a batsman deflects a ball by an angle of 45 without changing its initial speed which is equal to 54km/h. what is the impulse imparted to the ball? mass of the ball is 0.15kg

the impulse imparted to the ball is 4.158 kg.m/s

Explanation:

Please find attached image for diagram.

Given;

initial speed of the ball from point A, u = 54km/h = 15 m/s

angle of deflection of the ball along BOP, θ = 45⁰/2 = 22.5⁰

mass of the ball, m = 0.15 kg

the final speed of the ball along B,  = u (in reverse direction)

The initial momentum of the ball, P₁ = mucosθ

The final momentum of the ball is P₂ in reverse direction to P₁.

the impulse imparted to the ball = change in momentum of the ball

J = ΔP = P₂ - P₁

J = mucosθ - (-mucosθ)

J = mucosθ + mucosθ

J = 2mucosθ

J = 2(0.15)15cos(22.5⁰)

J = 4.158 kg.m/s

Thus, the impulse imparted to the ball is 4.158 kg.m/s

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