Question

A golfer attempts a birdie putt, sending the 0.050-kg ball toward the hole with a speed of 1.4 m/s. Unfortunately, the ball does a "horseshoe" and goes halfway around the rim of the hole. The ball ends up heading back toward the golfer with a speed of 1.3 m/s.If we take the positive direction to be from the golfer to the hole, what is the impulse delivered to the ball by the hole?

Answer 1

Answer:

I = - 0.135 kg m/s

Explanation:

The formula to calculate the impulse of delivered to the ball is:

I = Ft

where,

I = Impulse delivered to the ball by hole = ?

F = Force = ma (By Newton's Second Law)

t = time

Therefore,

I = mat

where,

m = mass of ball = 0.05 kg

a = acceleration of ball = (Vf - Vi)/t

Therefore,

I = m[(Vf - Vi)/t]t

I = m(Vf - Vi)

where,

Vf = Final velocity = - 1.3 m/s

Vi = Initial Velocity = 1.4 m/s

Therefore,

I = (0.05 kg)(- 1.3 m/s - 1.4 m/s)

I = - 0.135 kg m/s