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

We are given:
Mass of ball, m = 0.15 kg
`v= 54 km//h = 54 xx (5)/(18)= 15 m s^(-1)`
We know that,
Intial momentum `p_(1)= mv`
`p_(1)= 0.15 xx 15= 2.25 kgms^(-1)` along DA
Final momentum `p_(2)= 2.25 kgms^(-1)` along AC
Since, Impulse = change in momentum
Impulse (I)=` vecR= vecp_(2) - vecp_(1)= vecp_(2)+ (-vecp_(1))`

Using parallelogram law for vector addition
`AE= sqrt(p_(1)^(2) +p_(2)^(2) + 2p_(1)p_(2)cos45^(@))`
`= sqrt((2.25)^(2)+ (2.25)^(2)+ 2 xx 2.25 xx 2.25 xx cos45^(@))`
`= sqrt(17.31)= 4.16 kgms^(-1)`
Therefore, Impulse= `4.16 kgms^(-1)`