When a golfer tees off, the head of her golf club which has a mass of 151 g is traveling 43.9 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 28.2 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact

51.54 m/s

Explanation:

Applying,

Law of conservation of momentum,

Total momentum before collision = Total momentum after collision

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

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

From the question,

Given: m = 151 g = 0.151 kg, u = 43.9 m/s, m' = 46 g = 0.046 kg, u' = 0 m/s (at rest), v = 28.2 m/s

Substitute these values into equation 1

0.151(43.9)+0.046(0) = 0.151(28.2)+0.046(v')

solve for v'

6.6289+0 = 4.2582+0.046v'

0.046v' = 6.6289-4.2582

0.046v' = 2.3707

v' = 2.3707/0.046

v' = 51.54 m/s

i think the answer would be reserve.