A body of mass $M$ hits normally a rigid wall with velocity $v$ and bounces back with the same velocity. The i

Question

A body of mass $M$ hits normally a rigid wall with velocity $v$ and bounces back with the same velocity. The impulse experienced by the body is:

Accepted Answer

1. **Concept:** Impulse is defined as the change in momentum of a body ($Δ\vec{p} = \vec{p}_f - \vec{p}_i$). This is a direct application of Newton's Second Law involving time duration [NCERT Physics Class 11, Laws of Motion, Section 4.5/5.7].
2. **Setup:** 
   - Let the initial velocity be directed towards the wall. Taking this direction as positive, $\vec{v}_i = +v$.
   - The body bounces back normally (perpendicularly) with the same speed. The direction is reversed, so $\vec{v}_f = -v$.
3. **Calculation:**
   - Initial Momentum ($p_i$) = $Mv$
   - Final Momentum ($p_f$) = $M(-v) = -Mv$
   - Impulse ($J$) = Change in Momentum = $p_f - p_i$
   - $J = -Mv - Mv = -2Mv$
4. **Magnitude:** The magnitude of the impulse imparted by the wall on the body is $|-2Mv| = 2Mv$.
   (Reference: NCERT Physics Class 11, Laws of Motion, Example 4.5 deals with a similar case of billiards balls striking a wall).

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