A sled slides along a horizontal surface on which the coefficient of kinetic fri

Question

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.38. Its velocity at point A is 8.6 m/s and at point B is 4.0 m/s. Use the impulse-momentum theorem to find how long the sled takes to travel from A to B.

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 23 m/s. The ball rebounds at 41 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 10 ms, what is the average force that the racket exerts on the ball?
N

How does this compare to the gravitational force on the ball?
Favg / Wball =

Explanation / Answer

1:-
F(t-t0) = m(v - v0)
F = 0.38W = 0.38mg
=> 0.38*9.81*t = 8.6 m/s - 4 m/s (since m got cancelled)
t = 4.6/(0.38*9.81) sec
t = 1.234 sec

2:-
A) Use conservation of linear momentum
m1u1 + m2u2 = m1v1 + m2v2
m1 = mass of racket = 1000g
u1 = initial vel of racket = 12 m/s
m2 = mass of ball = 60 g
u2 = initial velo of ball = -23 m/s (negative since ball is moving opposite direction of racket.)
v1 = final vel of racket = ?
v2 = final vel of ball = 41 m/s

plug values into above equation and solve for v1
1000 g * 12 m/s + 60 g * (-23 m/s) = 1000 g * v1 + 60 g * 35 m/s
solve for v1
v1 = 8.52 m/s

B) To calculate average force use impulse-momentum theorem
I = F* (delta t)
I = m *(delta v)
equate the above and solve for F :-
F * (delta t) = m * (delta v)
F = m * (delta v)/(delta t)
F = 0.060 kg * (41 m/s + 23 m/s)/0.010 s
F = 384 N

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