t x=1/2(2+10)5=30. m

A woman runs 450 m at 30.o. Assuming the angle is measured from normal compass points how far North did she get?

390 m North would be the x component and standard angle values are given from positive x or East. Ay=Asin() Ay=450sin(30)=220 if you got 69 then your calculator is in radians and should be in degrees

A student carefully measures the time it takes for a turtle to move 10.0 m. He gets a value of 4218.6 s. However he discovers that the stopwatch had already been running for about 300 s before the turtle started to move. How long did it take the turtle to go 10.0 m?

3900 s

What is the circumference of a circle with a 7.38 cm radius? (carefully consider the number of significant figures)

46.4 cm

A woman swims directly across a river at 3.0 m/s but the water is running downstream at 5.0 m/s. What is the magnitude of the woman's velocity with respect to an observer on the shore?

5.8 m/s The two velocity values are at right angles to each other so they can be treated as the component of the resultant vector. R=sqrt(Rx2+Ry2) R=sqrt(32+52)=5.8 m/s

Three sticks of length 9.0 cm, 12 cm and 15 cm are assembled into a right triangle. What is the angle of the where the 9.0 cm stick connects to the 15 cm stick?

53º 9.0 cm becomes the adjacent side so cosθ=(9/15) θ=cos-1(9/15)=53o or 3-4-5 triangle so Pythagorean triple, with angles of 53º and 37º.

A rock is thrown vertically upward at an initial velocity of 70. m/s. How long does it take to reach its maximum height?

7.1s v=v0+at at max height v=0 0=70+(-9.8)t t=-70/-9.8=7.1s

A spaceship, far from any planet, with an initial velocity in the x direction of 21.0 m/s, an acceleration in the x direction of 15.0 m/s2, an initial velocity in the y direction of 17.0 m/s and a y directed acceleration of 13.0 m/s2. After 4.00 seconds, find (a) x (b) vx (c) y (d) vy (e) the magnitude and direction of the velocity at t= 4 s

A. x=1/2axt2+voxt 1 point = 1/2*(15)*42+21*4 = 204 m 1 point B. vx=axt+vox 1 point = 15*5+21 = 81 m/s 1 point C. y=1/2ayt2+voyt 1 point = 1/2*(13)*42+17*4 = 172m 1 point D. vx=axt+vox 1 point = 13*4+17 = 69 m/s 1 point E. sqrt(vx2+vy2) = 106 m/s 1 point tan-1(vy/vx) = 40 degrees above +x 1 point

Find the resultant vector (magnitude and direction) by adding the following set of vectors: A = 7.4m due South B = 4.8m at 30° South of East C = 6.2m due East

Ax=0, Ay=-7.4 1 Point Bx=4.8cos(30)=4.15 1 Point By=-4.8sin(30)=-2.4 1 Point Cx=6.2, Cy=0 1 Point Rx=0+4.15+6.2=10.35 1 Point Ry=-7.4+(-2.4)+0=-9.8 1 Point R=√(10.352+9.82)=14m 1 Point θ=tan-1(9.8/10.35)=43o 1 Point Or x is positive, y is negative so fourth quadrant-angle is below x which is 43o South of East For Correct Unit on Answer 1 Point For Reasonable Sig Figs (1-3) 1 Point

A dog walks 25.0 m East and 7.50 m South. What is the angle of the sum of these two vectors?

Correct answer: -16.7o

What is the x component of a jet flying 211 m/s at 215o?

Correct answer: -172 m/s Ax=Acos() Ax=211cos(215)=-172 if you got 41.7 your calculator is in radians and must be switched to degrees

Suppose that an object experiences constant acceleration. Which of the following is true?

In equal times its velocity changes by equal amounts. a = Δv /Δt = equal velocity change per equal time change

In the graph below, find a. the acceleration for each of the 3 segments i. t=0s to 3s ii. t=3s to 5s iii.t=5s to 7s b. the displacement covered during the entire trip.

Indication that student was considering slopes to determine acceleration. 2 points a1 = -1 m/s2. 1 point a2 = 0 m/s2 1 point a1 = 3 m/s2 1 point Indication that student was considering area to find displacement. 2 points Correct positive area from 6-10s (OK if student cancelled 0-3 s with 7-10 s areas): 6m 1 point Correct negative areas: -13.5 m 1 point Correct total displacement of -7.5 m 1 point

Which of the following is given in MKS unit?

