PSII Review (Rosenmund)

Impulse

A change in momentum. The product of force and the time through which the force acts.

elastic collision

A collision in which colliding objects "bounce". 2->2

Inelastic collision

A collision in which the colliding objects "stick". 2->1

Explosion collision

A collision in which the colliding objects break apart. 1->2

Friction Force

A force that acts to resist motion of objects that are in contact

Momentum

A vector quantity that is the product of mass and velocity of an object

Force

A vector quantity that tends to accelerate an object; a push or pull

Newton's 3rd Law

Action and reaction; for every force on an object there is am equal and opposite force from that object

Free-Body Diagrams

Are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation

Kinetic Energy

Energy of an object due to its motion

Potential Energy

Energy that is stored in an object based on its position

Equation for force

F=ma

Newton's 2nd Law

F=ma; the force of an object is equal to the mass of the object multiplied by its acceleration.

Equation for impulse

F∆t

Equation for impulse-momentum theorem

F∆t= Pf-Pi

Acceleration due to gravity

G= 9.8 m/s/s

Newton's 1st Law

Inertia, an object at rest want to stay at rest, an object in motion tends to stay in motion unless acted on by an outside force

Equation for kinetic energy

KE= 1/2 mv/v

Work

Measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement

Free-Fall

Occurs whenever an object is acted upon by gravity alone. - ignore air resistance - acceleration due to gravity g= -9.8 m/s squared

Equation for momentum

P=mv

Equation for potential energy

PE= mgh

Equation for Conservation of Energy

PEi+KEi = PEf + KEf

Equation for conservation of momentum

Pf=Pi

Velocity

Rate of change of displacement with time. (Vector) (measured in m/s)

Speed

Rate of change of distance with time. (Scalar) (measured in m/s)

Acceleration

Rate of change of velocity with time. (Measured in m/s squared)

Example: A force of 6.0 N acts on n objects for 10 seconds. The mass of the object is 3 kg. What is the object's change in momentum (impulse)?

Remember: F*t=∆p F= 6.0 T= 10 seconds 6*10= ∆p = 60 N*s

Example: What is the net force acting on a 1.0 kg ball in free-fall?

Remember: In free-fall only gravity acts on the object. Therefore acceleration is g. F=mg M= 1.0 kg G= 9.8 m/s/s F= 1.0(9.8) = 9.8 N

Example: A 90.0 kg man and a 55 kg man have a tug-of-war. The 90.0 kg man pulls on the rope such that the 55 kg man accelerates at 0.025 m/s/s. What force does the rope exert on the 90.0 kg man?

Remember: Newton's third law; equal forces. Find the force on the 55 kg man using F=ma M= 55 kg A= 0.025 m/s/s F= 55(.025)= 1.375 N The force exerted on the 90.0 kg man is 1.375 because of Newton's 3rd law.

Example: A bowling ball is dropped from the top of a building. If it hits the ground with a speed of 37.0 m/s, how tall was the building?

Remember: PE=KE (mgh=1/2mv/v); masses cancel each other out Gh=1/2v/v (9.8)h=1/2(37)squared 9.8h=1/2(1369) = 9.8h= 684.5= 69.85 meters

UNFINISHED Example: A skydiver jumps from a hovering helicopter that's 3,000 meters above the ground. If air resistance can be ignored, how fast will he be falling when his altitude is 2,000 m?

Remember: PE=PE + KE; PE= gh + 1/2v/v ; masses cancel each other out

Examples: The potential energy of an apple is 6.12 joules. The apple is 3.46 meters high. What is the mass of the apple?

Remember: PE=mgh PE= 6.12 j H= 3.46 6.12= m(9.8)3.46 = .180 kg

Example: If a 1.8 kg brick falls to the ground from a chimney that is 6.7 meters high. What is the change in its potential energy?

Remember: PEf - PEi; PE=mgh PE= 1.8(9.8)(6.7) = 118.188 j Change in PE= 0-118.188= -118.188 j

Example: A 450 newton gymnast jumps a distance of 0.55 meters. How much work did she do?

Remember: W=fd; 450 is the Fg F= 450 N D= 0.55 meters W= 450(.55) = 247.5 j

Example: A pool cue striking a stationary billiard ball (m= 0.25 kg) gives the ball a speed of 2 m/s. If the average force of the cue on the ball was 200 N, over what distance did this force act?

