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When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 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 44 m/s. Neglect the mass of the club handle and determine the speed (in m/s) of the golf ball just after impact.

User MmmHmm
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2 Answers

2 votes

Final answer:

The speed of the golf ball just after impact, calculated using the conservation of momentum, is approximately 20.87 m/s. This result is derived from the initial and final velocities of the golf club and the initial velocity and mass of the golf ball.

Step-by-step explanation:

The question involves a collision between a golf club and a golf ball and asks to determine the speed of the golf ball just after impact. This is a classic example of a conservation of momentum problem. Since no external forces are acting on the system, the total momentum before the collision is equal to the total momentum after the collision. The formula for the conservation of linear momentum in a two-object system is m1×u1 + m2×u2 = m1×v1 + m2×v2.

We are given the following information:

Initial speed of the golf club (u1) = 50 m/s

Mass of the golf club (m1) = 160 g = 0.160 kg (since 1g = 0.001 kg)

Initial speed of the golf ball (u2) = 0 m/s (at rest)

Mass of the golf ball (m2) = 46 g = 0.046 kg

Final speed of the golf club (v1) = 44 m/s

Applying conservation of momentum:

0.160 kg × 50 m/s + 0.046 kg × 0 m/s = 0.160 kg × 44 m/s + 0.046 kg × v2

Solving for v2 (the speed of the golf ball after collision):

8 kg×m/s = 7.04 kg×m/s + 0.046 kg × v2

0.046 kg × v2 = 8 kg×m/s - 7.04 kg×m/s

0.046 kg × v2 = 0.96 kg×m/s

v2 = 0.96 kg×m/s / 0.046 kg

v2 = 20.87 m/s

Therefore, the speed of the golf ball just after impact is approximately 20.87 m/s.

User Minoru
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3 votes

Answer:

v1=21.81m/s

Step-by-step explanation:

When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 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 44 m/s. Neglect the mass of the club handle and determine the speed (in m/s) of the golf ball just after impact.

According to the law of conservation of momentum, if the net external force on a system is zero, then the linear momentum of the system is conserved.

During collision of two particles, the external force on the system of two colliding particles is zero (only internal force acts between the colliding particles), therefore, the momentum is conserved during the collision.

Answer and Explanation:

Given :

head of the golf club=160g

velocity of the golf club=50 m/s

golf ball mass=46g

velocity=om/s

m1u1+m2u2=m1v1+m2v2.........................................1

160*50 +46*0=160*44+46*v1

8000=7040+46v1

960=46v1

v1=960/46

v1=21.81m/s

User KubiRoazhon
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