Physics Old Tests 1,2,3

A football player of mass m1 = 10 kg is running at a speed of v1 = 5 m/s down the field as shown in the figure. A second player of mass m2 = 20 kg, running at a speed of v2 = 3 m/s, tackles the first player so that they move together after the collision. What is the velocity of the two players immediately after the collision? (Express your answer in vector form).

(10+20) vf = (20)(3) Vf = 2 (10+20)vf = (10)(5) Vf = 5/3 vf = (2i+5/3j)

A uniform beam of mass M and length L pivoted at its end (as illustrated) has a moment of inertia of I = (1/3)ML2. What is the angular acceleration of the beam at the moment shown in the figure?

(L/2i)*(-mgj N) = -Lmg/2k M.n a = torque/I = (-LMg / 2k)(ML^2/3) = -3g/2L k

A ball rolls off a horizontal roof at an initial speed v. If the height of the building is h, find an expression for how far away in the x-direction the ball lands from the base of the building. g is the gravitational acceleration.

-h=0+1⁄2(-g)tf2 tf = sqrt(2h/g)

An object of mass m = 2.0 kg is attached to a massless string that is wrapped around a M = 6.0 kg, 5.0-cm-radius cylinder, as shown in the figure below. The cylinder rotates on an axle through the center. The object is released from rest 1.0 m above the floor. What is the speed of the object when it hits the floor? Use g = 10 m/s2 and the moment of inertia of the solid cylinder is (1/2)MR2.

1/2(1/2M+m)v^2f = ugh v^2f = 2mgh/(1/2M+m) =2(2kg)(10)(1)/(1/2(6)+2) =8 vf = sqrt(8)

Two astronauts, each having a mass of 70 kg, are connected by a d = 12.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at angular speed of 2.0 rad/s. By pulling on the rope, one astronaut shortens the distance between them to 6.0 m. What is the new angular speed of the system? The moment of inertia of a point mass is mr2 with m is the mass and r is the radius of the point mass from the axis of rotation.

2m(3m)^2wf = 2m(6m)^2(2rad/s) wf = 8rad/s

What is the correct y-component of vector A illustrated below?

A = A cos φ i ‒ A sin φ j Ay = ‒ A sin φ

Vector A has a negative x component 3.00 units in length and a positive y component 4.00 units in length. Determine the expression for A in unit vector notation and its magnitude.

A=‒3i+4j |A|= sqrt(3^2+4^2)=5

A bridge of length 80.0 m and mass 1.0x105 kg is supported on a smooth pier at each end as shown in the figure below. A truck of mass 4.0x104 kg is located 20.0 m from end point A. What is the force on the bridge at the support point B? Use g = 10 m/s2.

Choosing point A as the pivot: static equilibrium condition tnet =tNA + rWT + tWB + tNB = 0 0-(20m)(4*10^4 kg 10m/s2)-(40m)(1*10^5 kg 10m/s2)+80mNB = 0 Nb(80N) = 8*10^6 N+4*10^7 N NB = 6*10^5 N

Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline. Which has the larger speed at the bottom?

Conservation of mechanical energy: Ki + Ug,i + Ue,i = Kf + Ug,f + Ue,f 0 + mgH + 0 = Kf + 0 + 0 Kf = mgH Kf for the two blocks are the same, therefore their speed is the same when they hit the bottom.

What is a true statement about escape speed?

Escape speed depends on the mass of the planet from which the object is launched

A block of mass m = 1.0 kg slides on a horizontal frictionless surface as shown below. The speed of the block before it touches the spring is 5 m/s. How fast is the block moving at the instant the spring has been compressed 0.2 m? The spring constant k is 400 N/m.

Conservation of mechanical energy: Ki + Ug,i + Ue,i = Kf + Ug,f + Ue,f 1⁄2 m vi^2 + mgyi + 1⁄2 k (xi)^2 = 1⁄2 m vf^2 + mgyf + 1⁄2 k (xf)^2 1⁄2 (1 kg)(5 m/s)^2 + 0 + 0 = 1⁄2 (1 kg) vf^2 + 0 + 1⁄2 (400 N/m) (0.2 m)^2 25 J = vf^2 + 16 vf = 3 m/s

A 2.0-m-long, 300 g rod is hinged at one end and connected to a wall. Initially, it is held out horizontally, then released. What is the angular speed of the rod as it hits the wall? The moment of inertia of the rod with the rotation axis at the end of the rod is (1/3)ML2. Use g = 10 m/s2.

Ef = Ei 1/2Iw^2 +Mgy = 1/2 I w^2 +Mgy 1/2 Iw^2 +Mg(-L/2)L = 0+0 w^2f = MgL/I = MgL//(1/3ML^2) = 3g/l wf = sqrt(3g/L) = sqrt(3*10/2) = sqrt(15)

A rope pulls a 10 kg wooden crate 5.0 m across a wood floor. What is the change in thermal energy? The coefficient of kinetic friction is 0.30. Use g = 10 m/s2.

