Question

The head of a golf club (of mass 250 g) is traveling at 44 m/s just before it strikes a 50 g golf ball (which is initially at rest). After the collision, the club head travels (in the same direction) at 36 m/s.

a) Assuming momentum is conserved in the collision, find the speed of the golf ball just after the impact. (5 points)

b) How much energy was lost in this collision? (5 points)

Answer 1

(a) Mass of head of golf club, m1 = 250 g = 0.250 kg

Mass of golf ball, m2 = 50 g = 0.050 kg

Initial velocity of head of golf club, u1 = 44 m/s

Initial velocity of golf ball, u2 = 0

Final velocity of club head, v1 = 36 m/s

Final velocity of golf ball, v2 = ?

Apply conservation of momentum -

m1*u1 + m2*u2 = m1*v1 + m2*v2

=> 0.250*44 + 0.050*0 = 0.250*36 + 0.050*v2

=> 0.050*v2 = 11.0 - 9.0 = 2.0

=> v2 = 2.0 / 0.050 = 2.0 / 0.050 = 40.0 m/s in the same direction of the club head.

(b) Energy lost in collision = Initial kinetic energy - Final kinetic energy

= (1/2)*m1*u^2 - [(1/2)*m1*v1^2 + (1/2)*m2*v2^2]

= 0.5*0.250*44^2 - [(0.5)*0.250*36^2 + (0.5)*0.050*40^2]

= 242 - [162 + 40] = 242 - 202 = 40.0 Joule (Answer)