Question

A golf club strikes a 0.031-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6240 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club

Answer 1

66.5 m/s

Step-by-step explanation:

m = mass of the golf ball = 0.031 kg

F = magnitude of force applied to the ball = 6240 N

Acceleration experienced by the ball is given as

`a = (F)/(m)`

`a = (6240)/(0.031)`

a = 201290.32 m/s²

d = distance for which the ball is in contact with the golf club = 0.011 m

v₀ = initial speed of the ball = 0 m/s

v = final speed of the ball = 0 m/s

Using the kinematics equation

v² = v₀² + 2 a d

v² = 0² + 2 (201290.32) (0.011)

v = 66.5 m/s