A train car with mass m 1 = 646 kg is moving to the right with a speed of v 1 =

Question

A train car with mass m1 = 646 kg is moving to the right with a speed of v1 = 8.4 m/s and collides with a second train car. The two cars latch together during the collision and then move off to the right at vf = 5.3 m/s.

1)

What is the initial momentum of the first train car?
kg-m/s

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2)

What is the mass of the second train car?
kg

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3)

What is the change in kinetic energy of the two train system during the collision?
J

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4)

Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v1 = 8.4 m/s and collides with the second train car that is now moving to the left with a velocity v2 = -6.5 m/s before the collision. The two cars latch together at impact.

What is the final velocity of the two-car system? (A positive velocity means the two train cars move to the right a negative velocity means the two train cars move to the left.)
m/s

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5)

Compare the magnitude of the momentum of train car 1 before and after the collision:

p1 initial = p1 final

p1 initial > p1 final

p1 initial < p1 final

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Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v1 = 8.4 m/s and collides with the second train car that is now moving to the left with a velocity v2 = -6.5 m/s before the collision. The two cars latch together at impact. What is the final velocity of the two-car system? (A positive velocity means the two train cars move to the right ½ a negative velocity means the two train cars move to the left.) m/s You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. 5) Compare the magnitude of the momentum of train car 1 before and after the collision: p1 initial = p1 final p1 initial > p1 final p1 initial

Explanation / Answer

1) Momentum = Ma*Va=646*8.4=5426.4 kgm/s

2) applying conservation of momentum

M1V1+M2V2=(M1+M2)Vf

=>646*8.4+0=(646+M2)5.3

=>M2=377.84 kg

3)change in k.e.=1/2(M1+M2)Vf^2-1/2M1*V1^2-1/2M2*V2^2=-8410.92 joules

4)

applying conservation of momentum

m1v1+m2v2=(m1+m2)Vf

=>646*8.4-377.84*6.5=1023.84*Vf

=>2.9012 m/s to the right

5)

p1 initial > p1 final

is true since m1 will be same but speed before collision is greater in compare to after the collision

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