Question

A ball of mass 8.1 g with a speed of 22.6 m/s strikes a wall at an angle 11.0o and then rebounds with the same speed and angle. It is in contact with the wall for 44.0 ms. What is the magnitude of the impulse associated with the collision force?

Answer 1

Given that :

mass of the ball, m = 8.1 g = 8.1 x 10^-3 kg

speed of the ball, v = 22.6 m/s

angle, = 11 degree

time taken, t = 44 x 10^-3 sec

(a) Magnitude of the impulse associated with the collision force which is given as :

using an equation,    I = p = F t                                 { eq.1 }      

according to conservation of momentum, we have

| p_initial | = | p_final | = m v

p = 2 m v sin                                                                                           { eq.2 }

equating eq.1 and 2, we get

I = 2 m v sin                                                                 { eq.3 }

inserting the values in eq.3,

I = 2 (8.1 x 10^-3 kg) (22.6 m/s) sin 11⁰

I = (366.1 x 10^-3 kg.m/s) (0.1908)

I = 69.8 x 10^-3 kg

I = 0.0698 kg.m/s

(b) The average force exerted by the ball on the wall which will be given as :

using an equation, I = F_avgt                                                              { eq.4 }

inserting the values in eq.4,

(69.8 x 10^-3 kg) = F_avg (44 x 10^-3 s)

F_avg = (69.8 x 10^-3 kg) / (44 x 10^-3 s)

F_avg = 1.58 N