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iren2701 [21]
4 years ago
7

The mass of a regulation tennis ball is 57 g. The ball is in contact with the tennis racket for 30 ms. The ball was served at 73

.14 m/s. If the opponent returned the serve with a speed of 55 m/s, what force did he exert on the ball, assuming only horizontal motion?
Physics
1 answer:
yaroslaw [1]4 years ago
8 0

Answer:

Correct answer:  F₂ = 104.5 N

Explanation:

Given:

m = 57 g = 57 · 10⁻³ kg

Δt = 30 ms = 30 · 10 ⁻³ seconds

V₁ = 73.14 m/s   service speed

V₂ = 55 m/s  returned speed

M = m · V  Momentum or Impulse

You forgot to indicate what time the ball contact when returning.

We will assume that the time is the same Δt = 30 ms = 30 10 ⁻³ seconds.

The formula for calculating force is according to Newton's second law is:

F = ΔM / Δt = m · ΔV / Δt

Force during service is:

F₁ = 57 · 10⁻³ · 73.14 / 30 · 10 ⁻³ = 138.97 N

F₁ = 138.97 N

Returned force:

F₂ =  57 · 10⁻³ · 55 / 30 · 10 ⁻³ = 104.5 N

F₂ =  104.5 N

God is with you!!!

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