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!!!

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s