Question

A ball of mass $0.5\mathrm{Kg}$ moving with a velocity of $2{\mathrm{ms}}^{-1}$ strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:

• A. $2000\mathrm{N}$
• B. $1000\mathrm{N}$
• C. $5000\mathrm{N}$
• D. $125\mathrm{N}$

Answer 1

The correct option is A

$2000\mathrm{N}$

Step 2: Given data

We are given with,

Mass of a ball $\mathrm{m}=0.5\mathrm{Kg}$

Final velocity of a ball, $\mathrm{v}=2{\mathrm{ms}}^{-1}$

Initial velocity of a ball, $\mathrm{u}=-2{\mathrm{ms}}^{-1}(\because \mathrm{direction}\mathrm{of}\mathrm{velocity}\mathrm{changes}\mathrm{after}\mathrm{striking}\mathrm{the}\mathrm{wall})$

Time, $\mathrm{t}=1\mathrm{ms}={10}^{-3}\mathrm{s}$

Step 2: Calculating the average force exerted by a wall

Now we know that force is equal to rate of change of linear momentum, hence

$$\begin{gathered} \mathrm{F}=\frac{\mathrm{mv}-\mathrm{mu}}{\mathrm{t}}=\frac{(0.5\times 2)-(0.5\times (-2\left)\right)}{{10}^{-3}} \\ \Rightarrow \mathrm{F}=\frac{1+1}{{10}^{-3}}=2\times {10}^3=2000 \\ \Rightarrow \mathrm{F}=2000\mathrm{N} \end{gathered}$$

Hence, average force exerted by a wall is equal to $2000\mathrm{N}$.

Therefore, option(A) is correct answer.