Question

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. Find the speed of Car B after the collision (in terms of v).

Answer 1

Vb = v/2

Half of the speed of Car A before the collision

Step-by-step explanation:

This problem can be solved by using conservation of momentum.

`m_(A)*v_(A1) + m_(B)*v_(B1) = m_(A)*v_(A2) + m_(B)*v_(B2)`

Since car B was stopped before the collision and its mass is twice the mass of A:

`v_(B1) = 0\\m_(B) = 2*m_(A)`

Also, since it is an elastic collision, car A stops after hitting car B. Rewriting the equation:

`m_(A)*v_(A1) + 2m_(A)*0 = m_(A)*0 + 2m_(A)*v_(B2)\\m_(A)*v_(A1) = 2m_(A)*v_(B2)\\v_(B2)=(v_(A1))/(2)`

Therefore, the speed of car B after the collision is half of the speed of car A before the collision (v)

Vb = v/2