Consider the elastic collision of two bodies A and B of equal mass. Initially B is at rest and A moves with velocity v. After the collision :

the body A traces its path back with the same speed

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the body A comes to rest and B moves away in the direction of A's approach with the velocity v

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both the bodies stick together and are at rest

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B moves along with velocity

v / 2

and A retraces its path with velocity

v / 2

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Solution

The correct option is A the body A comes to rest and B moves away in the direction of A's approach with the velocity v
Given : $m_{1} = m_{2} = m$ $u_{1} = v$ $u_{2} = 0$ $e = 1$

Let the velocities of A and B be $v_{a}$ and $v_{b}$ respectively after the collision.

Using $v_{a} = \frac{(m_{1} - e m_{2}) u_{1} + (1 + e) m_{2} u_{2}}{m_{1} + m_{2}}$

$v_{a} = \frac{(m - 1 \times m) v + (1 + 1) m (0)}{m + m} ⟹ v_{a} = 0$

Using $v_{b} = \frac{(m_{2} - e m_{1}) u_{2} + (1 + e) m_{1} u_{1}}{m_{1} + m_{2}}$

$v_{b} = \frac{(m - 1 \times m) 0 + (1 + 1) m v}{m + m} ⟹ v_{b} = v$

Thus body A comes to rest whereas B moves with velocity $v$ in the same direction as that of A initially.

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