### Q.20:- 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.)

**Answer:-**

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

AO = Incident path of the ball

OB = Path followed by the ball after deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = *θ*

Initial and final velocities of the ball =* v*

Horizontal component of the initial velocity = *v*cos θ along RO

Vertical component of the initial velocity = *v*sin θ along PO

Horizontal component of the final velocity = *v*cos θ along OS

Vertical component of the final velocity = *v*sin θ* *along OP

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

∴ Impulse imparted to the ball = Change in the linear momentum of the ball

= *mvCos*θ – (-*mvCos*θ) = 2*mvCos*θ

Mass of the ball, *m* = 0.15 kg

Velocity of the ball, *v* = 54 km/h = 15 m/s

∴ Impulse = 2 × 0.15 × 15 *cos* 22.5° = 4.16 kg m/s