A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 5.0\times10^{-3} s. Find (a) the impulse imparted to the golf ball,and (b) the average force exerted on the ball by the golf club.

Rowan Bray

Rowan Bray

Answered question

2023-03-31

We have: A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 5.0×103 s. Find (a) the impulse imparted to the golf ball,and (b) the average force exerted on the ball by the golf club

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Answer & Explanation

duairceasrxtg

duairceasrxtg

Beginner2023-04-01Added 6 answers

(a) The impulse imparted to the golf ball can be calculated using the equation:
Impulse=Change in momentum=Final momentumInitial momentum
The momentum of an object is given by the product of its mass and velocity.
Given:
Mass of the golf ball (m) = 0.045 kg
Initial velocity of the golf ball (u) = 0 m/s (assuming it was initially at rest)
Final velocity of the golf ball (v) = 45 m/s
Time of contact (Δt) = 5.0 * 10^-3 s
The initial momentum (pinitial) is given by:
pinitial=m·u=0.045kg·0m/s=0kg·m/s
The final momentum (pfinal) is given by:
pfinal=m·v=0.045kg·45m/s=2.025kg·m/s
Therefore, the impulse imparted to the golf ball is:
Impulse=pfinalpinitial=2.025kg·m/s0kg·m/s=2.025kg·m/s
(b) The average force exerted on the ball by the golf club can be calculated using the equation:
Average Force=ImpulseΔt
Given:
Impulse = 2.025 kg*m/s
Time of contact (Δt) = 5.0 * 10^-3 s
Substituting these values into the equation, we get:
Average Force=2.025kg·m/s5.0×103s=405N
Therefore, the average force exerted on the ball by the golf club is 405 Newtons (N).

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