Remember: W=∆KE; Fd= 1/2mv/v M= 0.25 kg V= 2 m/s F= 200 N 200d= 1/2(0.25)(2)squared 200d= 0.5 D= 400 m

Example: A car starts from rest and accelerates up to 22.0 m/s in 34.0 seconds. What is its acceleration?

Remember: a= ∆v/t ∆V= 22.0 m/s T= 34.0 seconds A= 22/34= .647 m/s/s

Example: An armadillo is rolling up a hill at 8.0 m/s, with an acceleration of -12.0 m/s/s down the hill. What is the velocity of the ball after 2.0 seconds? Assume a coordinate system of down as negative.

Remember: a= ∆v/t; a= Vf-Vi/t; be careful with negative and positive signs. Solve for Vf. A= -12.0 m/s/s Vi= 8.0 m/s T= 2.0 seconds Vf= -12= Vf-8/2 = -8 m/s

Example: A ball is dropped from a cliff and has an acceleration of -9.8 m/s/s. How long will it take for the ball to reach a speed of -58.8 m/s?

Remember: a= ∆v/t; a= Vf-Vi/t; initial velocity is zero; solve for time A= -9.8 Vf= -58.8 -9.8= -58.8/t —> t= ∆v/a T= -58.8/-9.8= 6 seconds

Example: A 873 kg dragster, starting from rest, attains a speed of 26.3 m/s in 0.59 seconds. Find the average acceleration of the dragster during this time interval, and the magnitude of the average net force on the dragster during this time, assume that the driver has a mass of 68 kg and find its horizontal force seat exert on the driver.

Remember: a=Vf-Vi/t; started at rest, so initial velocity is zero, f=ma, same acceleration for the driver as the car, but different mass, use f=ma Average acceleration= 26.3-0/.59 = 44.58 m/s Average net force= 873(44.58)= 38918.34 N Horizontal force= 68(44.58)= 3031.44 N

Example: A car of mass 700 kg travels at 20 m/s. The car collides with a stationary truck of mass 1400 kg. The two vehicles interlock as a result of the collision. What is the velocity of the car-truck system?

Remember: inelastic collision; conservation of momentum; Mcar V car= MtotalV M= 700 kg V= 20 m/s 700(20)= 1400 N*s

Example: What is the potential energy of a 15.24 kg ball on the ground?

Remember: no height= no PE

Example: If Mr. Rosenmund around a 400 meter track twice in 300 seconds what his average speed, average velocity, and justify why.

Remember: speed= distance/time; velocity= displacement/time; velocity is based on displacement no distance, so if you end where you started there is no displacement. Average speed= 400/300 = 1.33 m/s Average velocity= 0/300 = 0 m/s

Example: How far could a paper airplane go if it has a velocity of 42.0 m/s north and stays in the air for 4.00 seconds?

Remember: v= ∆x/t, solve for ∆x, ∆=v*t V= 42.0 m/s T= 4.00 ∆x= 42*4= 168m

Example: A ball is kicked east. It takes the ball 3.00 seconds to travel a distance of 24.0 meters. What is its velocity?

Remember: v=∆x/t X= 24.0 meters T= 3.00 seconds V= 24/3 = 8 m/s EAST

Example: Your new motorcycle weighs 2450 N. What is its mass in kilograms?

Remember: weight is the force caused by gravity. Fg=mg; m=Fg/G F= 2450 N G= 9.8 m/s/s M= 2450=m(9.8) = 250 kg

Distance

Scalar measure of the interval between two locations meassured along the actual path connecting them. (Measured in meters)

Conservation of Energy

States that the total energy of an isolated system remains constant

Impulse Momentum Theorem

The impulse on an object is equal to the object's final momentum minus the object's initial momentum.

conservation of momentum

The momentum of a system will remain constant. Momentum isn't created or destroyed unless an outside force is acting on the system.

Equation for velocity

V= ∆x/t AND v= ∆d/∆t

Displacement

Vector measure of the interval between two locations along the shortest path connecting them. (Measured in meters)

Equation for work

W= fd

Equilibrium

When net force is equal to zero

Can an object be moving when its acceleration is zero? If so, give an example. If not explain.

Yes, constant velocity. And example is cruise control.

Can an automobile with a velocity toward the north simultaneously have an acceleration toward the south?

Yes, if the velocity and acceleration have the same sign (both positive or negative) then it will speed up; if they have opposite signs it will slow down.

Scalar

magnitude

Vector

magnitude and direction

Normal Force

support force that acts perpendicular to a surface

Equation for displacement

∆x = xf - xi