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Chapter 9: Problem 48

(II) Car A hits car B (initially at rest and of equal mass) from behind while going \(35 \mathrm{~m} / \mathrm{s}\). Immediately after the collision, car B moves forward at \(25 \mathrm{~m} / \mathrm{s}\) and car \(\mathrm{A}\) is at rest. What fraction of the initial kinetic energy is lost in the collision?

Short Answer

Expert verified
About 49% of the initial kinetic energy is lost in the collision.

Step by step solution

01

Identify the Initial Kinetic Energy

We start by calculating the initial kinetic energy of the system. Since Car B is initially at rest, only Car A contributes to the initial kinetic energy. The formula for kinetic energy is \( KE = \frac{1}{2} m v^2 \). For Car A, this is \( KE_{initial} = \frac{1}{2} m (35)^2 \).
02

Calculate Initial Kinetic Energy Value

Substitute the values into the kinetic energy equation for Car A: \( KE_{initial} = \frac{1}{2} m (1225) = 612.5m \) (where \( m \) is the mass of the cars).
03

Identify the Final Kinetic Energy

The final kinetic energy of the system includes only Car B, since Car A is at rest. Therefore, \( KE_{final} = \frac{1}{2} m (25)^2 \).
04

Calculate Final Kinetic Energy Value

Substitute the values into the kinetic energy formula for Car B: \( KE_{final} = \frac{1}{2} m (625) = 312.5m \).
05

Determine the Change in Kinetic Energy

The kinetic energy lost in the collision is the difference between the initial and final kinetic energy: \( \Delta KE = KE_{initial} - KE_{final} = 612.5m - 312.5m = 300m \).
06

Calculate the Fraction of Energy Lost

The fraction of initial kinetic energy lost is \( \frac{\Delta KE}{KE_{initial}} = \frac{300m}{612.5m} = \frac{300}{612.5} \).
07

Simplify the Fraction

Divide 300 by 612.5 to find the fraction of the energy lost: \( \frac{300}{612.5} \approx 0.49 \).

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Key Concepts

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

Collision Physics
In the realm of collision physics, we explore what happens when two objects meet and interact. A collision occurs between two bodies when they exert forces on each other for a brief period of time. In our exercise, Car A hits Car B, resulting in a certain exchange of momentum and energy.
  • Types of Collisions: Collisions can be classified based on how energy is conserved as elastic, inelastic, or perfectly inelastic.
  • Elastic collisions maintain both kinetic energy and momentum, while inelastic ones do not preserve kinetic energy, but still conserve momentum.
  • Our scenario is most likely inelastic as some kinetic energy is lost.
The principles governing these interactions center around forces and how they're applied during the contact phase. This basic understanding helps us explore deeper into the mechanics of the collision, providing insights into the energy and momentum shifts that occur during such interactions.
Energy Conservation
The concept of energy conservation states that within a closed system, energy cannot be created or destroyed, but only transformed from one form to another.
In our problem, the focus is on kinetic energy, which is energy due to motion. Before the impact, Car A possesses kinetic energy due to its velocity, while Car B is stationary with zero kinetic energy.
  • Initial Kinetic Energy: Car A's energy depends solely on its speed and mass: \( KE_{initial} = \frac{1}{2} m (35)^2 \).
  • Post-Collision: After the collision, only Car B moves forward, resulting in a new kinetic energy value.
  • Energy Transformation: The total system energy includes other forms post-collision, hinting at an energy loss in the kinetic form.
This change in energy forms the basis of calculating the fraction of energy transformed during the collision, providing insight into the dynamism of the event.
Momentum Conservation
Momentum conservation is a fundamental principle stating that the total momentum of a closed system remains constant if no external forces act upon it. Momentum is defined by the product of mass and velocity:
  • Before Collision: Car A's momentum results from its motion, while Car B, being at rest, contributes none to the initial momentum.
  • After Collision: The system's combined momentum equates to the altered states of the cars, involving Car B moving forward.
  • Conserved Quantity: Despite a change in individual velocities, overall momentum is conserved, confirming the unchangeability principle.
The exercise showcases that although kinetic energy is not entirely conserved in the collision, momentum remains intact, making it crucial for understanding the underlying mechanics of interacting bodies.

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Most popular questions from this chapter

(III) A particle of mass \(m_{\mathrm{A}}\) traveling with speed \(v_{\mathrm{A}}\) collides elastically head-on with a stationary particle of smaller mass \(m_{\mathrm{B}} \cdot(a)\) Show that the speed of \(m_{\mathrm{B}}\) after the collision is $$ v_{\mathrm{B}}^{\prime}=\frac{2 v_{\mathrm{A}}}{1+m_{\mathrm{B}} / m_{\mathrm{A}}} $$ (b) Consider now a third particle of mass \(m_{\mathrm{C}}\) at rest between \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) so that \(m_{\mathrm{A}}\) first collides head on with \(m_{\mathrm{C}}\) and then \(m_{\mathrm{C}}\) collides head on with \(m_{\mathrm{B}}\). Both collisions are elastic. Show that in this case, $$ v_{\mathrm{B}}^{\prime}=4 v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{\left(m_{\mathrm{C}}+m_{\mathrm{A}}\right)\left(m_{\mathrm{B}}+m_{\mathrm{C}}\right)} $$ (c) From the result of part (b), show that for $$ \operatorname{maximum} v_{\mathrm{B}}^{\prime}, m_{\mathrm{C}}=\sqrt{m_{\mathrm{A}} m_{\mathrm{B}}} $$ \((d)\) Assume \(m_{\mathrm{B}}=2.0 \mathrm{~kg}\) \(m_{\mathrm{A}}=18.0 \mathrm{~kg}\) and \(v_{\mathrm{A}}=2.0 \mathrm{~m} / \mathrm{s}\). Use a spreadsheet to calculate and graph the values of \(v_{\mathrm{B}}^{\prime}\) from \(m_{\mathrm{C}}=0.0 \mathrm{~kg}\) to \(m_{\mathrm{C}}=50.0 \mathrm{~kg}\) in steps of \(1.0 \mathrm{~kg} .\) For what value of \(m_{\mathrm{C}}\) is the value of \(v_{\mathrm{B}}^{\prime}\) maximum? Does your numerical result agree with your result in part \((c) ?\)

A massless spring with spring constant \(k\) is placed between a block of mass \(m\) and a block of mass \(3 m .\) Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount \(D\) from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring.

(III) Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always \(90^{\circ} .\)

(II) A 0.060 -kg tennis ball, moving with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\), has a head-on collision with a \(0.090-\mathrm{kg}\) ball initially moving in the same direction at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision.

The gravitational slingshot effect. Figure \(9-55\) shows the planet Saturn moving in the negative \(x\) direction at its orbital speed (with respect to the Sun) of \(9.6 \mathrm{~km} / \mathrm{s}\). The mass of Saturn is \(5.69 \times 10^{26} \mathrm{~kg} .\) A spacecraft with mass \(825 \mathrm{~kg}\) approaches Saturn. When far from Saturn, it moves in the \(+x\) direction at \(10.4 \mathrm{~km} / \mathrm{s}\). The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn's gravitational pull.
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