Question

A cricket ball of mass 150 𝑔 moving with a speed of 126 𝑘𝑚/ℎ hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0. 001 𝑠, the force that the batsman had to apply to hold the bat firmly at its place would be

• A. $2.1\times {10}^{4}N$
• B. $21N$
• C. $10.5N$
• D. $1.05\times {10}^{4}N$

Answer 1

The correct option is D $1.05\times {10}^{4}N$

Formula used: $F=\frac{\Delta \overrightarrow{p}}{\Delta t}$

Given,

Mass of the ball $\left(m\right)=150g=0.15kg$

As collision is completely elastic

Speed of Ball before collision= 𝑆peed of Ball before collision

$\left[\overrightarrow{v}\right]=\left[\overrightarrow{u}\right]=126\frac{km}{h}=126\times \frac{5}{18}ms=35\frac{m}{s}$

Impulse imparted by ball on bat = Impulse imparted by bat on ball

$J_{bat}=-J_{bal}$

$J_{ball}=\Delta \overrightarrow{p}=m\left(v\right)-(-u)=0.15\times \left(70\right)=10.5kgm/s$

$J_{bat}=-J_{ball}=-10.5kgm/s$

Force applied by batsman to hold the bat firmly at its place would be:

$F=\left|\begin{array}{c}\frac{\Delta \overrightarrow{p}}{\Delta t}\end{array}\right|=\frac{10.5}{0.001}$

$F=1.05\times {10}^{4}N$

Final Answer: (c)