a. The mass of a golf ball is .046 kg; the mass of a club head is .28 kg; and, t

Question

a. The mass of a golf ball is .046 kg; the mass of a club head is .28 kg; and, the velocity of the club head is 55 m/s before it impacts the ball. If the coefficient of restitution of the ball is .82, how fast is the ball moving after impact? (1pt)

         Use this formula for part a.: v1 = ((CoR + 1)m2u2 + u1(m1 – CoRm2)) / m1 + m2

                  Where     v = velocity after impact

                                    u = velocity before impact

b. What is the momentum of the club head before impact? (1pt)

c. What is the momentum of the ball after impact? (1pt)

d. What is the momentum of the club head after impact? (1 pt)

e. Is the total momentum before impact equal to the total momentum after impact? If not, why not?

Explanation / Answer

Velocity after impact of the ball:

V= (0.82+1)0.28*55 / 0.046+0.28

V=85.98 m/s

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b) Momentum = mass * Velocity

Before impact = 0.28 (55) = 15.4 Kg m/s

c) Ball after impact = 0.046 (85.98) = 3.956 Kg m/s

d) The coefficient of restitution is:

CR= (V2-V1) / - U2

With this we find the value of V2 (Velocity of club after impact)

V2=-CR*U2 +V1 = 40.88 m/s

Momentum of club after impact:

M= 0.28(40.88) = 11.446 Kg m/s

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e) If you add up all of the momentums, you will see that it is the same before and after the impact.

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