Physics chapter 6

Conservation of momentum equation

m1v1,I + m2v2,i=m1v2,f+m2v2f

Perfectly inelastic collision equation

m1v1,i + m2v2,i = (m1 + m2)(vf)

Collisions

-total momentum remains constant in any type of collision -KE is generally not conserved in a collision (it may be converted sound or another form of energy)

Sample problem F (kinetic energy in perfectly inelastic collisions) Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.00 m/s to the right. The second ball has a mass of 0.250 kg and an initial velocity of 3.00 m/s to the left. What is the decrease in kinetic energy during the collision?

1. m1=0.500 kg v,i=4.00 m/s m2=0.250 kg v2,i= 3.00 m/s to the left initial KE = 1/2(m1)(v1,i)^2 + 1/2(m2)(v2,i)^2 final KE = 1/2(m1+m2)(vf)^2 m1v1,i + m2v2,i = (m1 +m2)(vf) (0.5)(4) + (0.25)(-3) = (0.5 + 0.25)(vf) 1.25 = 0.75(vf) 1.67 m/s to the right = vf initial KE= 1/2(.5)(4)^2 + 1/2(.25)(-3)^2 4 + 1.125 = 5.125 J final KE = 1/2(.5 + .25)(1.67)^2 0.375(2.7889)=1.05 J ΔKE= KEf - KEi ΔKE= 1.05 - 5.125 ΔKE= -4.075 J

Sample problem E (perfectly inelastic collisions): A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision?

1. m1=1850 kg m2=975 kg m1,i=0 m/s m2,f=22 m/s vf=? 2. m1v1,i + m2v2,i = (m1 + m2)(vf) 0+975(22)=(1850 + 975)vf 21450=(2825)vf 7.59 m/s north =vf

Sample problem D (conservation of momentum): A 76 kg boater, initially at rest in a stationary 45, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

1. m1=76 kg v1,i=0 m/s v1,f= 2.5 m/s m2=45 kg v2,i=0 m/s v2,f= ? 2. m1v1,i + m2v2,i = m1v1,f + m2v2,f 76(0)+45(0)=76(2.5)+45(v2,f) 0=190+45(v2,f) -190=45(v2,f) -4.22 m/s to the right 4.22 m/s to the left

Sample problem B (impulse momentum theorem): a 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the force exerted on the car during the collision.

1. m=1400 vi=15 vf=0 Δt= 0.30 2. FΔt=mvf-mvi F(0.30)=1400(0)-1400(15) F(0.30)=-21000 F=21000/.30 F= 70,000 N to the east

Sample problem C (stopping distance): A 2240 kg car traveling to the west slows down uniformly from 20.0 m/s to 5 m/s. How long does it take the car to decelerate if the force on the car is 8410 N to the east? How far does the car travel during deceleration?

1. m=2240 kg vi=20 m/s west (-20 east) vf=5 m/s west (-5 east) F=8410 N 2. FΔt=mvf-mvi 8410Δt=2240(-5)-2240(-20) 8410Δt=-11200+49280 8410Δt=38080/8410 Δt=4 s Δx=1/2(vi+vf)Δt Δx=1/2(20+5)(4) Δx=12.5(4) Δx=50.00 m west

Sample problem A (momentum): a 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

1. m=2250 kg v= 25 m/s 2. p=mv p= (2250)(25) p=56,250 kg•m/s to the east

Momentum & Kinetic energy conserved in elastic collision equations

1/2(m1)(v1,i)^2 + 1/2(m2)(v2,i)^2 = 1/2(m1)(v1,f)^2 + 1/2(m2)(v2,f)^2 m1(v1,i) + m2(v2,i) = m1(v1,f) + m2(v2,f)

Elastic collision

A collision in which the total momentum and the total kinetic energy are conserved

Perfectly inelastic collision

A collision in which two objects stick together after colliding (i.e. the objects collide and move together as one)

Momentum

A quantity defined as the product of mass and velocity of an object

Momentum is conserved in...

Collisions

Kinetic energy is NOT...

Conserved in inelastic collisions

Most collisions are neither...

Elastic nor perfectly inelastic

Formulas

Equation of momentum p=mv Impulse-momentum theorem FΔt=mvf- mvi Conservation of momentum equation m1v1,I + m2v2,i=m1v2,f+m2v2f Perfectly inelastic collision equation m1v1,i + m2v2,i = (m1 + m2)(vf) initial KE 1/2(m1)(v1,i)^2 + 1/2(m2)(v2,i)^2 final KE 1/2(m1+m2)(vf)^2 Change in Kinetic Energy ΔKE= KEf - KEi Momentum & Kinetic energy conserved in elastic collision equations 1/2(m1)(v1,i)^2 + 1/2(m2)(v2,i)^2 = 1/2(m1)(v1,f)^2 + 1/2(m2)(v2,f)^2 m1(v1,i) + m2(v2,i) = m1(v1,f) + m2(v2,f)

Alternate form of F=ma

F=Δp/Δt

A change in momentum requires:

Force and time

Impulse-momentum theorem

FΔt=mvf- mvi

Elastic material

Material in which the work done that deforms the material during a collision is equal to the work the material does to return to its original shape

Momentum is conserved for....

Objects punching away from each other

Momentum conservation

One object with momentum collided with another object whose momentum is zero. The first object loses momentum during the collision and the other object gains momentum. (Pa,i+Pb,i=Pa,f+Pb,f)

Stopping times and distances depend on:

The impulse momentum theorem (i.e. a truck with 2x the mass of another truck requires 2x more time to stop and 2x more distance to stop than the other truck)

Impulse

The product of force and time over which the force acts on the object (FΔt)

Forces in real collisions are not...

constant during the real collisions

Momentum is measured in:

kg•m/s

Equation of momentum

p=mv