Question

A ball falls from $20\, m$ height on floor and rebounds to $5 \,m$. Time of contact is $0.02\, s$. Find acceleration during impact.

• A. $ 1200\text{ }m/{{s}^{2}} $
• B. $ 1000\text{ }m/{{s}^{2}} $
• C. $ 2000\text{ }m/{{s}^{2}} $
• D. $ 1500\text{ }m/{{s}^{2}} $

Answer 1

Answer: (D)

According to Newtons second law of motion force acting on a body is equal to the rate of change of momentum during impact.
$ F=\frac{\Delta p}{\Delta t} $
Also, $ F=ma $
$ \therefore $ $ ma=\frac{{{p}_{2}}-{{p}_{1}}}{\Delta t} $
Or $ a=\frac{m{{v}_{2}}-(-m{{v}_{1}})}{m\Delta t} $
Or $ a=\frac{{{v}_{2}}+{{v}_{1}}}{\Delta t} $
$ \therefore $ $ a=\frac{\sqrt{2\times 10\times 20}+\sqrt{2\times 10\times 2}}{0.02} $
Or $ a=\frac{20+10}{0.02} $
$ =1500\,m/{{s}^{2}} $