A `"150 g"` ball, moving horizontally at `"20 m/s"` was hit straight back to bowler at `"35 m/s"`. If contact with bat lasted for `0.04 sec`, than average force exerted by the bat on the ball is:

Explanation

The impulse momentum diagram of the ball is dhown in the figure below. Here F, mg, and Deltat represent the average value of the force exerted by the bat, weight of the ball and time interval.

Applying principle of impulse and momentum in x-direction, we have ltbr gt `p_(fx)=p_(ix)+sum l_(mp,x) rarr " " -2+F_xDeltat=2.8`
`F_x=4.8/(Delta t)N`...(i)
Applying principle of impulse and momentum in y-direction, we have
`p_(fy) = p_(iy)+sum I_(mp.y) rarr " " 0.0 + F_(y)Deltat - 1.0Delta t = 2.1`
`F_(y) = ((2.1)/(Deltat)+1.0)N`.....(ii)
Applying principle of impulse and momentum in y-direction, we have
(a) Substituting `Delta t = 0.30s`, on equations (i) and (ii), we find `vec(F) = 16hat(i) + 8hat(j) N`
(b) Substituting `Deltat = 0.03s`, in equations (i) and (ii), we find `vec(F) = 160hat(i) + 71hat(j) N`
Substituting `Deltat = 0.003s`, in equations 1 and 2, we find `vec(F) = 1600hat(i) + 701hat(j) N`
(d) It is clear from the above results that as the duration of contact between the ball and the bat decreases, effect of the weight of the ball also decreases as compared with that of the force of the bat and for sufficiently short time interval, it can be neglected.