Train A, mA = 150,000 kg is traveling west at 70 km/h . Train B, mB = 100,000 kg

Question

Train A, mA = 150,000 kg is traveling west at 70 km/h . Train B, mB = 100,000 kg, behind train A on the same track, is traveling west at 83 km/h and so is gaining on train A. Because the engineer driving B fails to slow down, B runs into the back of A. The two trains stick together and then move as a single unit after the collision. Let the west direction be the +x direction.

1.) What is the x-component of the momentum of train A before the collision according to an observer standing alongside the tracks?

2.) What is the x-component of the momentum of train B before the collision according to an observer standing alongside the tracks?

3.) What is the x-component of the momentum of train A after the collision according to an observer standing alongside the tracks?

4.) What is the x-component of the momentum of train B after the collision according to an observer standing alongside the tracks?

5.) What is the x-component of the momentum of train A before the collision according to an observer in a westbound automobile traveling at 100 km/h on a road that parallels the tracks?

6.) What is the x-component of the momentum of train B before the collision according to an observer in a westbound automobile traveling at 100 km/h on a road that parallels the tracks?

7.) What is the x-component of the momentum of train A after the collision according to an observer in a westbound automobile traveling at 100 km/h on a road that parallels the tracks?

8.) What is the x-component of the momentum of train B before the collision according to an observer in a westbound automobile traveling at 100 km/h on a road that parallels the tracks?

Explanation / Answer

mA = mass of train A = 150,000 kg

VA = speed of train A = 70 km/h = 19.44 m/s

mB = mass of train B = 100,000 kg

VB = speed of train B = 83 km/h = 23.06 m/s

1)

x-component of the momentum of train A = PAix = mA VA = (150,000 ) (19.44 ) = 2.92 x 106 kgm/s

2)

x-component of the momentum of train B = PBix = mB VB = (100,000 ) (23.06 ) = 2.31 x 106 kgm/s

3)

V = final common speed

Using conservation of momentum

initial total momentum = final total momentum

mA VA + mB VB = (mA + mB ) V

2.92 x 106 + 2.31 x 106 = (250,000) V

V = 20.92 m/s

after collision :

x-component of the momentum of train A = mA V = (150,000 ) (20.92 ) = 3.14 x 106 kgm/s

4)

after collision :

x-component of the momentum of train B = mB V = (100,000 ) (20.92 ) = 2.09 x 106 kgm/s

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