Question

If a ball of mass $0.1kg$ hits the ground from the height of $20m$ and bounce back to the same height, then find out the force exerted on the ball if the time of impact is $0.04s$. (Take, $g=10m/s^2$ )

• A. $100(+\hat{j})N$
• B. $200(+\hat{j})N$
• C. $100(-\hat{j})N$
• D. $1000(\hat{j})N$

Answer 1

Answer: (A)

Given, mass of ball, $m=0.1kg$
Velocity attained by the ball before hitting the ground,
$v=\sqrt{2gh}(-\hat{j})$
$=\sqrt{2\times 10\times 20}(-\hat{j})=-20\hat{j}m/s$
Velocity of ball when bounce back to the same height after hitting the ground,
$v^{\prime }=-v=-(-20\hat{j})=20\hat{j}m/s$
$\therefore$ Change in velocity,
$\Delta v=v^{\prime }-v=20\hat{j}-(-20\hat{j})=40\hat{j}m/s$
$\therefore$ Force exerted on the ball,
$F=\frac{\Delta p}{\Delta t}=\frac{m\Delta v}{0.04}$ [given, $\Delta t=0.04s$]
$=\frac{0.1\times 40\hat{j}}{0.04}=100(\hat{j})N$