A batsman deflects a ball by an angle of 45° without changing its initial speed which equal to 54 km/h. What is the impulse imparted to the ball? (Mass of  the ball is 0.15 kg).

Solution:

Mass of the ball m = 0.15 kg

Velocity of the ball v = 54 km/h

$=54\times \frac{5}{18}m/s(\because 1kmh=\frac{5}{18}m/s)$

= 15 m/s

Let the ball be incident along path PO. Batsman deflects the ball by an angle of 45° along both OQ.

$\angle PON=\angle NOQ=\frac{{45}^0}{2}={22.5}^0$

The horizontal component of velocity $u\sin \theta$ remains unchanged while vertical component of velocity is just reversed.

$\therefore$Impulse imparted to the ball

= change in linear momentum of the ball

$=mu\cos \theta -(-mu\cos \theta )$

$=2mu\cos \theta$

$=2\times 0.15\times 15\times \cos {22.5}^0$

$=4.5\times 09239kg-m/s=4.16kg-m/s$