A rigid ball of mass $M$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed, as sho

Question

A rigid ball of mass $M$ strikes a rigid wall at $60^{\circ}$ and gets reflected without loss of speed, as shown in the figure. The value of the impulse imparted by the wall on the ball will be:

Accepted Answer

1. **Concept:** Impulse is defined as the change in momentum ($\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$). The force exerted by the wall acts only in the direction perpendicular to the wall (normal direction) [NCERT Class 11, Physics Part I, Sec 5.7].
2. **Resolve Components:** Let the angle of incidence with the normal be $	heta = 60^{\circ}$.
   - **Initial Momentum ($\vec{p}_i$):**
     $p_{ix} = Mv \cos 60^{\circ}$ (towards the wall)
     $p_{iy} = Mv \sin 60^{\circ}$ (parallel to the wall)
   - **Final Momentum ($\vec{p}_f$):**
     $p_{fx} = -Mv \cos 60^{\circ}$ (away from the wall)
     $p_{fy} = Mv \sin 60^{\circ}$ (parallel to the wall, unchanged)
3. **Calculate Impulse:**
   - Along the wall (y-axis): $\Delta p_y = p_{fy} - p_{iy} = 0$.
   - Perpendicular to the wall (x-axis): $\Delta p_x = p_{fx} - p_{ix} = (-Mv \cos 60^{\circ}) - (Mv \cos 60^{\circ}) = -2Mv \cos 60^{\circ}$.
4. **Magnitude:**
   $$ |\vec{J}| = |-2Mv \cos 60^{\circ}| = 2Mv \left(\frac{1}{2}ight) = Mv $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Laws of Motion, Example 4.5).

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