Question

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N

Answer 1

Answer:

(a) The speed of the racket immediately after the impact is -8.7 m/s

(b) The average force that the racket exerts on the ball is 0.471 N

Explanation:

Given;

mass of racket, m₁ = 1000 g = 1 kg

initial speed of racket, u₁ = -12 m/s (negative because it swings backwards)

mass of tennis, m₂ = 60g = 0.06 kg

initial speed of tennis, u₂ = 15 m/s

final speed of the tennis ball, v₂ = -40 m/s (negative because it moved backwards)

(a) How fast is her racket moving immediately after the impact?

Let the final speed of the racket  = v₁

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1 x (-12) + (0.06 x 15) = (1 x v₁) + 0.06 x (-40)

-12 + 0.9 = v₁ - 2.4

-11.1 = v₁ - 2.4

v₁ = -11.1 + 2.4

v₁ = -8.7 m/s

The speed of the racket immediately after the impact is -8.7 m/s

The final speed of the racket is still backwards but at a lower speed.

(b) The average force that the racket exerts on the ball during 7 s in contact with the ball;

$F = \frac{\delta P_{racket}}{\delta t} \\\\F = \frac{M_1(V_1 -U_1)}{t} \\\\F = \frac{1[-8.7-(-12)]}{7} \\\\F = \frac{3.3}{7} \\\\F = 0.471 \ N$

The average force the racket exerts on the ball is on the right