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  (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

God is with you!!!

it doesn't. it was imagined up in the middle ages.

which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?  

answer

a 75kg dining table