Eth = fkx = ukNx = ukmgx =(0.3)(10kg)(10m/s2)(5m)=150J

A rocket with an initial mass of 1000 kg accelerates in outer space while ejecting fuel at a speed of 40 km/s relative to the rocket. Its rate of mass ejection is 0.40 kg/s. What is the acceleration of the rocket at an instant when its mass is 800 kg (80% of the original mass)? Ignore any external forces on the rocket.

F(thrust) =v(ER) dMR/dt = (40*10^3 m/s)(0.4kg/s) =16&10^3 N Acceleration a =F(thrust) /m =(16*10^3 N)/(800kg) =20m/s2

A 65-kg woman descends in an elevator that briefly accelerates at 0.20 g downward when leaving a floor. She stands on a scale that reads in kg. During this acceleration, which is her weight and what does the scale read? (g=10m/s2)

FN =mg-ma=(65kg)(g-0.2g)=(65kg)(0.8g) FN =52kg Scale reads = 52kg, Weight = 650 N

The centers of 3 masses lie on a line as illustrated below. If the gravitational force on mass m2 is zero, what is the ratio m1/m3?

Fnet on m2 = 0 F12 = F32 Gm1m2/2^2 = Gm3m2/3^2 m1/m3 = 4/9

Suppose a planet with a mass of 3x10^25 kg is orbiting a star with a mass M in a circular orbit with a radius of 2x10^12 m and a period of 2pi x10^7 s. Using Newton's version of the Kepler's third law, determine the mass M of the star. Use G = 1x10^-10 N m2/kg2.

Fnet to center = mw^2r GMm/r^2 = m(2pi/T)^2 r M = (2pi/T)^2 r^3/G M = [2pi/(2pi x10^7 s)]^2 (2x10^12 m)^3 / (1x10^-10 N m2/kg2) = 8x10^32 kg

A filled treasure chest of mass m = 8 kg with a long rope tied around its center lies in the middle of a room. Dirk wishes to drag the chest, but there is friction between the chest and the floor with a coefficient of static friction μs = 0.5. If Dirk pulls horizontally on the rope, what is the magnitude of the tension required to just get the chest moving?

Fnet, y = m*ay N - mg = m*0 N = mg Fnet, x = m*ax T - fs, max = m*0 T = fs, max = usN = usmg = (0.5)(8 kg)(10 m/s2) = 40 N

The speed of a 2kg toy car at the bottom of a vertical circular portion of track is 3m/s. If the radius of curvature of this portion of the track is 1 m, what is the magnitude of the force the track exerts on the car?

Fnet,to center = m v2/r N - mg = mv2/r N = m(g+v2/r) = (2 kg)(10 + 3^2/1) = 38 N

The Earth (mass of 6x10^24 kg) is orbiting the Sun (mass of 2x1030 kg) with a radius of 2x10^11 m. Determine the speed of the Earth orbiting around the Sun in a circular orbit. Use G = 1x10^-10 N m2/kg2.

GMm/r^2 = mv^2/r v = √(GM/r) = √[(1x10^-10 N m2/kg2)(2x10^30 kg) / (2x10^11 m) = √(10^9) m = 10^4√10 m/s

The instantaneous power delivered to an object is P = 4t2 W. If the object has a force of 6 N acting in its direction of velocity, what is the instantaneous speed of the object at t = 3 s?

Instantaneous Power P = Fv cos0 4t^2 =4(3s)^2 =(6N)*vcos(0) v =6m/s

A 0.5-kg soccer ball is kicked at an initial speed of 20 m/s at an angle of 36.87o above the horizontal. What is the change of the potential energy of the soccer ball when it has reached the apex of its trajectory? Use sin(36.87) = 0.6, cos(36.87) = 0.8, and g = 10 m/s2.

Kf =Kapex =1⁄2m(vx,apex2 +vy,apex2) =1⁄2m[(vcos0)^2 +0]=1⁄2m(vcos0)^2 Conservation of mechanical energy: U = -K = -(Kf - Ki) = - [ 1⁄2 m (v cos0)^2 - 1⁄2 m { (v cos0)^2 + (v sin0)^2 } = 1⁄2 m (v sin0)^2 =1⁄2(0.5kg)(20m/s0.6)^2 =(1/4)12^2 =36J.

The equation that describes the angular momentum of a moving particle is L = I where I is the moment inertia of the particle with SI unit of kg m2 and is the angular velocity of the particle with SI unit of 1/s. Find the quantity that is consistent with the angular momentum. m is the mass of the particle, v is the velocity of the particle, a is the acceleration of the particle, r is the radius of the particle from the origin of the coordinate system, g is the gravitational acceleration, and y is the vertical position of the particle.

