question_answer

In a cricket match between India and South Africa, the Indian player defends a ball from South African player which is thrown with a speed of 20m/s. After striking the Indian player's bat, the ball goes in the opposite direction with the same speed. The time of impact was 0.5 sec and the mass of the ball is 4 kg. Find (i) force exerted by the Indian player's bat on the ball (ii) impulse.

Answer:

\[m=4kg\] \[u=-20m/s\] (\[-ve\]sign indicates that the ball goes in the opposite direction) \[v=20m/s\] Time of impact\[=t=0.5\,\,\sec \] Force exerted by the Indian player?s bat on the ball = F = ? When the defender strikes the ball, he is exerting the impulsive force on the ball. We know that impulsive \[(I)=F\times t\]       ?.. (1) \[\Rightarrow \,\,mv-mu=F\times t\] \[\Rightarrow \]Force \[=F=\frac{mv-mu}{t}=m\left( \frac{v-u}{t} \right)\] ?..(2) Substituting above values in (2), we have \[F=4\left( \frac{20+20}{0.5} \right)=4\left( \frac{40}{0.5} \right)=320N\] \[F=320N\] ???..(3) \[\therefore \] force exerted by the Indian player?s bat on the ball is 320 N. We also know that, impulse \[1=F\times t\] ?.. (4) [\[\because \] from (1)] Substituting the above values in (4), we have Impulse \[(I)=320\times 0.5\] \[=(320\times 0.5)kg\,\,m/s=160\,\,kgm/s.\] Therefore, impulse (I) is 160 kgm/s.