Question

An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00360 s. After the collision, the ball leaves the club at a speed of 34.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answer 1

Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

                    = 1.87

F = $\frac{1.87}{0.0036}$

 = 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.