A ball of mass \( m \) approaches a wall of mass \( M \) (\( M \gg m \)) with speed \( 4\text{ m/s} \) along the normal to the wall. The speed of the wall is \( 1\text{ m/s} \) towards the ball. The speed of the ball after an elastic collision with the wall is:
\( 5\text{ m/s} \) away from the wall
\( 9\text{ m/s} \) away from the wall
\( 3\text{ m/s} \) away from the wall
\( 6\text{ m/s} \) away from the wall
Solution:
Using the coefficient of restitution \( e = 1 \), the relative velocity of separation equals the relative velocity of approach. The approach velocity is \( 4 - (-1) = 5\text{ m/s} \). After collision, the relative separation velocity is \( v' - 1 = 5 \implies v' = 6\text{ m/s} \) away from the wall.
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