A train car with mass m_1 = 566 kg is moving to the right with a speed of v_1 =

Question

A train car with mass m_1 = 566 kg is moving to the right with a speed of v_1 = 8.8 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 v_f = 5.5 m/s. What is the initial momentum of the first train car? What is the change in kinetic energy of the two train system during the collision? Now the same two cars are involved in a second collision. The first car is again moving to the right with a speed of v_1 8.8 m/s and collides with the second train car that is now moving to the left with a velocity v_2 = -6.2 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.) Compare the magnitude of the momentum of train car 1 before and after the collision: P_1 initial = P_1 final P_1 initial > P_1 final P_1 initial

Explanation / Answer

mass of the train1, m1=566kg

speed, vi=8.8 m/sec

speed, vf=5.5 m/sec

1)

initial momentum of first train car,

P1=m1*v1

P1=566*8.8

P1=4980.8 kg.m/sec^2

2)

P1=P2

m1*u1=(m1+m2)*vf

566*8.8=(566+m2)*5.5

===> m2=339.6 kg

mass of the train, m2=339.6 kg

3)

change in K.E,

dK=K1-K2

=1/2*m1^vi2 -1/2*(m1+m2)*vf^2

=1/2*566*8.8^2 - (1/2*(566+339.6)*5.5^2)

=8218.32 J

4)

m1*u1+m2*u2=(m1+m2)*vf

566*8.8+339.6*(-6.2)=(566+339.6)*vf

===> vf=3.175 m/sec^2

5)

momentum is conserved,

P1(initial) = P1(final)

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