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](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%5Cdelta%20P_%7Bracket%7D%7D%7B%5Cdelta%20t%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7BM_1%28V_1%20-U_1%29%7D%7Bt%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B1%5B-8.7-%28-12%29%5D%7D%7B7%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B3.3%7D%7B7%7D%20%5C%5C%5C%5CF%20%3D%200.471%20%5C%20N)
The average force the racket exerts on the ball is on the right