Rowan Bray

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 $5.0×{10}^{-3}$ 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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(a) The impulse imparted to the golf ball can be calculated using the equation:

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 ($\Delta t$) = 5.0 * 10^-3 s
The initial momentum (${p}_{\text{initial}}$) is given by:
${p}_{\text{initial}}=m·u=0.045\phantom{\rule{0.167em}{0ex}}\text{kg}·0\phantom{\rule{0.167em}{0ex}}\text{m/s}=0\phantom{\rule{0.167em}{0ex}}\text{kg}·\text{m/s}$
The final momentum (${p}_{\text{final}}$) is given by:
${p}_{\text{final}}=m·v=0.045\phantom{\rule{0.167em}{0ex}}\text{kg}·45\phantom{\rule{0.167em}{0ex}}\text{m/s}=2.025\phantom{\rule{0.167em}{0ex}}\text{kg}·\text{m/s}$
Therefore, the impulse imparted to the golf ball is:
$\text{Impulse}={p}_{\text{final}}-{p}_{\text{initial}}=2.025\phantom{\rule{0.167em}{0ex}}\text{kg}·\text{m/s}-0\phantom{\rule{0.167em}{0ex}}\text{kg}·\text{m/s}=2.025\phantom{\rule{0.167em}{0ex}}\text{kg}·\text{m/s}$
(b) The average force exerted on the ball by the golf club can be calculated using the equation:

Given:
Impulse = 2.025 kg*m/s
Time of contact ($\Delta t$) = 5.0 * 10^-3 s
Substituting these values into the equation, we get:

Therefore, the average force exerted on the ball by the golf club is 405 Newtons (N).

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