Question
A batsman deflects a ball by an angle of $45^\circ$ 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 $0.15 \ kg).$
✓
Answer
Let the point $B$ represents the position of bat. The ball strikes the bat with velocity $v$ along the path $AB$ and gets deflected with same velocity along $BC.$ such that $\angle ABC = 45^\circ$
Initial momentum of the ball
$= mv \cos \left(\frac{\theta}{2}\right)$ along $NB$
Final momentum of the ball
$=m v \cos \left(\frac{\theta}{2}\right)$ along $BN$
Hence, Impulse $=$ change in momentum
$=m v \cos \left(\frac{\theta}{2}\right)-\left[-m v \cos \left(\frac{\theta}{2}\right)\right]$
$=2 mv \cos \left(\frac{\theta}{2}\right) \quad(\because v =54 \ km / hr =15 m / s )$
$=2 \times 0.15 \times 15 \times \cos \left(22.5^{\circ}\right)=4.157 \ kg m s ^{-1}$
Thus, impulse imparted to the ball is $4.157 \ kg ms^{-1}$
Initial momentum of the ball
$= mv \cos \left(\frac{\theta}{2}\right)$ along $NB$
Final momentum of the ball
$=m v \cos \left(\frac{\theta}{2}\right)$ along $BN$
Hence, Impulse $=$ change in momentum
$=m v \cos \left(\frac{\theta}{2}\right)-\left[-m v \cos \left(\frac{\theta}{2}\right)\right]$
$=2 mv \cos \left(\frac{\theta}{2}\right) \quad(\because v =54 \ km / hr =15 m / s )$
$=2 \times 0.15 \times 15 \times \cos \left(22.5^{\circ}\right)=4.157 \ kg m s ^{-1}$
Thus, impulse imparted to the ball is $4.157 \ kg ms^{-1}$
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