The head of a 245 g golf club traveling at 62 m/s just before it strikes a 48 g golf ball at rest on a tee. After the collision, the club travels in the same direction at 43 m/s. Use the law of conservation of momentum find the speed of the golf ball just after impact.

The speed of the golf ball just after impact is 96.98 m/s

Explanation:

Given;

mass of the club, m₁ = 245 g = 0.245 kg

initial velocity of the club, u₁ = 62 m/s

mass of the golf ball, m₂ = 48 g = 0.048 kg

initial velocity of the golf ball, u₂ = 0

final velocity of the club, v₁ = 43 m/s

let the final velocity of the golf ball

Apply the law of conservation of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.245(62) + (0.048 x 0) = 0.245(43) + 0.048v₂

15.19 = 10.535 + 48v₂

15.19 - 10.535 = 48v₂

4.655 = 48v₂

v₂ = 4.655 / 0.048

v₂ = 96.98 m/s

Therefore, the speed of the golf ball just after impact is 96.98 m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.