L = I w [kg m2] [1/s] = [ kg m2/s ] mvr =[kg][m/s][m] =[kgm2/s]

A person of mass m = 50 kg stands on a rope ladder that is hanging from a freely floating balloon of mass M = 100 kg. The balloon is initially at rest with respect to the ground. (The buoyant force on the person-balloon system is countering the force of gravity.) At what speed will the balloon move if the person starts to climb the rope ladder at constant velocity of v = 30 m/s relative to the ladder?

Pf=Pi m(p)v(pg)+m(b)v(bg) = 0 m(p)[v(pb)+v(bg)]+m(b)v(bg) =0 m(p)v(vb)+[m(b)m(p)]v(bg) = 0 -[50/(50+100)](30) = -10

Two blocks of masses m1 and m2 are connected by a massless cord that passes over a massless pulley as shown in the figure below. They are released from rest at the same height. What is the change of the total potential energy of the system at the moment the block m1 reaches the ground after traveling a distance h?

Setting x-axis at the dashed line, Ui = 0 Uf = -m1gh + m2gh sin0 U = Uf - Ui = -m1gh + m2gh sin0

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see figure below). Using the symbols below, including g, what is a correct expression for the tangential speed of the bob?

T cos0 - mg = m*0 T = mg/cos0 Fnet, center = mv2/r T sin0 = mv2/r (mg/cos0) sin0 = mv2/r v2 = gr tan0 V2 = g (L sin0) tan0 v = √(g L sin0 tan0)

Two planets with the same mass orbit a star. Planet #1 has an orbit radius of r1 and planet #2 has an orbit radius one half that of planet #1 such that r1 = 2r2. If planet #1 has an orbit period of T1, what is the period of planet #2, T2?

T1^2 = c r1^3 T2^2 = c r2^3 c = T1^2 / r1^3 = T2^2 / r2^3 (T2/T1)^2 = (r2/r1)^3 = (1/2)^3 = 1/8 T2/T1 = 1/√8

A v-vs-t graph is drawn for a ball moving in one direction. The graph starts at the origin and at t = 5 s the acceleration of the ball is zero. We know that at t = 5s,

The acceleration of the ball a = 0, therefore the velocity is constant or not changing.

A skater of mass 40 kg standing on ice throws a stone of mass 20 kg with a speed of 40 m/s in a horizontal direction. Find the distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between the skater and the ice is 1/2. Use g = 10m/s2.

The throwing of the stone: Pf =Pi ;m(skater)v(skater)+ m(stone) v(stone) =0 v(skater) = -[m(stone)/m(skater)]v(stone) =-(20 / 40)(40 m/s i) = -20 i m/s The slowing down of the skate due to friction: Kf +Uf +Eth = Ki+Ui+W(other) 0+mgy+ukmgx =1/2mv^2+mgy+0 ukgx = 1/2 v^2 (1/2)(10m/s2)x =(1/2)(20m/s)2 x =40m

A block slides down (starting from rest) a 3-4-5 incline with friction as illustrated below. If the coefficient of kinetic friction is k = 1/3, how long will it take to get to the bottom of the incline?

ax = g(sin0 - uk cos0) = (10 m/s2)(4/5 - 1/3 3/5) = 6 m/s2 x = vix t + 1⁄2 ax t2 5 = 0 + 1⁄2 (6 m/s2) t2 t = √(5/3) s.

Two Boy Scouts, Bobby and Jimmy, are carrying a uniform wooden tent pole 12.0 m in length and having weight W = 450 N. Bobby is holding the pole 1.0 m from the back, and Jimmy is holding the pole 2.0 m from the front. If the pole is held horizontally by the two boys, what is the force Jimmy exert on the beam?

choosing Bobby's hand as the pivot: Static Equilibrium condition tnet = tNB + tW + tNJ = 0 0-(-5m)(450N)+(9m)NJ = 0 NJ = 250N

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10,000 N. How far does the car travel before stopping?

m=2000kg v0 =30m/s Fbrake =10,000N Fnet,x = -max; ax = -Fbrake/m = -10000N/2000 kg = -5 m/s2 vf2 = v02+2ax x, 0 = (30m/s)2 +2(-5m/s2)x x=90 m

A car drives 2 miles east and then one mile north and then 4 miles south. What is the car's displacement?

r=(2i+j-4j)miles =(2i‒3j)miles |r|= sqrt(2^2 +3^2) = sqrt(13) miles tan(theta) = 3/2 theta = arctan(3/2) South of East

An object moves with an initial velocity vi = 3 i m/s and acceleration a = -10 j m/s2. Assume the object is initially at position ri =(2 i + 4 j ) m. What is the position vector of the object as a function of time?

