Question

A $0.5kg$ ball moving with a speed of $12m{s}^{-1}$ strikes a hard wall at an angle of $30°$ 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 $0.25s$ the average force acting on the wall is

• A. $96N$
• B. $48N$
• C. $24N$
• D. $12N$

Answer 1

The correct option is C

$24N$

Step 1: Given data

Mass of the ball is $m=0.5kg$

The speed of the ball is $v=12m{s}^{-1}$

The time of contact is $t=0.25s$

The angle with the wall is, $\theta =30°$

Step 2: Finding the change in momentum, $∆p$

Change in momentum is equal to the difference between initial momentum $\left(p_i\right)$ and final momentum $\left(p_f\right)$.

$∆p=p_f-p_i$

$∆p=OB\sin 30°-\left(-OA\sin 30°\right)$

[Where $OA,OB$ represents the magnitude of the momentum before striking after striking the wall respectively.]

As the magnitudes of $OA,OB$ are equal, the components of $OA,OB$ along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite.

Thus the change in momentum is only due to a change in direction of perpendicular components.

$∆p=\left(mv\sin 30°\right)-\left(-mv\sin 30°\right)$ [As the velocities are in opposite direction]

$∆p=2mv\sin 30°$ [$eq\left(1\right)$]

Step 3. Finding the force applied to the wall

By using the formula of force, $F$

$F=\frac{∆p}{t}$ [Where $t$ is time, $∆p$ is a change in momentum]

$F=\frac{2mv\sin 30°}{t}$

$F=\frac{2\times 0.5\times 12\times 1}{0.25\times 2}$ $\left[\because \sin 30°=\frac{1}{2}\right]$

$F=24N$

Hence, the correct option is (C).