Question

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes it. If the club and ball are in contact for 2.00 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

The average force on the ball is 562.5 N, and the effect of its weight is insignificant.

Step-by-step solution

Identify Given Values

First, we determine the values given in the problem. The mass of the golf ball is $m=0.0450\text{kg}$, the initial velocity $u=0\text{m/s}$, the final velocity $v=25.0\text{m/s}$, and the time of contact $\mathrm{\Delta }t=2.00\text{ms}=2.00\times {10}^{-3}\text{s}$.

Calculate Change in Velocity

The change in velocity $\mathrm{\Delta }v$ is computed as the difference between the final velocity and the initial velocity: $\mathrm{\Delta }v=v-u=25.0\text{m/s}-0\text{m/s}=25.0\text{m/s}$.

Use Impulse-Momentum Theorem

The impulse ($J$) is equal to the change in momentum, which can also be written as the average force $F$ times the change in time $\mathrm{\Delta }t$: $$J=F\mathrm{\Delta }t=m\mathrm{\Delta }v.$$

Solve for Average Force

Rearrange the impulse equation to find $F$: $$F=\frac{m\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{0.0450\text{kg}\times 25.0\text{m/s}}{2.00\times {10}^{-3}\text{s}}.$$

Calculate the Force

Calculate the average force by substituting the values: $$F=\frac{0.0450\times 25.0}{2.00\times {10}^{-3}}=\frac{1.125}{0.002}=562.5\text{N}.$$

Consider the Effect of the Ball's Weight

The weight of the ball is calculated using $W=mg=0.0450\text{kg}\times 9.8{\text{m/s}}^2=0.441\text{N}$. Since the average force is 562.5 N, much greater than the ball's weight, the effect of weight during contact is insignificant.

Key concepts

Impulse-Momentum Theorem

When we talk about the Impulse-Momentum Theorem, we're referring to how the force applied to an object and the time it acts determine the change in the object's momentum. This theorem is fundamental in understanding motion and collisions.
It's expressed as $J=\mathrm{\Delta }p=F\mathrm{\Delta }t$, where $J$ represents impulse, $\mathrm{\Delta }p$ is the change in momentum, $F$ is the force applied, and $\mathrm{\Delta }t$ is the time duration.
In simple terms, think of impulse as the product of a force and the time that force is applied. If you push a shopping cart gently for a long time, you provide a small force over a longer time, similar to one big push for a short time. Both scenarios can result in the same change in momentum.
This theorem helps us solve problems like the golf ball being struck by a club. The momentum change (mass times velocity change) can calculate the force exerted when we know the contact time.

Average Force Calculation

Average force gives us an insight into the overall effect of a force during a specific time interval. Especially in brief interactions like a golf club hitting a ball, calculating average force provides a clear picture of the dynamics involved.
The formula involved is $F=\frac{m\mathrm{\Delta }v}{\mathrm{\Delta }t}$. It shows us that the average force is the mass times velocity change ( also momentum change), divided by the time taken. In our golf ball example, this calculation provided 562.5 N of force, showing us how much effort was made to change the ball's speed.
Average force highlights the influence of both magnitude and time. Large forces for short times can match small forces over longer durations, giving engineers insight for safety in crash tests or other scenarios.

Newton's Laws of Motion

Newton's Laws of Motion are the bedrock for understanding mechanics. They relate the forces acting upon an object with its motion, making them essential for everything from casually throwing a ball to complex engineering designs.
In the context of the golf ball, Newton's First Law tells us that without a force, the ball would stay at rest. It's the force of the club that changes this status.
The Second Law, $F=ma$, directly relates to our calculations. It explains how a force (from the club) produces an acceleration (change in velocity) in the ball.
Lastly, Newton's Third Law underscores the action-reaction between the club and ball. As the club exerts force on the ball, the ball equally exerts force on the club. However, the effect on the club due to different mass is less noticeable.

• Understanding these laws helps break down the motion process and explains why and how forces like gravity are often insignificant during high-force interactions over short times.