A batsman deflects a ball by the 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 balls is `0.15` kg ).

Explanation

In fig . The fall hits the bat KL along AO and is deflected by the bat along OB . Where `angleAOB = 45^(@)` .ON is normal to the portion of the bat KL. Defelcting the ball .
` :. O = LNOA = 45^(@)//2 = 22.5 5^(@)`

Initial vel along AO = u = 54 km/h
= 15 `m//S_(1)` and mass of ball m = 0.15 kg
Initial velocity along AO has the two rectangular components : u cos `theta ` along NO produced and u sin `theta ` along the horizontal OL .
Finalvelocity along OB has the same magnitude = u
It is resolved in two rectangular components
` :. u cos theta ` along ON and `u sin theta ` along OL .
We observe that there is no change in velocity along vertical is just reserved.
` :. ` Impulse imparted to the ball
= Change in linear momentum of the ball
` = m cos theta - (- " m u cos "theta )`
` = 2 " m u cos " theta `
` = 2 xx 00.15 xx 15 cos 22.5 5^(@)`
` = 4.5 xx 0.9239`
` = 4.16 ` kg m/s