r=ri +vi t+1⁄2at2 =(2i+4j)+3it+1⁄2(‒10j)t2 =(2+3t)i+(4‒5t2)j

When a woman who is 1.5 m tall stands near a flagpole, the shadow of her head and the shadow of the top of the flagpole are superimposed as shown in the figure below. The woman's shadow is 10 m long. Determine the height of the flagpole.

tan(a) = h/(30 m) = (1.5 m)/(10 m) h = (1.5/10)*(30 m) = 4.5 m

A uniform disk has a moment of inertia about its center of mass of ICM = 1⁄2 MR2. The disk as a radius of R = 2 m and a mass of M = 1 kg. A force of N is applied to the disk as illustrated. If the disk rotates about its center, what is the magnitude of its angular acceleration?

torque =r*F = (-2j m)*(3i-4j)N = -6k N.m The magnitude of angular acceleration = |a| = |torque|/I 6N.m/[1kg(2m)^2/2]=3rad/s^2

The figure shows the y-position (in blue) of a particle versus time. What is the average velocity of the particle during the time interval t = 2.0 s to t = 4.0 s?

vavg = (yf - yi) j/(tf - ti) = (18 m - 14 m) j / (2 s) = 2 j m/s 18 and 14 are from the graph

An object of mass m = 1 kg that is moving to the right with a speed of 20 m/s collides head-on with another object, and the collision lasts 1.5 s. A graph showing the magnitude of the force during the collision versus time is shown in the figure below, where the force is exerted in the direction opposite the initial velocity. Find the x- component of the velocity of the 1-kg mass after collision.

vi = 20im/s pi = mvi = (1kg)(20im/s) =20ikgm/s The impulse I = -i area under F vs t graph = -40i N.s pf = pi + I = 20i kgm/s -40i N.s = -20i kgm/s vf = -20im/s

A river flows due east at 3.0 m/s. A boat crosses the 300-m-wide river by maintaining a constant velocity of 10 m/s due north relative to the water. If no correction is made for the current, how far downstream does the boat move by the time it reaches the far shore?

vwg = 3i m/s vbw = 10j m/s vbg =vbw +vwg =10jm/s+3im/s rbg = vbg t xi + yj = (3i + 10j) m/s t xi + 300j m = 3 t i + 10 t j 300j = 10tj t = 30 s. x = (3 m/s) t = (3 m/s) (30 s) = 90 m

An object begins to move along the y axis and its position is given by the equation y = -2 +10t - 2.5t2, with y in meters and t in seconds. What is the time t when the object's velocity is -5 j m/s?

vy =dy/dt=d(‒2+10t‒2.5t^2)/dt =+10-5t=‒5 t=3.0s

A ball rotates in uniform circular motion with a period of T = 10 s. If the tangential speed of the ball is v = 4pi m/s, what is the radius of the circle?

w = 2pi/T = 2pi/(10 s) = 0.2pi rad/s v = wr r = v/w = (4pi m/s) / (0.2pi rad/s) = 20 m

As you are walking along an alley in downtown toward a main street, you glimpse a particularly stylish Alpha Romeo pass by. Tall buildings on either side of the alley obscure your view, so you see car only as it passes between the buildings. The width of the alley between the two buildings is 36 m. The car was in view for 1.0 s. You also heard the engine rev when the car started from a red light, so you know the Alpha Romeo started from rest 4 s before you first saw it. Find the magnitude of the acceleration.

x1 =x0 +v0x t1 +1⁄2ax t12 =0+0+1⁄2ax (4s)2 =8ax x2 =x0 +v0x t2 +1⁄2ax t22 =0+0+1⁄2ax (5s)2 =12.5ax x2 -x1 =36m 12.5ax -8ax =36 ax =8m/s2

An automobile driver puts on brakes and decelerates from 30.0 m/s to zero in 10.0 s. What distance does the car travel?

x=1⁄2(vix +vfx)t =1⁄2(30m/s+0m/s)(10s)=150m

Find the center of mass (xCM, yCM) of a system with three particles of masses m1 = m2 = m3 = 1.0 kg kept at the vertices of an equilateral triangle of side 1.0 m (see figure below). Use sin(60) = 1⁄2√3, cos(60) = 1⁄2.

xcm = m1x1+m2x2+m3x3/(m1+m2+m3) X = (0+1/2+1)/3 = 1/2 Y = (0+1/2√(3) + 0)/3 = 1/2√(3) (1/2,1/2√3)

A rock is thrown straight down with an initial velocity of 15 m/s from a cliff. What is the rock's magnitude of displacement after 2.0 s? (Use acceleration due to gravity g = 10 m/s2)

y=viy t+1⁄2ayt2 =(-15m/s)(2.0s)+1⁄2(-10m/s2)(2.0s)2 =-30m-20m=-50m