A batsman deflects a ball by an angle of `45^(@)` without changing its initial speed which is equal to `54 km/h` . What is the impulse imparted to the ball ? ( Mass of the ball is 0.15 kg)

Explanation

The ball hits the bat KL along AO and is deflected by the bat along OB where `angle AOB = 45^(@)` ON is normal to the portion of the KL deflecting the ball
`:. theta = angle NOA = 45^(@) // 2 = 22.5^(@)`
Initial vel.along `AO = mu = 54 km h^(-1) = 15 ms^(-1)`
and mass of ball , `m = 0.15 kg `
Initial velocity along AO has two rectangular components : `u cos theta` along NO produced and `u sin theta` along the horizontal OL
Final velocity along OB has the same magnitude = u .It is resolved into two rectangular components : `u cos theta` along ON and `u sin theta` along OL We observe that there is no change in velocity along the horizontal but velocity along vertical is just - reversed .
`:.` Impulse imparted to the ball = change in linear momentum of the ball
`m u cos theta - (- m u cos theta) = 2 m u cos theta `
`2 xx 0.15 xx 15 cos 22.5^(@) = 4.5 xx 0.9239 = 4.16 kg ms^(-1)`
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