Question

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.

Answer 1

a) Final Momentum = P2 =2.025 kgm/s

b) Impulse = J = 2.025 kgm/s

c) Favg = 578.57 N

Step-by-step 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 2

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

Step-by-step 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