A golf club strikes a 0.057-kg golf ball in order to launch it from the tee. For
ID: 1510023 • Letter: A
Question
A golf club strikes a 0.057-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 6830 N, and is in contact with the ball for a distance of 0.0090 m. With what speed does the ball leave the club? A golf club strikes a 0.057-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 6830 N, and is in contact with the ball for a distance of 0.0090 m. With what speed does the ball leave the club?Explanation / Answer
net force F = m*a = 6830 N
accelaration a = 6830/m = 6830/0.057 = 119824.56 m/s^2
distance d = 0.0090 m
using v^2 - u^2 = 2*a*d
v^2-0^2 = 2*119824.56*0.0090
v = 46.44 m/sec