A golfer uses a club to hit a 45 g golf ball resting on an elevated tee so that the golf ball leaves the tee at the horizontal speed of + 38m/s. (1) What is the impulse of the golf ball? (2) What is the average force that the club exerts on the golf ball if they are in contact for 2.0x10^-3s? (3) What average force does the golf ball exert on the club during this time interval?

Correct answer: (1) I = 1.71 kg m/s, (2) F = 855 N

Explanation:

Given:

The mass of the ball m = 45 g = 45 · 10⁻³ kg

Initial velocity V = 38 m/s

Contact time t = 2 · 10⁻³ s

(1) I = ?

The impulse is calculated according to the formula:

I = m · V = 45 · 10⁻³ · 38 = 1,710 · 10⁻³ = 1.71 kg m/s

I = 1.71 kg m/s

(2) F = ?

The average force is calculated according to the formula:

F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N

F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N

F = 855 N

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