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Question
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.)
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
∠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
Initial velocity's horizontal component = vcos θ along RO
Initial velocity's vertical component = vsin θ along PO
Final velocity's horizontal component = vcos θ along OS
Final velocity's vertical component = vsin θ along OP
The horizontal velocity components remain unchanged. The vertical velocity components are directed oppositely.
∴Impulse imparted to the ball = Change in the linear momentum of the ball
`= mvcostheta - (- mvcos theta)`
`= 2mvcos theta`
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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