Problem 62 - College Physics

Question

\(\bullet\) A steel ball with a mass of 40.0 \(\mathrm{g}\) is dropped from a height of 2.00 \(\mathrm{m}\) onto a horizontal steel slab. The ball rebounds to a height of 1.60 \(\mathrm{m}\) . (a) Calculate the impulse delivered to the ball during the impact. (b) If the ball is in contact with the slab for \(2.00 \mathrm{ms},\) find the average force on the ball during the impact.

Step-by-Step Solution

Answer

Impulse: 0.4744 Ns, Average Force: 237.2 N

Step 1: Convert Mass to Kilograms

The mass of the steel ball is given as 40.0 g. First, we need to convert this into kilograms since the metric units ought to be consistent, especially for using formulas that involve mass and gravity. 1 g = 0.001 kg So, the mass is: \( m = 40.0 \, \mathrm{g} = 0.040 \, \mathrm{kg} \)

Step 2: Calculate Initial and Final Velocities

First, calculate the velocity of the ball just before it hits the ground using the formula for free fall: \( v_i = \sqrt{2gh} \)Here, \( g \) is acceleration due to gravity (9.81 m/s²) and \( h \) is the initial height (2.00 m). \( v_i = \sqrt{2\times 9.81 \times 2.00} = 6.26 \mathrm{m/s} \) (downward)Next, calculate the velocity of the ball just after the rebound using the same formula: \( v_f = \sqrt{2\times 9.81 \times 1.60} = 5.60 \mathrm{m/s} \) (upward)

Step 3: Determine Change in Velocity

The change in velocity (\( \Delta v \)) is the difference between the final and initial velocities, taking direction into account:\( \Delta v = v_f - (-v_i) = 5.60 + 6.26 = 11.86 \mathrm{m/s} \)

Step 4: Calculate Impulse

Impulse (\( J \)) can be found using the formula:\( J = m \times \Delta v \)Substituting the known values:\( J = 0.040 \times 11.86 = 0.4744 \mathrm{Ns} \)

Step 5: Calculate Average Force

The average force (\( F_{avg} \)) exerted on the ball can be calculated using the formula:\( F_{avg} = \frac{J}{\Delta t} \)Where \( \Delta t = 2.00 \times 10^{-3} \mathrm{s} \) (the time the ball is in contact with the slab).\( F_{avg} = \frac{0.4744}{2.00 \times 10^{-3}} = 237.2 \mathrm{N} \)

Key Concepts

Physics Problem Solving

Physics problem solving involves breaking down a complex problem into smaller, manageable steps. This process helps to apply theoretical concepts in practical scenarios. Here, by using formulas and logical reasoning, one can understand how physical quantities are interrelated.
For instance:

• First, identify the given information and what you need to find.
• Convert units if necessary to ensure consistency.
• Use the appropriate formulas to connect what you know with what you need to discover.
• Calculate step by step while checking if results make sense in the context.

These steps not only lead to the correct solution but also deepen your understanding of physics concepts.

Impact Forces

The concept of impact forces deals with the interactions that occur when two objects come into contact and exert force on one another. In this exercise, when the steel ball hits the steel slab, it undergoes a rapid change in motion, which results in a force. When examining impact forces: - **Impulse and Impact:** Impulse is the product of the force applied and the time duration of the force. It links the force of impact directly to the change in momentum of the object. - **Materials and Force Response:** Different materials behave differently upon impact. Steel, being rigid, reflects much of the energy without deformation.
Understanding these interactions allows us to calculate things like impulse, which quantifies the change in momentum during the impact.

Velocity Calculation

Velocity calculation is crucial for determining how objects move and change speed or direction. For falling objects, the velocity right before impact can be calculated using kinematic equations.In the example:- **Free Fall Initial Velocity:** The velocity just before the steel ball hits the ground is found using the equation \( v = \sqrt{2gh} \). The gravitational constant \( g \) (9.81 m/s²) and the height \( h \) from which the object falls are used to calculate this.- **Rebound Velocity:** After rebounding, the velocity is calculated similarly, indicating how much kinetic energy was retained.
By accurately calculating these velocities, one can determine the change in speed upon rebounding, crucial for finding impulse.

Average Force

The average force is a measure of how much force is applied during the impact duration. It can be determined using the impulse measured and the time of contact.To calculate average force:- **Link with Impulse:** Use the relation \( F_{avg} = \frac{J}{\Delta t} \), where \( J \) is the impulse and \( \Delta t \) is the duration of contact.- **Importance of Contact Time:** A short contact time results in a high average force, reflecting how quickly the momentum changed.
Understanding average force helps grasp how impactful forces are distributed over the time they are applied.