Kilogram meters per second

Study of an automobile collision requires which of the following topics in physics?

Newtonian mechanics

When adding two vectors with magnitudes A and B that are pointing along the same axis which of the following are possible values for the magnitude of the resultant vector? (Please choose all correct answers)

R=A-B R=A+B

Which of the following is the correct relationship between distance and the magnitude of displacement as an object moves from point A to point B along the curved path shown?

The distance is greater than the magnitude of the displacement

Which of the following is the NOT a correct relationship between distance and the magnitude of displacement as an object moves from point A to point B along the curved path shown?

The distance is less than the magnitude of the displacement

Cars A and B start at rest and travel with the same acceleration. Compare the speed of of car A to car B, after car A has traveled nine times the distance of B.

The problem said nothing about time, so you should think of v2=vo2+2ax, but vo=0. So with the same acceleration, v2 = 2ax = (constant) * x. v=√(con)x, square root of 9=3

Which of the following is/are vectors? (Choose all that are correct)

The speed and direction of a bullet headed for a target The information needed to catch a moving ball

A rock is dropped from the top of a building. At the same time, a second rock is thrown straight down from the same building. Compare the accelerations of the two rocks after being released.

The two rocks have equal rates of acceleration.

An object traveling in a straight line at 14 m/s when it begins to undergo an acceleration. The object has a displacement of 25 m in the next 5.0 s. What is the acceleration of the car?

-3.6 m/s2 v0=14, x=25, t=5, a=? x=v0t+1/2at2 25=14(5)+1/2a(5)2 a=-3.6 m/s2

Using the v-t graph above, determine the average acceleration between t=4 and t=6s.

-4.5 m/s2 on a v-t graph the average acceleration is the slope of the line between the two points a=rise/run a=(-7-2)/(6-4)=-4.5 m/s2

A rock is thrown vertically downward at an initial velocity of 50. m/s. What is the displacement of the rock after 7.0 s?

-590 m x=v0t+1/2at2 x=-50(7)+1/2(-9.8)(7)2 x=-350-240=-590 m

A rock is thrown vertically upward at an initial velocity of 50. m/s. What is the displacement of the rock when it has a velocity of -20. m/s?

110 m v2=v02+2ay y=(v2-v02)/2a y=(202-502)/2(-9.8) y=110 m

A sprinter running in a straight line passes through two gates in 2.50 seconds. If the first gate is a displacement of 10.0 meters from the starting line, and the second gate is a displacement of 30.0 meters from the starting, what was the sprinters average velocity?

8.00 m/s vavg = Δx/Δt = (30-10) / 2.5 = 8 m/s

Given a particle with the displacement versus time graph shown above, during which interval(s) is the velocity minimum?

C only Direction matters for velocity, delta v / delta t = (-3) / (+1) at C, which is less than (+2) / (+1) at D.

Given a particle with the displacement versus time graph shown above, during which interval(s) is the velocity minimum?

C only Direction matters for velocity, delta v / delta t = (-3) / (+1) at C, which is less than (+2) / (+1) at D.

A man is running 12 m/s at 120o. What is the y component of his velocity?

10. m/s vy=vsin() vy=12sin(120)=10. m/s If your calculator shows 7.0 you are in radians and need to be in degrees

In which one of the following cases does the car have a southward acceleration?

Traveling northward and slowing down.

A submarine is travels 32500 m but only goes 9250 South. What is the angle of the subs displacement relative to East?

-16.5o Ay=Asin() ()=sin-1(Ay/A) ()=sin-1(-9250/32500)=-16.5o

A dog walks 25.0 m East and 7.50 m South. What is the angle of the sum of these two vectors?

-16.7o ()=tan-1(Ay/Ax) ()=tan-1(-7.5/25)=-16.7o

A rock is thrown vertically downward at an initial velocity of 30.0 m/s. What is the approximate maximum velocity after it has fallen 40.0 m?

-41.0 m/s v2=v02+2ay v2=-302+2(-9.8)(-40) v2=1684 v=41.0 down so v=-41.0 m/s

Given a particle with the displacement versus time graph shown above, during which interval(s) is the object moving toward the negative x direction?

