Question

Two train carts are moving in the same direction, such that one cart catches up with and collides with the second cart. That is, a 480-kg train cart moving at 14.4 m/s hits from behind a 570-kg train cart that is moving at 13.3 m/s in the same direction. If the new speed of the 570-kg cart is 14.0 m/s after the collision, what is the speed of the lighter cart after the collision

Answer 1

Answer:

14m/s

Explanation:

According to the law of conservation of momentum, the sum of momentum of the bodies before collision is equal to the sum of momentum of the bodies after collision. The two bodies move with the same velocity after collision.

If the new speed of the 570-kg cart is 14.0 m/s after the collision, then the speed of the lighter cart (480kg cart) will also be 14.0m/s since both carts moves with a common velocity after their collision according to the law.

Answer 2

Answer:

13.57 m/s

Explanation:

According to the law of momentum conservation, the total momentum must be conserved before and after the collision.

Before the collision the total momentum is:

480*14.4 + 570*13.3 = 14493 kg m/s

After the collision, the total momentum is

480v + 570*14 = 480v + 7980

As the 2 momentum are the same, we can equate the 2 equations above:

480v + 7980 = 14493

v = (14493 - 7980)/480 = 13.57 m/s