Question

a) A tennis racket a 450 N force on a 0.05 kg tennis ball over a 0.006 s time period. What is the impulse applied to the ball?

b) A ball (0.5 kg) is thrown with a velocity of 15 m/s. A receiver catches the ball and brings it to rest is 0.02 s.
a) What is the change in momentum?

c) A ball (0.5 kg) is thrown with a velocity of 15 m/s. A receiver catches the ball and brings it to rest is 0.02 s.
a) What is the change in momentum of the ball?

b) What is the impulse delivered to the ball?

c) What force is exerted on the ball?

d) A ball (0.40 kg) is moving with a velocity of 18 m/s. A player strikes the ball and causes it to move in the opposite direction with a velocity of 22 m/s. The player exerts a force of 5500 N.

a) What is the change in momentum of the ball?

b) What is the impulse delivered to the ball?

c)How long is the foot in contact with a ball?

~formula's~
J= Ft
J= P
Ft=mvf-mvi

PLease help me!! :)

Answer 1

These are all the same so I am only going to do d

initial momentum = .4 * 18 = + 7.2 kg m/s

final momentum = - .4 * 22 = - 8.8 kg m/s

change in momentum = -8.8 - 7.2 = -16 kg m/s

impulse = change in momentum = -16 kg m/s

Force = change in momentum / change in time
so
-5500 = -16 /t
t = .003 seconds