When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.

56.5 m/s²

Explanation:

From the law of conservation of momentum,

mu+m'u' = mv+m'v' Equation 1

Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.

From the question,

The golf ball is at rest, Hence u' = 0 m/s

mu = mv+m'v'

Make v' the subject of the equation

v' = (mu-mv)/m' Equation 2

Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.

Substitute into equation 2

v' = (0.152×44.8+0.152×27.7)/0.046

v' = (6.8096-4.2104)/0.046

v' = 2.5992/0.046

v' = 56.5 m/s²

the answer would be c