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

• A

  $96 \mathrm{~N}$
• B

  $48 \mathrm{~N}$
• Correct answer
  C

  $24 \mathrm{~N}$
• D

  $12 \mathrm{~N}$

Solution

Applying law of conservation of momentum. Change in final momentum = initial momentum $m v \sin \theta$ after collision $-(-m v \sin$ $\theta$) before collision $\begin{aligned} & f \times t=\text { change in momentum }=2 m v \sin \theta \\ & \Rightarrow \quad f=\frac{2 m v \sin \theta}{t} \end{aligned}$ Putting the values we get, $\begin{aligned} f= & \frac{2 \times 0.5 \times 12 \times \sin 30^{\circ}}{0.25} \\ & =24 \mathrm{~N} . \end{aligned}$

Hence, the correct answer is option: 'C'.