A ball of mass $m$ approaches a wall of mass $M (> > m)$ with the speed of $4 m / s$ along normal to the wall. The speed of the wall is $1 m / s$ towards the ball. The speed of the ball after an elastic collision with the wall is

$5 m / s$ away from the wall

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$3 m / s$ away from the wall

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$9 m / s$ away from the wall

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$6 m / s$ away from the wall

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Solution

The correct option is D
$6 m / s$ away from the wall

Step 1: Given Data
Speed of the wall $v_{w} = 1 m / s$
Mass of the wall $= M$
Speed of the ball $v_{b} = 4 m / s$
Mass of the ball $= m$
$M > > m$
The collision is elastic in nature.
Step 2: Calculate the required velocity
Since $M > > m$ the velocity of the wall will not change after the collision.
Let the velocity of the ball after collision be $v$ .
Relative velocity of approach $=$ Relative velocity of separation
$∴ v_{b} - (- v_{w}) = v - v_{w}$
$\Rightarrow 4 + 1 = v - 1$
$\Rightarrow v = 6 m / s$
Hence, the correct answer is option (D).

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