Question

A batsman deflects a ball by an angle of ${90}^o$ without changing its initial speed, which is equal to $54km/hr$. What is the impulse to the ball? Mass of the ball is $0.15kg$

Answer 1

Let the point $O$ represnet the position of the bat. Draw a line $XY$ through the point bat $O$ and $ON$, normal to the line $XY$. The of mass $M$ initially moving along the path $AO$ with speed $u$ is deflected by the batsman along $OB$ (Without change in speed of the ball), such that $AOB={90}^o$ the $\angle AON=BON={45}^o$
Then initial momentum of the ball can be written as
(i) $Mu\cos {45}^o$ along $NO$ and
(ii) $Mu\sin {45}^o$
Also, the final momentum of the ball can be resolved inot the following components:
(i) $=Mv\cos {45}^o$ along $ON$ and
(ii) $M\sin {45}^o$ along $XY$
The component of the momentum of the ball along XY remains unchanged (both in magnitude and direction). However, the components of the momentum of the ball along normal are equal in magnitude but opposite in direction. SInce the impulse imparted by the batsman to the ball is equal to the change in momentum of the ball along normal
Normal impulse $=Mv\cos {45}^o-(-Mu\cos {45}^o)=2Mu\cos {45}^o$
There impulse $=2\times 0.15\times 15\times \cos {45}^o=3.18kgm/s$