In a square cut, the speed of the cricket ball changes from `30 ms^9-1)` to `40 ms^(-1)` during the time of its contact `Delta t = 0 .01 s` with the bat. If the ball is deflected by the bat through an angle of `theta = 90^@`, find the magnitude of the average acceleration `("in" xx 10^(2) ms^(-2))` of the ball during the square cut.

Answer

50

Explanation

As the velocity of the ball changes from `vecv_1 to vecv_2`, the change in velocity `Deltavecv` is given by
`|Deltavecv| = sqrt(v_(1)^(2) + v_(2)^(2) - 2v_1v_2cos theta)`
where `v_1 = 30ms^(-1), v_2 = 40ms^(-1) and theta = 90^@`.
Then, `|Deltavecv| = 5 ms^(-1)`,
`a_(av) = |Deltavecv|/(Deltat) = 5/(0.01)`
`= 5 xx 10^2 ms^(-2)`.