Question

A tennis ball traveling horizontally at a speed of 40 m/s hits a wall and rebounds in the opposite direction. The time Interval for the collision is about 0.013 s, and the mass of the ball Is 0.059 kg. Assume that the ball rebounds at the same speed.
(a) Determine the initial and final momenta of the ball. Assume the initial velocity direction is positive.

(b) Determine the change In momentum of the ball.

(c) Determine the average force exerted by the wall on the ball.

Answer 1

Answer

given,

initial velocity of the ball, u = 40 m/s

final velocity of the ball, v= -40 m/s

time of contact = 0.013 s

mass of the ball = 0.059 Kg

a) initial momentum

   P₁ = m u = 0.059 x 40 = 2.36 kg.m/s

  final momentum  

   P₂ = m v = 0.059 x (-40) = -2.36 kg.m/s

b) change in momentum

      Δ P = P₂- P₁

      Δ P = -2.36 - 2.36

     Δ P = -4.72 kg.m/s

c) Average force

 average force exerted by the ball is equal to change in momentum per unit time.

     $F =\dfrac{\Delta P}{t}$

     $F =\dfrac{-4.72}{0.013}$

            F = -363 N