Question

High-speed stroboscopic photographs show that the head of a 180 g golf club is traveling at 47 m/s just before it strikes a 46 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 35 m/s. Find the speed of the golf ball just after impact.

Answer 1

47.17 m/s

Step-by-step explanation:

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

Note: The gulf ball was at rest before impact.

Therefore,

mu = mv+m'v'

Make v' the subject of the equation

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

Given: m = 180 g = 0.180 kg, m' = 46 g = 0.046 kg, u = 47 m/s, v = 35 m/s

Substitute into equation 2

v' = [(0.18×47)-(0.18×35)]/0.046

v' = (8.47-6.3)/0.046

v' = 2.17/0.046

v' = 47.17 m/s

Hence the speed of the gulf ball just after impact = 47.17 m/s

Answer 2

47 m/s

Step-by-step explanation:

golf club mass, mc = 180 g

golf ball mass, mb = 46 g

initial golf club speed, vc1 = 47 m/s

final golf club speed, vc2 = 35 m/s

initial golf ball speed, vb1 = 0 m/s

final golf ball speed, vb2 = ? m/s

The total momentum is conserved, then:

mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2

Replacing with data and solving (dimension are omitted):

180*47 + 46*0 = 180*35 + 46*vb2

vb2 = (180*47 - 180*35)/46

vb2 = 47 m/s