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Chapter 8: Problem 7

Force of a Golf Swing. A \(0.0450-\mathrm{kg}\) golf ball initially at rest is given a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) when a club strikes. If the club and ball are in contact for 2.00 \(\mathrm{ms}\) , what average force acts on the ball? Is the effect of the ball's weight during the time of contact significant? Why or why not?

Short Answer

Expert verified
The average force is 562.5 N. The ball's weight is negligible compared to the swing force.

Step by step solution

01

Identify Given Values

In the problem, we are given the mass of the golf ball as \( m = 0.0450 \, \text{kg} \), the final speed of the ball as \( v_f = 25.0 \, \text{m/s} \), and the time during which the club is in contact with the ball as \( \Delta t = 2.00 \, \text{ms} = 2.00 \times 10^{-3} \, \text{s} \). The initial speed of the ball \( v_i \) is 0 m/s since the ball starts from rest.
02

Use the Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse on an object is equal to the change in momentum of the object: \( F_{\text{avg}} \Delta t = m \Delta v \). The change in velocity \( \Delta v = v_f - v_i = 25.0 \, \text{m/s} - 0 \, \text{m/s} \).
03

Solve for Average Force

Rearranging the impulse-momentum equation for average force gives \( F_{\text{avg}} = \frac{m \Delta v}{\Delta t} \). Substitute the given values: \( F_{\text{avg}} = \frac{0.0450 \, \text{kg} \times 25.0 \, \text{m/s}}{2.00 \times 10^{-3} \, \text{s}} \). Calculate the force: \( F_{\text{avg}} = 562.5 \, \text{N} \).
04

Analyze the Significance of Weight

The average force exerted by the club (562.5 N) is much greater than the weight of the ball, which is:\( W = mg = 0.0450 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.44145 \, \text{N} \). Since 0.44145 N (the force due to gravity) is negligible compared to 562.5 N, the effect of the ball's weight during the contact is not significant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Force Calculation
When a golf club strikes a ball, it exerts a force over a short period of time, changing the ball's momentum. To find this force, we use the Impulse-Momentum Theorem. This theorem relates impulse (force applied over time) to the change in momentum of an object.

In our example, the ball starts from rest and reaches a speed because of the club鈥檚 impact. Impulse is calculated using the formula:
  • \( F_{\text{avg}} \Delta t = m \Delta v \)
Here, \( F_{\text{avg}} \) is the average force, \( \Delta t \) is the contact time, \( m \) is mass, and \( \Delta v \) is the change in velocity. For calculation, rearrange the formula:
  • \( F_{\text{avg}} = \frac{m \Delta v}{\Delta t} \)
Plugging in the values:
  • \( F_{\text{avg}} = \frac{0.0450 \, \text{kg} \times 25.0 \, \text{m/s}}{2.00 \times 10^{-3} \, \text{s}} \)
  • \( F_{\text{avg}} = 562.5 \, \text{N} \)
Thus, the average force exerted by the club on the ball is 562.5 N.
Golf Ball Dynamics
Golf ball dynamics during a swing involve quick acceleration and deceleration. At first, the ball is stationary on the ground. Upon impact, the golf club imparts a force, setting the ball into motion. This interaction is swift, lasting merely milliseconds.

The force applied by the club is what determines the speed of the ball post-impact. The greater the force during this brief contact, the faster the ball travels. However, the total momentum transferred is dependent on both the magnitude of the force and the time duration of the impact.
  • A strong but quick hit translates to a large change in speed.
  • A weaker or prolonged force can have similar or different impacts depending on the specific interaction.
This interaction encapsulates the core concepts of momentum and force, demonstrating how key variables like time and speed are interlinked in golf ball dynamics.
Significance of Ball's Weight
When considering the significance of a golf ball's weight during the brief contact with a club, we compare it with the average force applied. The weight of the ball is calculated using its mass and gravity:
  • \( W = mg = 0.0450 \, \text{kg} \times 9.81 \, \text{m/s}^2 \)
  • Weight \( = 0.44145 \, \text{N} \)
In this scenario, the club's impact exerts an average force of 562.5 N, considerably larger than the gravitational weight acting on the ball. This contrast illustrates that during the split-second impact, the ball鈥檚 weight is negligible. The force of gravity is too small to significantly influence the ball's immediate acceleration and velocity.

Therefore, while gravity always affects the ball's behavior over a longer span (like when in flight), its influence during the dynamic strike is minimal. Understanding this helps in appreciating how external forces surpass gravitational effects in very short-force applications.

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Most popular questions from this chapter

A wagon with two boxes of gold, having total mass 300 \(\mathrm{kg}\) . is cut loose from the horses by an outlaw when the wagon is at rest 50 \(\mathrm{m}\) up a \(6.0^{\circ}\) slope (Fig. 8.50\() .\) The outlaw plans to have the wagon roll down the slope and across the level ground. and then fall into a canyon where his confederates wait. But in a tree 40 \(\mathrm{m}\) from the canyon edge wait the Lone Ranger (mass 75.0 \(\mathrm{kg}\) ) and Tonto (mass 60.0 \(\mathrm{kg}\) ). They drop vertically into the wagon as it passes beneath them. (a) If they require 5.0 \(\mathrm{s}\) to grab the gold and jump out, will they make it before the wagon goes over the edge? The wagon rolls with negligible friction. (b) When the two heroes drop into the wagon, is the kinetic energy of the system of the heroes plus the wagon conserved? If not, does it increase or decrease, and by how much?

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A stationary object with mass \(m_{B}\) is struck head-on by an object with mass \(m_{A}\) that is moving initially at speed \(v_{0}\) , (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases \((1) m_{A}=m_{B}\) and \((i i) m_{A}=5 m_{B} ?(c)\) For what values, if any, of the mass ratio \(m_{A} / m_{B}\) is the original kinetic energy shared equally by the two objects after the collision?

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