/*! 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 26 A billiard ball moving at a spee... [FREE SOLUTION] | 91Ó°ÊÓ

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A billiard ball moving at a speed \(u_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}^{2}=v_{\mathrm{lf}}^{2}+v_{2 f}^{2} $$

Short Answer

Expert verified
In a completely elastic collision, the initial speed squared equals the sum of the squares of the final speeds.

Step by step solution

01

Understand Elastic Collision

In a completely elastic collision, both momentum and kinetic energy are conserved. Here, we have two identical billiard balls. The first ball, with initial velocity \(u_{1i}\), strikes a second ball at rest. After the collision, the first ball moves with velocity \(v_{1f}\) and the second ball moves with velocity \(v_{2f}\).
02

Write the Momentum Conservation Equation

For the system, the total momentum before and after the collision must be equal. Since the second ball is at rest initially, the initial momentum is simply \(m u_{1 i}\). Therefore, \[ m u_{1 i} = m v_{1 f} + m v_{2 f} \]Here, \(m\) is the mass of each ball, which can be canceled out to simplify.\[ u_{1 i} = v_{1 f} + v_{2 f} \]
03

Write the Kinetic Energy Conservation Equation

Conservation of kinetic energy gives:\[ \frac{1}{2} m u_{1 i}^2 = \frac{1}{2} m v_{1 f}^2 + \frac{1}{2} m v_{2 f}^2 \]Cancel \(\frac{1}{2} m\) from both sides:\[ u_{1 i}^2 = v_{1 f}^2 + v_{2 f}^2 \]
04

Combine Equations

Since \(u_{1 i} = v_{1 f} + v_{2 f}\), squaring both sides gives:\[ u_{1 i}^2 = (v_{1 f} + v_{2 f})^2 = v_{1 f}^2 + 2v_{1 f}v_{2 f} + v_{2 f}^2 \]From kinetic energy conservation, we have \(u_{1 i}^2 = v_{1 f}^2 + v_{2 f}^2\), so:\[ v_{1 f}^2 + v_{2 f}^2 = v_{1 f}^2 + 2v_{1 f}v_{2 f} + v_{2 f}^2 \]Subtracting \(v_{1 f}^2 + v_{2 f}^2\) from both sides confirms that:\[ 0 = 2v_{1 f}v_{2 f} \]Since \(v_{1 f}v_{2 f} = 0\), the equation \(u_{1 i}^2 = v_{1 f}^2 + v_{2 f}^2\) holds.

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Key Concepts

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

Momentum Conservation
In physics, the principle of momentum conservation is a fundamental law and holds true for isolated systems where no external forces act. In the context of an elastic collision, such as a billiard ball hitting another, the total momentum of the system before the collision is equal to the total momentum after the collision.
This means if one object exerts a force on another, there will be an equal and opposite reaction, making the net momentum change zero.

For a better understanding, let's consider our example: a billiard ball with initial velocity \(u_{1i}\) collides with a second ball at rest. Their masses are equal, simplifying the equation:
  • Before the collision: Momentum = \(m \times u_{1i} + m \times 0 = m \times u_{1i}\)
  • After the collision: Momentum = \(m \times v_{1f} + m \times v_{2f}\)
By arranging these momentum equations, we have:
\[ u_{1i} = v_{1f} + v_{2f} \]
This relationship shows that the initial velocity of the striking ball equals the sum of both velocities after impact, respecting the conservation of momentum.
Kinetic Energy Conservation
The concept of kinetic energy conservation in elastic collisions is equally crucial. An elastic collision is defined where there is no loss of kinetic energy in the system. Energy doesn't diminish but transforms between participants, in this case, between the billiard balls.

To delve deeper using our example, the kinetic energy of the moving billiard ball prior to collision is entirely transferred between itself and the stationary ball during the elastic collision. The mathematical representation of this conservation is:
  • Initial kinetic energy: \( \frac{1}{2} m u_{1i}^2\)
  • Final kinetic energy: \( \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2\)
By cancelling out common terms, we simplify the equation:
\[ u_{1i}^2 = v_{1f}^2 + v_{2f}^2 \]
This formula shows that all the kinetic energy before the collision is perfectly distributed afterwards, maintaining the energy balance as dictated by the law of conservation of energy.
Billiard Balls Collision
Billiard ball collisions are excellent practical examples of elastic collisions, typically showcasing both momentum and energy conservation principles. This makes them a popular choice for physics experiments and exercises.

When two billiard balls collide head-on, like in our exercise, several interesting dynamics occur:
  • One moving ball can indeed impart all its energy to another identical ball at rest, bringing itself to rest post-collision while setting the other in motion at the previous speed of the first.
  • This scenario perfectly illustrates how energy and momentum distribute across the objects involved.
Understanding these principles can help us predict and analyze outcomes in various settings, from educational experiments to professional billiard games.
Billiard balls, due to their design and material, minimize energy loss through heat or sound, closely aligning with the ideal conditions of an elastic collision and providing an intuitive understanding of theoretical physics concepts.

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Most popular questions from this chapter

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10-g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(0.90\).

Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm}\). Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and z-coordinates of the center of mass are zero.

A force of \(1000 \mathrm{~N}\) is applied to a small space satellite for a time of \(10.0\) minutes. If the craft has a mass of \(200 \mathrm{~kg}\), what will be its final speed? [Hint: Be careful with those exponents when using a calculator.]

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-6? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, (Impulse) \(_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.

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