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\(\bullet\) 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.

Short Answer

Expert verified
Impulse: 0.4744 Ns, Average Force: 237.2 N

Step by step solution

01

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: \( m = 40.0 \, \mathrm{g} = 0.040 \, \mathrm{kg} \)
02

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)
03

Determine Change in Velocity

The change in velocity (\( \Delta v \)) is the difference between the final and initial velocities, taking direction into account:\( \Delta v = v_f - (-v_i) = 5.60 + 6.26 = 11.86 \mathrm{m/s} \)
04

Calculate Impulse

Impulse (\( J \)) can be found using the formula:\( J = m \times \Delta v \)Substituting the known values:\( J = 0.040 \times 11.86 = 0.4744 \mathrm{Ns} \)
05

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).\( F_{avg} = \frac{0.4744}{2.00 \times 10^{-3}} = 237.2 \mathrm{N} \)

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Key Concepts

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

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.

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Most popular questions from this chapter

\(\bullet\) A ball with a mass of 0.600 \(\mathrm{kg}\) is initially at rest. It is struck by a second ball having a mass of 0.400 \(\mathrm{kg}\) , initially moving with a velocity of 0.250 \(\mathrm{m} / \mathrm{s}\) toward the right along the \(x\) axis. After the collision, the 0.400 \(\mathrm{kg}\) ball has a velocity of 0.200 \(\mathrm{m} / \mathrm{s}\) at an angle of \(36.9^{\circ}\) above the \(x\) axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the 0.600 kg ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

Your little sister (mass 25.0 \(\mathrm{kg}\) ) is sitting in her little red wagon (mass 8.50 \(\mathrm{kg} )\) at rest. You begin pulling her forward and continue accelerating her with a constant force for 2.35 \(\mathrm{s}\) at the end of which time she's moving at a speed of 1.80 \(\mathrm{m} / \mathrm{s}\) . (a) Calculate the impulse you imparted to the wagon and its passenger. (b) With what force did you pull on the wagon?

\(\bullet\) On an air track, a 400.0 g glider moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) collides elastically with a 500.0 g glider moving in the opposite direction at 3.00 \(\mathrm{m} / \mathrm{s}\) . Find the velocity of each glider after the collision.

\(\bullet\) Some useful relationships. The following relationships between the momentum and kinetic energy of an object can be very useful for calculations: If an object of mass \(m\) has momentum of magnitude \(p\) and kinetic energy \(K,\) show that (a) \(K=\left(p^{2} / 2 m\right),\) and \((b) p=\sqrt{2 m K}\) . (c) Find the momen- tum of a 1.15 kg ball that has 15.0 J of kinetic energy. (d) Find the kinetic energy of a 3.50 kg cat that has 0.220 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) of momentum.

\(\bullet\) Tennis, anyone? Tennis players sometimes leap into the air to return a volley. (a) If a 57 g tennis ball is traveling horizontally at 72 \(\mathrm{m} / \mathrm{s}\) (which does occur), and a 61 kg tennis player leaps vertically upward and hits the ball, causing it to travel at 45 \(\mathrm{m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{ms},\) what force does her racket exert on the ball? What force does the ball exert on the racket?

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