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

Answer:

Vb = v/2

Half of the speed of Car A before the collision

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:

$m_{A}*v_{A1} + 2m_{A}*0 = m_{A}*0 + 2m_{A}*v_{B2}\\m_{A}*v_{A1} = 2m_{A}*v_{B2}\\v_{B2}=\frac{v_{A1}}{2}$

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

Vb = v/2