In cricket match, a bowler throws a ball of 0.5 kg with a speed of 20 m/s. When a batsman swings the bat the ball strikes with the bat normal to it, and returns in opposite direction with speed of 30 m/s. If the time of contact of the ball with the bat is 0.1s, then the force acting on the bat is......N.

Options

A. 50
B. 125
C. 250
D. 25

Answer

C

Explanation

change in momentum of ball
`Deltap=p_(2)-p_(1)`
`=m(v_(2)-v_(1))`
`=0.5(-3-20)`
`=-0.5xx50`
`=-25kgm//s`
Momentum gained by bat `Deltap=+25kgm//s`
`therefore` The force acting on bat `F=(Deltap)/(Deltat)=(25)/(0.1)`
`=250N`