Problem 7
Question
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?
Step-by-Step Solution
Verified Answer
The average force on the ball is 562.5 N. The weight is insignificant compared to the force from the swing.
1Step 1: Identify the Given Values
Identify the mass of the golf ball as \( m = 0.0450 \, \text{kg} \), the final speed \( v = 25.0 \, \text{m/s} \), and the time of contact \( t = 2.00 \, \text{ms} = 2.00 \times 10^{-3} \, \text{s} \).
2Step 2: Determine the Change in Velocity
As the golf ball is initially at rest, its initial velocity \( u = 0 \). Thus, the change in velocity \( \Delta v = v - u = 25.0 \, \text{m/s} \).
3Step 3: Calculate the Acceleration
Using the formula for acceleration, \( a = \frac{\Delta v}{t} \). Substitute \( \Delta v = 25.0 \, \text{m/s} \) and \( t = 2.00 \times 10^{-3} \, \text{s} \): \[ a = \frac{25.0}{2.00 \times 10^{-3}} = 12500 \, \text{m/s}^2 \].
4Step 4: Calculate the Average Force
Use the formula for force, \( F = ma \), where \( m = 0.0450 \, \text{kg} \) and \( a = 12500 \, \text{m/s}^2 \): \[ F = 0.0450 \times 12500 = 562.5 \, \text{N} \].
5Step 5: Consider the Ball's Weight Effect
Compare the gravitational force on the ball, \( F_g = mg \), where \( g = 9.81 \, \text{m/s}^2 \): \[ F_g = 0.0450 \times 9.81 = 0.44145 \, \text{N} \]. Since \( 0.44145 \, \text{N} \) is negligible compared to \( 562.5 \, \text{N} \), the weight is insignificant during contact.
Key Concepts
Average forceConstant accelerationMomentum
Average force
Average force is an important concept in dynamics, which helps us understand the effect of a force over a short duration. When a force is applied to an object, it usually doesn’t remain constant over time. Instead, we see a variation in magnitude. To make things simpler, we calculate the average force, which gives us a useful approximation of the overall impact during the time of contact.
In the exercise, the average force was calculated when a golf club hits a ball. We used the equation for force:
In the exercise, the average force was calculated when a golf club hits a ball. We used the equation for force:
- Average Force (\( F \)) = Mass (\( m \)) x Acceleration (\( a \))
Constant acceleration
In many physics problems, especially those involving motion, constant acceleration plays a critical role. It assumes that acceleration does not change over time.In the context of the golf ball problem, after determining the ball speeds up to 25 m/s in 2 milliseconds, we calculated acceleration using the formula:
- Acceleration (\( a \)) = Change in velocity (\( \Delta v \)) / Time (\( t \))
Momentum
Momentum is a key concept in dynamics representing the quantity of motion an object has. It depends on two factors: mass and velocity. In formula terms:
- Momentum (\( p \)) = Mass (\( m \)) x Velocity (\( v \))
- Impulse = Change in momentum (\( \Delta p \))
Other exercises in this chapter
Problem 5
One 110 -kg football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another 125 -kg lineman is running directly toward him at 2.60 \(
View solution Problem 6
BIO Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis
View solution Problem 8
Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\)
View solution Problem 9
A 0.160 -kg hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0,\) the puck is moving to the right at 3.00 \(\mathrm{m} / \mathrm{s} .\)
View solution