A cricket ball of mass `150g` moving with a speed of `126km//h` hits at the middle of the bat, held firmly at its position by the batman. 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 `0.001s`, the force that the batsman had to apply to hold the bat firmly at its place would be

Options

A. `10.5 N`
B. `21N`
C. `1.05 xx 10^(4) N`
D. `2.1 xx 10^(4) N`

Answer

C

Explanation

Given, `m = 150 g = (150)/(1000) kg=(3)/(20) kg`
`Delta t` = time of constant `= 0.001 s`
`u = 126 km//h = (126 xx 1000)/(60 xx 60) m//s = 35 m//s`
`v = -126 km//h = -25 m//s`
Change in momentum of the ball
`Delta p=m(v-u)=(3)/(20)(-35-35)kg-m//s = (3)/(20)(-70) = -(21)/(2)`
We know that force `F = (Delta p)/(Delta t)`
=`(-21//2)/(0.001) N=-1.05 xx 10^(4) N`
Here, `-ve` sign shown that force will be opposite to the direction of movement of the ball before hitting.