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 `0.001s` s , the force that the batsman had to apply tohold the bat firmly at its place would be

Options

A. `10.5` N
B. 21 N
C. `1.05 xx10^(4) N `
D. `2.1 xx10^(4)`N

Answer

C

Explanation

`m = 150 g = (150)/(1000) kg = 3/20 ` kg
`Deltat ` = time of contact = `0.001` s
u = 126 km/h = `(126 xx1000)/(60 xx60)` m/s = 35 m/s
`v = - 126 km//h = - 35 m//s`
Change in momentum of the ball
`Deltap = m Deltav = m(v-u)`
` =3/20 (-35 -35)`kg-m/s
` = 3/20 (-70)`
`=-21/2 `
Force F = `(Deltap)/(Deltat)`
`= ((-21/2)/(0.001))N `
` = - 0.05 xx10^(4)` N
Here , -ve sign shown that force will be opposite to the directionof motion of the ball before collision .