A train car with mass m1 = 661 kg is moving to the right with a speed of v1 = 8.

Question

A train car with mass m1 = 661 kg is moving to the right with a speed of v1 = 8.7 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 = 5.6 m/s.

1) What is the initial momentum of the first train car?

2) What is the mass of the second train car?

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

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= 8.7 m/s and collides with the second train car that is now moving to the left with a velocity v2= -6.1 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

Explanation / Answer

1. total momentum = m1v1 = 661*8.7= 5750.7 2. because m1v1 + m2v2 = m3 v3.... 661*8.7 + m*0 = (m1+m2)(5.6) so 5750.7 = (661+m2)(5.6) 5750.7 = 3701.6 +5.6m2 2049.1 = 5.6m2 m2 = 365.9kg 3. initial KE = (.5)m1(v^2) + (.5)(m2)(v^2) = (.5)(661)(8.7)^2 + 0 = initial = 25015.545 final KE = (.5)(m1+m2)(vf^2) = (.5)(1026.9)(5.6)^2 = 16101.792 KE difference = KE final - KE initial = 16101.792 - 25015.545 = -8913.753 4. m1 v1 + m2v2 = (m1+m2)v3 (661)(8.7) + (365.9)(-6.1) = (661+365.9)(v3) 3518.71 = (1026.9)v3 v3 = 3.42 m/s 5. momentum of train car 1 before collision = mv = 661*8.7 = 5750.7 momentum of train car 1 after collision = mv2 = 661*3.42 = 2260.62

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