Question

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

• 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, $m=0.1\, kg$
Velocity attained by the ball before hitting the ground,
$v =\sqrt{2 g h}(-\hat{ j })$
$=\sqrt{2 \times 10 \times 20}(-\hat{ j })=-20\, \hat{ j } m / s$
Velocity of ball when bounce back to the same height after hitting the ground,
$v'=-v=-(-20 \hat{ j })=20\, \hat{ j } m / s$
$\therefore$ Change in velocity,
$\Delta v=v'-v=20 \hat{ j }-(-20 \hat{ j })=40\, \hat{ j } m / s$
$\therefore$ Force exerted on the ball,
$F =\frac{\Delta p}{\Delta t}=\frac{m \Delta v}{0.04}$ [given, $\Delta t=0.04 \,s$]
$=\frac{0.1 \times 40 \hat{ j }}{0.04}=100(\hat{ j }) N$