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Chapter 7: Problem 11

A \(0.500-\mathrm{kg}\) ball is dropped from rest at a point \(1.20 \mathrm{~m}\) above the floor. The ball rebounds straight upward to a height of \(0.700 \mathrm{~m}\). What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

Short Answer

Expert verified
The impulse is 4.275 kg鈰卪/s upward.

Step by step solution

01

Calculate the initial velocity before impact

The ball is dropped from a height of \(1.20\,\mathrm{m}\). Using the formula for the velocity of a falling object, \(v^2 = u^2 + 2gh\), where \(u = 0\,\mathrm{m/s}\), \(g = 9.8\,\mathrm{m/s^2}\) and \(h = 1.20\,\mathrm{m}\). We find the initial velocity \(v\) right before impact: \[v_1 = \sqrt{0 + 2 \times 9.8 \times 1.20} = 4.85\,\mathrm{m/s}\] downward.
02

Calculate the final velocity after the rebound

The ball rebounds to a height of \(0.700\,\mathrm{m}\). At maximum height, final velocity is 0, and upward velocity \(v_2\) right after rebound can be found using \(v^2 = u^2 + 2gh\), where \(v = 0\), \(h = 0.700\,\mathrm{m}\). We find: \[v_2 = \sqrt{0 + 2 \times 9.8 \times 0.700} = 3.70\,\mathrm{m/s}\] upward.
03

Determine the change in velocity

The change in velocity takes into account both the reversal and change in magnitude of direction. Initial velocity \(v_1\) is \(-4.85\,\mathrm{m/s}\) and final velocity \(v_2\) is \(3.70\,\mathrm{m/s}\). Thus, the change in velocity \(\Delta v = v_2 - v_1 = 3.70 - (-4.85) = 8.55\,\mathrm{m/s}\).
04

Calculate the impulse

Impulse \(J\) is the product of mass \(m\) and change in velocity \(\Delta v\). Thus, \(J = m \cdot \Delta v = 0.500\,\mathrm{kg} \times 8.55\,\mathrm{m/s} = 4.275\,\mathrm{kg \cdot m/s}\). The direction is upward.

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Key Concepts

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

Collision Dynamics
Collision dynamics refer to how objects behave and interact during an impact. When two objects collide, various forces come into play and affect their motion. In our example, we have a ball hitting the floor and then rebounding. This event includes substantial changes in velocity and direction. During a collision, it's essential to observe factors like:
  • The time period of the contact
  • The velocity of the objects before and after the impact
  • The forces acting during the impact
In the ball's situation, the collision dynamics are about how the ball changes direction upon hitting the floor. The instant change in speed and direction shows that the floor exerts a force on the ball during the short contact time, leading to the formation of impulse. Understanding these dynamics helps predict motions from collisions in various settings, from sports to vehicle safety.
Momentum
Momentum is a fundamental concept in physics and can be described as the quantity of motion an object has. It depends on two factors: mass and velocity. The formula for momentum (p) is defined as:\[p = m \cdot v\]where \(m\) is mass, and \(v\) is velocity.In our context, when the ball hits the floor, it has a momentum that results from its mass (0.500 kg) and its velocity before impact (4.85 m/s downward). After rebounding, the ball gains new momentum with a velocity of 3.70 m/s upward. The change in momentum during the impact or collision contributes to the impulse. Specifically, the impulse is equal to the change in momentum, and it tells us how much the velocity of an object changes after the force is applied for a certain period. Often, preserving momentum is crucial in collision dynamics, unless external forces act, which is key in analyzing how interactions like the bounce of a ball respond to forces.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed in an isolated system, but it can change forms. This principle plays a crucial role in analyzing collisions like the ball-to-floor scenario.As the ball falls under the influence of gravity, its potential energy (due to height) converts to kinetic energy (due to velocity). At the point of impact, most of the initial potential energy has become kinetic energy. After the rebound, the kinetic energy shifts back partly into potential energy since the ball rises again.The equation for potential energy (PE) is:\[PE = m \cdot g \cdot h\]or kinetic energy (KE) as:\[KE = \frac{1}{2} m \cdot v^2\]Despite some energy loss due to sound and heat upon collision, most of the energy travels back into the ball鈥檚 upward motion. The analysis shows how predictable energy transformations help us understand the outcome after a collision. Knowing how energy balances in such events is key in conservation and efficiency studies.

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Most popular questions from this chapter

At provides a review of the concepts that are involved in this problem. A \(62.0-\mathrm{kg}\) person, standing on a diving board, dives straight down into the water. Just before striking the water, her speed is \(5.50 \mathrm{~m} / \mathrm{s}\). At a time of \(1.65 \mathrm{~s}\) after she enters the water, her speed is reduced to \(1.10 \mathrm{~m} / \mathrm{s}\). What is the net average force (magnitude and direction) that acts on her when she is in the water?

ssm Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m}\), a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60\) -kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

A lumberjack (mass \(=98 \mathrm{~kg}\) ) is standing at rest on one end of a floating log (mass \(=230 \mathrm{~kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{~m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

At presents a method for modeling this problem. The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of \(1.13 \times 10^{-10} \mathrm{~m}\). The mass \(m_{\mathrm{C}}\) of the carbon atom is 0.750 times the mass \(m_{\mathrm{O}}\) of the oxygen atom, or \(\mathrm{m}_{\mathrm{c}}=0.750 \mathrm{~m}_{0} .\) Determine the location of the center of mass of this molecule relative to the carbon atom.

A volleyball is spiked so that its incoming velocity of \(+4.0 \mathrm{~m} / \mathrm{s}\) is changed to an outgoing velocity of \(-21 \mathrm{~m} / \mathrm{s}\). The mass of the volleyball is \(0.35 \mathrm{~kg}\). What impulse does the player apply to the ball?
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