A $0.5 ext{ kg}$ ball moving with a speed of $12 ext{ m/s}$ strikes a hard wall at an angle of $30^{\circ}

Question

A $0.5 	ext{ kg}$ ball moving with a speed of $12 	ext{ m/s}$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for $0.25 	ext{ s}$, the average force acting on the wall is:

Accepted Answer

1. **Identify the Angle:** The problem states the ball strikes at an angle of $30^{\circ}$ *with the wall*. This corresponds to a glancing angle. The angle with the normal (perpendicular to the wall) is $	heta = 90^{\circ} - 30^{\circ} = 60^{\circ}$. Alternatively, one can resolve components directly using the angle with the wall.
2. **Change in Momentum (Impulse):** The force exerted by the wall changes the momentum of the ball only in the direction perpendicular to the wall.
   - **Component perpendicular to the wall:**
     $v_{\perp} = v \sin(30^{\circ})$ (using angle with wall) or $v \cos(60^{\circ})$ (using angle with normal).
   - **Calculation of Impulse ($\Delta p$):**
     $\Delta p = m(v_{\perp, final} - v_{\perp, initial})$
     $\Delta p = m(v \sin 30^{\circ} - (-v \sin 30^{\circ})) = 2mv \sin 30^{\circ}$
     Substituting values: $\Delta p = 2 	imes 0.5 	imes 12 	imes 0.5 = 6 	ext{ kg m/s}$.
3. **Calculate Average Force:** According to Newton's Second Law:
   $$ F_{avg} = \frac{\Delta p}{\Delta t} $$
   $$ F_{avg} = \frac{6}{0.25} = 24 	ext{ N} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.7 regarding Impulse).

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