High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45.9 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 30.7 m/s. Find the speed of the golf ball immediately after impact. Answer in units of m/s.

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

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