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.)

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Ans:  The given situation can be represented as shown in the following figure given below:

AO = Incident path of the ball 

OB = Path followed by the ball after deflection by the ball

∠ 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 and

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

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

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

= m v cos θ - (- m v cos θ) 

= m v cos 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

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