/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A 685-g block is sliding on a fr... [FREE SOLUTION] | 91Ó°ÊÓ

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A 685-g block is sliding on a frictionless surface when it collides elastically and head-on with a stationary block of mass \(232 \mathrm{~g}\). What percentage of the more massive block's kinetic energy is transferred to the lighter block?

Short Answer

Expert verified
The percentage of kinetic energy transferred from the more massive block to the lighter block is given by: Percentage = \(\frac{ KE_{2f}}{ \frac{1}{2} m_1 v_{1i}^2 } \times 100\), where \(v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i}\), \(m_1\) and \(m_2\) are the masses of the blocks, and \(v_{1i}\) is the initial velocity of the more massive block.

Step by step solution

01

Determine the initial conditions

Firstly, let's denote the more massive block as block 1 and the lighter block as block 2. From the problem, we have the following initial condition that block 1 of mass \(m_1 = 685 \, \mathrm{g}\) (we will convert this mass to kg for standard unit, so \(m_1 = 0.685 \, \mathrm{kg}\) ) is moving while block 2 of mass \(m_2 = 232 \, \mathrm{g}\) (or \(m_2 = 0.232 \, \mathrm{kg}\) ) is initially at rest. So, the initial velocity of block 1 is \(v_{1i}\) (unknown), and the initial velocity of block 2 is \(v_{2i} = 0\).
02

Apply conservation of momentum

Since we're dealing with an elastic collision and no external forces are acting upon the system, we can say that the total momentum before the collision is equal to the total momentum after. This can be written as \(m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}\) where \(v_{1f}\) and \(v_{2f}\) are the final velocities of block 1 and 2, respectively.
03

Apply conservation of kinetic energy

For an elastic collision, total kinetic energy before the collision is equal to the total kinetic energy after the collision. It can be written as \(\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2\)
04

Solve for final velocities

By solving the equations from step 2 and 3 together, we can express the final velocities in terms of the initial velocity of the more massive block. The final velocity of block 1, \(v_{1f}\), can be found as \(v_{1f} = \frac{(m_1 - m_2)}{(m_1 + m_2)} v_{1i}\). The final velocity of block 2, \(v_{2f}\), can be found as \(v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i}\)
05

Find the percentage of kinetic energy transferred

We initially calculate the kinetic energy of both blocks after the collision: \(KE_{1f} = \frac{1}{2} m_1 v_{1f}^2\) and \(KE_{2f} = \frac{1}{2} m_2 v_{2f}^2\). To find the percentage of the kinetic energy transferred from block 1 to block 2, we divide the final kinetic energy of block 2 by the initial kinetic energy of block 1 and multiply by 100. It can be written as: Percentage = \(\frac{ KE_{2f}}{ \frac{1}{2} m_1 v_{1i}^2 } \times 100\)

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Key Concepts

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

Conservation of Momentum
In an elastic collision, the overall momentum of the system remains unchanged, consistent with the principle of conservation of momentum. Momentum is defined as the product of an object's mass and velocity, and it can be visualized as the 'quantity of motion' an object possesses. There is no net force acting on the system, meaning momentum prior to the collision equals momentum following the collision.

To ensure momentum conservation in our example, we begin by identifying the total initial momentum. The heavier block, with an initial momentum of \(m_1v_{1i}\), moves towards the stationary block. The lighter block, having no motion initially, doesn't contribute to the total initial momentum. Thus, the total initial momentum is \(m_1v_{1i}\).

After the collision, the momentum shared between both blocks is hence expressed as:
  • \(m_1v_{1f}\) for the more massive block
  • \(m_2v_{2f}\) for the less massive block
The final formula for conservation of momentum therefore becomes:
\[m_1v_{1i} = m_1v_{1f} + m_2v_{2f}\]
Kinetic Energy Transfer
Kinetic energy gives us a measure of the energy an object has due to its motion. It is expressed through the formula \(KE = \frac{1}{2}mv^2\). An elastic collision not only preserves momentum but also ensures that the total kinetic energy of the system remains constant. This means that the kinetic energy before the collision and after the collision is equal.

In our exercise, the kinetic energy initially is all contained in the moving block given by \(\frac{1}{2}m_1v_{1i}^2\). Post-collision, it is spread out between both blocks. For the more massive block, the final kinetic energy is \(\frac{1}{2}m_1v_{1f}^2\), and for the less massive block, it's \(\frac{1}{2}m_2v_{2f}^2\).

The rule for an elastic collision requires these energies to be equal, thus:
\[\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2\] This conservation offers insights not only into how energy is shifted between the blocks but also helps calculate the resulting velocities after the collision.
Final Velocities Calculation
The final velocities of two colliding bodies in an elastic collision are determined using the simultaneous application of conservation principles - momentum and kinetic energy. These final velocities provide insight into the system dynamics post-collision.

By rearranging the equations obtained from conservation laws, we derive the final velocities as follows:
  • The final velocity of the more massive block: \(v_{1f} = \frac{(m_1 - m_2)}{(m_1 + m_2)}v_{1i} \)
    • This expression indicates a reduction in speed due to mass redistribution.
  • The final velocity of the less massive block: \(v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} \)
    • This formula shows an increase in speed, as the stationary block gains momentum.
These calculations offer a precise measure of how each block's motion changes after the collision. They also illustrate the shift of kinetic energy from the moving block to the stationary one, ensuring an intuitive and mathematically sound grasp of collision dynamics.

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Most popular questions from this chapter

Two objects, one initially at rest, undergo a one-dimensional elastic collision. If half the kinetic energy of the initially moving object is transferred to the other object, what is the ratio of their masses?

You're working in quality control for a model rocket manufacturer, testing a class-D rocket whose specifications call for an impulse between 10 and \(20 \mathrm{~N} \cdot \mathrm{s}\). The rocket's burn time is \(\Delta t=2.8 \mathrm{~s}\), and its thrust during that time is \(F(t)=a t(t-\Delta t)\), where \(a=-4.6 \mathrm{~N} / \mathrm{s}^{2}\). Does the rocket meet its specs?

A mass \(m_{1}\) collides elastically and head-on with a stationary mass \(m_{2}\), and three-fourths of \(m_{1}\) 's initial kinetic energy is transferred to \(m_{2}\). How are the two masses related?

Model rocket motors are specified by giving the impulse they a provide, in \(\mathrm{N} \cdot \mathrm{s}\), over the entire time the rocket is firing. The table below shows the results of rocket-motor tests with different motors used to launch rockets of different masses. Determine two data-based quantities that, when plotted against each other, should give a straight line and whose slope should allow you to determine \(g\). Plot the data, establish a best-fit line, and determine \(g .\) Assume that the maximum height is much greater than the distance over which the rocket motor is firing, so you can neglect the latter. You're also neglecting air resistance-but explain how that affects your experimentally determined value for \(g\). $$ \begin{array}{|l|r|r|r|r|r|} \hline \text { Impulse, } J(\mathrm{~N} \cdot \mathrm{s}) & 4.5 & 7.8 & 4.5 & 7.8 & 11 \\ \hline \text { Rocket mass }(\mathrm{g}) \text { (including motor) } & 180 & 485 & 234 & 234 & 485 \\ \hline \text { Maximum height achieved }(\mathrm{m}) & 22 & 13 & 19 & 51 & 23 \\\ \hline \end{array} $$

A block of mass \(m_{1}\) undergoes a one-dimensional elastic collision with an initially stationary block of mass \(m_{2}\). Find an expression for the fraction of the initial kinetic energy transferred to the second block, and plot your result for mass ratios \(m_{1} / m_{2}\) from 0 to 20 .

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