Question

A batsman deflects a ball by an angle of 60⁰ without changing its initial speed which
is 54km/h. What is the impulse imparted to the ball if the mass of the ball is 0.15kg? If the
ball remains in contact with the ball for 0.1s, then find the force exerted by the bat on
the ball in Newton.

(PLEASE ANSWER WITH STEPS)

Answer 1

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

Where,

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 = vcos θ along RO

Vertical component of the initial velocity = vsin θ along PO

Horizontal component of the final velocity = vcos θ along OS

Vertical component of the final velocity = vsin 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

= mvCosθ – (-mvCosθ) = 2mvCosθ

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°

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