Answer:
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