C only x final - x initial < 0 only within C

In the trajectory shown below, the y component of the velocity vector at point A is best represented by which of the following vectors?

north

A man goes out for a walk. He first walks 120 m and then rests. He then continues on his walk for another 64.8 m. What is the total length of his walk?

185 m

A cannon sits on the edge of a castle wall that is 5.0 m above the ground. A cannon ball leaves the cannon with an initial velocity of 25 m/s at 30o to the horizontal. When it is traveling upward and the vertical (y component) velocity has reached 4.0 m/s what is its horizontal velocity?

19 m vy=25sin(30)=12.5 vy=v0y+at 4=12.5-(9.8)t t=0.867 vx=vcos() vx=25cos(30)=21.7 x=vxt=(21.7)(0.867)=19 m

An airplane is traveling in the x axis with a velocity of 60.0 m/s. The plane accelerates in the y to avoid a storm. After 3.00 s the velocity is now 80. m/s at 35.0o. What is the magnitude of the displacement of the airplane during this time period?

193 m y displacement y=vt=(60)(3)=180 m x displacement x=1/2(v+v0)t v0y=0 (initial velocity was in the y axis) v=80sin(35)=45.9 y=1/2(45.9+0)(3)=68.8 d=sqrt(1802+98.22)=193 m

Given a particle with the displacement versus time graph shown above, what is the velocity of the particle at t = 4.5 s?

2.0 m/s Δx/Δt = 2/1 m/s = 2.0 m/s

Using the v-t graph above, what is the average acceleration from t=0 to t=2?

2.0 m/s2 average acceleration is the slope of a v-t graph a=rise/run a=(6-2)/(2-0)=2.0 m/s2

A car traveling in a straight line when the driver puts on the brakes slows to 14m/s in 35 s. If the car travels 620 m as it slows, what was the initial velocity of the car?

21 m/s v0=0, v=14, t=35, x=620 x=1/2(v+v0)t 620=1/2(14+v0)(35) 1240=(14+v0)(35) 35.4=14+v0 v0=35.4-14=21 m/s

A dog walks 25.0 m East then turns and walks 7.50 m North. What is the magnitude of the sum of these two vectors?

26.1 m

For a car accelerating from rest, assuming all other variables remain unchanged, if you increased the acceleration from a to 2a you would expect the final velocity to increase from v to

2v v=v0+at v=at since v0=0 t=v/a v/a=2v/2a so the final velocity would be 2v

Subtract 46.73 - 11 and determine the answer in significant figures.

36 46.73 - 11 = 35.73 In addition and subtract we determine the significant digits by the significant columns in the original numbers. 46.73 is significant to the 1/100th's column and 11 to the 1's column. The answer can only be significant to the 1's column. 36

Divide 46.73/11 and determine the answer in significant figures.

4.248

Express .052 g in engineering notation.

52 mg

Cars A and B start at rest and travel with the same acceleration. Compare the speed of car A to car B, after car A has traveled for twice the time of car B.

A is twice as fast as B The problem said nothing about displacement, so you think of v=at+vo, but vo=0. So with the same acceleration, v = at. Doubling time doubles velocity.

A squirrel dashes across a 240 cm wide road in 0.075 minutes. (You must always show work on any Free Response Question. Most points are awarded for showing the work and not for the actual answer.) A. Convert both values to standard units. B. Using the equation speed=distance/time calculate the speed using significant figures. C. Convert the answer to Engineering Notation.

A. 240cm(1m/100cm)=2.4m 2 Points 0.075min(60s/1min)=4.5s 2 Points B. 2.4/4.5=0.5333 2 Points 0.53 m/s 2 Points C. 530 x 10-3 2 Points

A cat covers a distance of 31800 mm in 0.0450 hours. A. Convert the distance to standard units. B. Convert the time to standard units. C. What is the cat's average speed in meters per second during this time period?

A. d=31800 mm(1m/1000mm) 1 Point d=31.8m 1 Point B. t=0.0450 hr(60min/hr)(60s/min) 1 Point t=162s 1 Point C. s=d/t 1 Point s=31.8/162 1 Point s=0.196m/s 2 Point Reasonable Sig Figs (2-4) 1 Point Correct units for all 3 answers 1 Point

A boat traveling with an initial velocity of 10.0 m/s accelerates at 2.00 m/s2 for 15.0s. A. What distance does the boat cover in this period of time? B. What is the final velocity of the boat?

