Question

Please please help me, don't guess please my test is timed !!

A tennis player hit a .057-kg ball with a force of 40 N. The duration of the force was .05 s.

1. Calculate the impulse delivered to the ball. (1pt)


2. What was the change in momentum of the ball? (1pt)


3. What was the velocity of the ball as it left the tennis racket? (1pt)


4. What would happen to the ball's velocity if it remained on the tennis racket longer than .05 s? (2pts)

Answer 1

a) J = F t = 40 * .05 = 2 N-s

b) J = 2 N-s momentum changed by 2 N-s

c) Initial momentum appears to be zero

J = change in momentum = m v2 - m v1 = m v2 = 2 N-s

v2 = J / m = 2 / .057 = 35 m/s

d) if the impulse time was increased and the average force remained the same then the change in momentum would increase with a corresponding increase in velocity attained - note the increase in v2 in part c)