# A batsman deflects a ball by an angle of $45^{\circ}$ without changing

Question.
A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54 \mathrm{~km} / \mathrm{h}$. What is the impulse imparted to the ball? (Mass of the ball is $0.15$ kg.)

solution:

The given situation can be represented as shown in the following figure.

Where,

AO = Incident path of the ball

OB = Path followed by the ball after deflection

$\angle A O B=$ Angle between the incident and deflected paths of the ball $=45^{\circ}$

$\angle \mathrm{AOP}=\angle \mathrm{BOP}=22.5^{\circ}=\theta$

Initial and final velocities of the ball $=v$

Horizontal component of the initial velocity $=v \cos \theta$ along RO

Vertical component of the initial velocity $=v \sin \theta$ along PO

Horizontal component of the final velocity $=v \cos \theta$ along $O S$

Vertical component of the final velocity $=v \sin \theta$ along $O P$

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

$\therefore$ Impulse imparted to the ball $=$ Change in the linear momentum of the ball

$=m v \cos \theta-(-m v \cos \theta)$

$=2 m v \cos \theta$

Mass of the ball, $m=0.15 \mathrm{~kg}$

Velocity of the ball, $v=54 \mathrm{~km} / \mathrm{h}=15 \mathrm{~m} / \mathrm{s}$

$\therefore$ Impulse $=2 \times 0.15 \times 15 \cos 22.5^{\circ}=4.16 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$