A. v0=10.0, a=2.00, t=15.0, x= x=v0t+½at2 1 Point x=(10)(15)+½(2)(15)2 1 Point a=375m 1 Point for the correct unit 1 Point for a reasonable number of significant digits (2-4) 1 Point B. v0=10.0, a=2.00, t=15.0, v= v=v0+at 1 Point v=(10)+(2)(15) 1 Point v=40.0m/s 1 Point for the correct unit 1 Point for a reasonable number of significant digits (2-4) 1 Point

You are driving your car at a speed of 23.0 m/s and you hit the brakes. You skid to a stop. Upon measuring the skid marks, you find they are 85.0 meters long. (a) Determine the acceleration of the car. (b) Determine the elapsed time during braking.

A. v0=23, v=0, x=85 1 Point v2=v02+2ax 1 Point a=(v2-v02)/2x 1 Point a=(02-232)/2(85) 1 Point a=-3.11m/s2 1 Point B. Using original variables x=1/2(v+v0)t 1 Point t=2x/(v+v0) 1 Point t=2(85)/(0+23) 1 Point t=7.39s 1 Point for both correct units and reasonable sig figs (2-4) 1 Point

Using the following graph A. Determine the acceleration during the time periods i. t=0.0s to t=2.0s ii. t=5.0s to t=9.0s B. Determine the Displacement during the time periods i. t=2.0s to t=5.0s ii. t= 5.0s to t=9.0s

Ai. for any indication of an attempt to calculate the slope of the graph 1 Point v=slope=rise/run=(v-v0)/(t-t0) 1 Point v=(-12.5-0)/(2-0)=-6.2 m/s2 1 Point Aii. v=(17.5-(12.5))/(9-5)=7.5 m/s2 1 Point Bi. For any indication that the student determined the area under the line 1 Point x=(base x height) =(-12.5-0)(5-2)=-38 m 1 Point Bii. For correctly calculating the displacement from 5s to 6.5s 1 Point x1=1/2(-12.5-0)(6.5-5)=-9.4m For correctly calculating the displacement from 6.5s to 9s x2=1/2(17.5-0)(9-6.5)=21.9m 1 Point x=(-9.4)+(21.9)=12m 1 Point For correct units on all answers 1 Point

What is the circumference of a circle with a 7.38 cm radius? (carefully consider the number of significant figures)

Correct answer: 46.4 cm

Given the vector relationship A + B = C. If the magnitudes of the vectors are related by A + B = C which of the following are true?

Correct answer: A and B point in the same direction.

Given a particle with the displacement versus time graph shown above, during which interval(s) is the velocity maximum?

D only Direction matters for velocity, delta v / delta t = (+2) / (+1) at D, which is greater than (-3) / (+1) at C.

An object traveling at a constant velocity v has a displacement of x in a time t. If over the same period of time an acceleration a was applied to the object

The displacement could increase or decrease

A rock is dropped from the top of a building. At the same time, a second rock is thrown straight down from the same building. Compare the accelerations of the two rocks after being released.

The two rocks have equal rates of acceleration. Acceleration is -9.8 m/s2 (approximately -10 m/s2 ) regardless of whether the rock starts from rest, or has a positive or negative initial velocity.

Given the vector relationship A + B = C. If the magnitudes of the vectors are related by A + B > C which of the following statements are true? (Choose all that are correct)

The vectors A and B could NOT be pointing in the same direction

A ball is thrown upward with an initial velocity of 40 m/s. Three seconds after the ball was thrown, the ball's velocity vector is ___, and the acceleration vector is ___.

Upward, downward.

Which of the following statements about vectors is NOT true.

Vectors must originate at the point 0,0 by definition.

An object moves initially in the x axis with a velocity vx and accelerates in the y axis with an acceleration ay.

With an acceleration in the y axis the overall value of the velocity will increase as a y velocity component is added and the initial x velocity value remains constant. v=sqrt(vx2+vy2)

Which equation represents the angle θ in the drawing?

cos^-1(3/7) -3 is adjacent, 7 is hypotenuse, cos θ = adjacent over hypotenuse. θ = cos-1(3/7)=65o

If x is measured in meters, v and vo in m/s and a in m/s2, which of the following is NOT dimensionally consistent?

AP Physics Final exam

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