how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 * 10-3 s. Find (a) the final momentum of the ball, (b) the impulse imparted to the golf ball, and (c) the average force exerted on the ball by the golf club.

a) Final Momentum = P2 = 2.025 kgm/s

b) Impulse = J = 2.025 kgm/s

c) Favg = 578.57 N

Explanation:

Mass of the golf ball = m = 0.045 kg

Initial speed = V1 = 0 m/s

Final Speed = V2 = 45 m/s

Time = t = 3.5 * 10^-3

a) Final Momentum = P2 = mV2 = (0.045) (45) = 2.025 kgm/s

b) Impulse = J = change in momentum = ∆P = mV2 - mV1

J = (0.045) (45) - (0.045) (0) = 2.025 kgm/s

c) Impulse = Favg * t

Favg = J/t = 2.025/3.5 * 10^-3

Favg = 578.57 N

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 /.0035 = 12857.1 m/s/s

(b).045 x 12857.1 = 578.6 N