Question

A $3\text{kg}$ steel ball strikes a wall with a speed of $10{\text{ms}}^{-1}$ at an angle of ${60}^{\circ }$ with the surface of the wall. The ball bounce off with the same speed and same angle. If the ball in contact with the wall for $0.25\text{s}$. The average force exerted by the wall on the ball is

• A. $75\sqrt{3}\text{N}$
• B. $30\sqrt{3}\text{N}$
• C. $150\sqrt{3}\text{N}$
• D. $60\sqrt{3}\text{N}$

Answer 1

The correct option is C $150\sqrt{3}\text{N}$




From impulse-momentum theorem,

$\overrightarrow{\Delta P}=\overrightarrow{F}\cdot \Delta t$

$\Rightarrow \text{Change in momentum of the ball}=\text{Impulse on the ball}$

$\Rightarrow P_f-P_i=F\cdot \Delta t$

$\Rightarrow [mv\cos {60}^{\circ }\hat{j}-mv\sin {60}^{\circ }\hat{i}]-[mv\cos {60}^{\circ }\hat{j}+mv\sin {60}^{\circ }\hat{i}]=F\cdot \Delta t$

$\Rightarrow -2mv\sin {60}^{\circ }\hat{i}=F\cdot \Delta t$

$\Rightarrow \left|\begin{array}{c}-2mv\sin {60}^{\circ }\end{array}\right|=F\cdot \Delta t$

$\Rightarrow F=\frac{2mv\sin {60}^{\circ }}{\Delta t}$

Given,

$m=3\text{kg};v=10{\text{ms}}^{-1}\theta ={60}^{\circ };\Delta t=0.25\text{s}$

$F=\frac{2\times 3\times 10\times \frac{\sqrt{3}}{2}}{0.2}=150\sqrt{3}\text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question ? Caution: Average force is calculated as ratio of impulse to time of collision.