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 imparted to the ball? (Mass of the ball is $0.15kg$).

Answer 1

Given,

Mass of the ball $m=0.15kg$.

The initial speed of the ball $u=54km/h=\frac{54\times 1000}{60\times 60}=15m/s$

The final speed of the ball $v_f=ucos{45}^0$

The momentum change takes place in the ball in the vertical direction only because the initial and final momentum in the horizontal direction is canceled out.

$v_i=-ucos{45}^0$

According to the impulse-momentum theorem,

$Impulse=Changeinmomentum$

$\Rightarrow I=m(v_f-v_i)$

$I=mucos{45}^0-(-mucos{45}^0)$

$I=2mucos{45}^0$

Substitute the value of mass and velocity.

$I=2\times 0.15\times 15\times 0.7071$

$I=3.19kgm/s$