Question

A ball of $2\text{kg}$ strikes a wall with a velocity $1\text{m/s}$ at an angle ${30}^{\circ }$ with the horizontal as shown in the figure below. It reflects at the same angle from the wall in $0.1\text{s}$. Force exerted on the ball will be:

• A. $20\sqrt{3}\text{N}$
• B. $15\sqrt{3}\text{N}$
• C. $10\sqrt{3}\text{N}$
• D. $5\sqrt{3}\text{N}$

Answer 1

The correct option is A $20\sqrt{3}\text{N}$



Change in momentum of ball perpendicular to the wall $=m(v\cos \theta -(-v\cos \theta \left)\right)$
(taking rightwards +ve)
$=2mv\cos \theta =2\times 2\times 1\times \cos {30}^{\circ }=2\times 2\times \frac{\sqrt{3}}{2}=2\sqrt{3}\text{kg m/s}$
We know that force is equal to rate of change of momentum
So, Rate of change of momentum
$=\frac{\text{change in momentum}}{\text{time}}=\frac{2\sqrt{3}}{0.1}=20\sqrt{3}\text{N}\therefore \text{Force on the ball}=20\sqrt{3}\text{N}$