Question

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.

Answer 1

A) pbin = 1.535 Kgm/s (+)

B) pbf = 1.696 Kgm/s (-)

C) Δp = 3.3925 Kgm/s

D) Δvr = 10.249 m/s

Step-by-step explanation:

Given

Mass of the ball: m = 57.5 g = 0.0575 Kg

Initial speed of the ball: vbi = 26.7 m/s

Mass of the racket: M = 331 g = 0.331 Kg

Final speed of the ball: vbf = 29.5 m/s

A) We use the formula

pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)

B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)

C) We use the equation

Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s

D) Knowing that

Δp = 3.3925 Kgm/s

we can say that

Δp = M*Δvr

⇒ Δvr = Δp / M

⇒ Δvr = 3.3925 Kgm/s / 0.331 Kg

⇒ Δvr = 10.249 m/s