Question

A ball of mass 250 g moving with a speed of hits a wall normally . it rebounds with the same speed . If the two were in contact for ${10}^{-3}s$ , the average force exerted by the wall on the ball is

• A. $4\times {10}^{-3}N$
• B. $2\times {10}^{-3}N$
• C. ${10}^{-3}N$
• D. ${10}^{-2}N$

Answer 1

The correct option is B $2\times {10}^{-3}N$

$\begin{array}{c}\text{Correct option is}\left(B\right)\\ \text{As we know;}\\ F=\frac{dP}{dt}=\frac{mdV}{dt}\\ \therefore \frac{250}{1000}=1/4kg=\frac{1/4(V-(-v\left)\right)}{{10}^{-3}}\\ \text{Here,}V=(4\times {10}^{-6})\frac{m}{B}=\frac{2V}{4\times {10}^{-3}}\\ =\frac{2\times (4\times {10}^{-6})}{4\times {10}^{-3}}\\ =2\times {10}^{-3}N\end{array}$