Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker point with a cross-court passing shot. The 57.5-gram ball hit her racket with a northward velocity of 26.7 m/s. Upon impact with her 331-gram racket, the ball rebounded in the exact opposite direction (and along the same general trajectory) with a speed of 29.5 m/s.
A. Determine the pre-collision momentum of the ball.
B. Determine the post-collision momentum of the ball.
C. Determine the momentum change of the ball.
D. Determine the velocity change of the racket

Question

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wagonbelleville wagonbelleville · Answer 1 · 2020-01-23T09:59:48+01:00

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s