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Chapter 7: Problem 52

Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of \(0.50 \mathrm{m} / \mathrm{s},\) someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at $1.30 \mathrm{m} / \mathrm{s} .$ (a) What velocity does the other glider have? (b) Is the total kinetic energy of the two gliders after the collision greater than, less than, or equal to the total kinetic energy before the collision? If greater, where did the extra energy come from? If less, where did the "lost" energy go?

Short Answer

Expert verified
Answer: The second glider is moving to the left with a velocity of -0.30 m/s after the separation. The total kinetic energy after the separation is greater than the initial kinetic energy. The extra kinetic energy comes from the potential energy stored in the compressed spring.

Step by step solution

01

Understand the given information and identify the unknowns

We are given that the two gliders are initially moving together with a common speed of \(0.50\,\text{m/s}\) to the right. When the string is burned, the spring forces the two gliders apart, and consequently, one of the gliders is now moving at a speed of \(1.30\,\text{m/s}\) to the right. We need to find the velocity of the other glider after the separation and assess the variation of kinetic energy.
02

Apply the conservation of linear momentum principle

Before the separation, both gliders have a total momentum of \(M (0.50\,\text{m/s})\) to the right, where \(M\) is the combined mass of both gliders. After the string is burned, we know that one glider has a velocity of \(1.30\,\text{m/s}\) to the right. Let's denote the velocity of the other glider as \(v_2\). Applying the conservation of linear momentum, we have: $$ M(0.50\,\text{m/s}) = \frac{1}{2}M(1.30\,\text{m/s}) + \frac{1}{2} M v_2 $$. We are interested in finding \(v_2\).
03

Solve for the unknown velocity \(v_2\)

Divide both sides of the equation by \(\frac{1}{2}M\): $$ 2(0.50\,\text{m/s}) = 1.30\,\text{m/s} + v_2 $$. Now, solve for \(v_2\): $$ v_2 = 2(0.50\,\text{m/s}) - 1.30\,\text{m/s} = 1.00\,\text{m/s} - 1.30\,\text{m/s} = -0.30\,\text{m/s}. $$ The second glider is moving to the left with a velocity of \(-0.30\,\text{m/s}\) after the separation.
04

Analyze the kinetic energy before and after the separation

Before the separation, the total kinetic energy is: $$ KE_{initial} = \frac{1}{2}M(0.50\,\text{m/s})^2 $$. After the separation, the total kinetic energy is: $$ KE_{final} = \frac{1}{2}\left(\frac{1}{2}M\right)(1.30\,\text{m/s})^2 + \frac{1}{2}\left(\frac{1}{2}M\right)(-0.30\,\text{m/s})^2 $$. Comparing the initial and final kinetic energies: $$ KE_{initial}=\frac{1}{2}M(0.50\,\text{m/s})^2=0.125M $$, $$ KE_{final} = \frac{1}{8}M(1.30\,\text{m/s})^2 + \frac{1}{8}M(-0.30\,\text{m/s})^2 = 0.210625M $$. Since \(KE_{final}> KE_{initial}\), the total kinetic energy after the separation is greater than the initial kinetic energy.
05

Identify the source of the extra kinetic energy

The extra kinetic energy comes from the potential energy stored in the compressed spring. When the string is burnt and the spring is released, this potential energy is converted into kinetic energy, causing the gliders to move apart with different velocities.

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