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Physics Principles with Applications
Found in: Page 170
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

A neon atom \(\left( {m = 20.0\;{\rm{u}}} \right)\) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 55.6° angle from its original direction and the unknown atom travels away at a \( - {50.0^ \circ }\) angle. What is the mass (in u) of the unknown atom? [Hint: You could use the law of sines.]

The mass of the unknown atom is \(39.95\;{\rm{u}}\).

See the step by step solution

Step by Step Solution

Step 1: Given data

First, find the vector representation for the momentum conservation. Then using the laws of sines, you can get the relation between the velocities. Finally, using kinetic energy conservation, you get the result.

The mass of the neon atom is \(m = 20.0\;{\rm{u}}\).

The direction of the neon atom after the collision is \(\theta = {55.6^ \circ }\) anticlockwise (as positive), and the direction of the unknown atom is \(\phi = {50.0^ \circ }\) clockwise (as negative) with the initial direction of the neon atom. Let \(\alpha \) be the angle of the third angle of the triangle of momentum vectors.

Step 2: Use of law of sines

From the figure, you get:

\(\begin{array}{c}\alpha = {180^ \circ } - \left( {\theta + \phi } \right)\\ = {180^ \circ } - \left( {{{55.6}^ \circ } + {{50.0}^ \circ }} \right)\\ = {74.4^ \circ }\end{array}\)

Using momentum conservation, you get:

\(m{\vec v_1} + {m_2}{\vec v_2} = m\vec v\)

Now, using the laws of sine, you get:

\(\begin{array}{l}\frac{{m{v_1}}}{{\sin \phi }} = \frac{{mv}}{{\sin \alpha }}\\{v_1} = v\frac{{\sin \phi }}{{\sin \alpha }}\end{array}\)

And also,

\(\begin{array}{l}\frac{{{m_2}{v_2}}}{{\sin \theta }} = \frac{{mv}}{{\sin \alpha }}\\{v_2} = v\frac{{m\sin \theta }}{{{m_2}\sin \alpha }}\end{array}\)

Step 3: Use of kinetic energy conservation

As the collision is elastic, the kinetic energy is conserved. Then,

\(\begin{array}{c}\frac{1}{2}m{v^2} = \frac{1}{2}mv_1^2 + \frac{1}{2}{m_2}v_2^2\\{m_2}v_2^2 = m{v^2} - mv_1^2\\{m_2}{\left( {v\frac{{m\sin \theta }}{{{m_2}\sin \alpha }}} \right)^2} = m{v^2} - m{\left( {v\frac{{\sin \phi }}{{\sin \alpha }}} \right)^2}\\\frac{{{m^2}{v^2}}}{{{m_2}}}{\left( {\frac{{\sin \theta }}{{\sin \alpha }}} \right)^2} = m{v^2} - m{v^2}{\left( {\frac{{\sin \phi }}{{\sin \alpha }}} \right)^2}\\\end{array}\)

After further simplification, you get:

\(\begin{array}{c}\frac{m}{{{m_2}}}{\left( {\frac{{\sin \theta }}{{\sin \alpha }}} \right)^2} = 1 - {\left( {\frac{{\sin \phi }}{{\sin \alpha }}} \right)^2}\\\frac{{m{{\sin }^2}\theta }}{{{m_2}}} = {\sin ^2}\alpha - {\sin ^2}\phi \\{m_2} = \frac{{m{{\sin }^2}\theta }}{{{{\sin }^2}\alpha - {{\sin }^2}\phi }}\end{array}\)

Now, substituting the values in the above equation, you get:

\(\begin{array}{c}{m_2} = \frac{{\left( {20.0\;{\rm{u}}} \right){{\sin }^2}{{55.6}^ \circ }}}{{{{\sin }^2}{{74.4}^ \circ } - {{\sin }^2}{{50.0}^ \circ }}}\\ = 39.95\;{\rm{u}}\end{array}\)

Hence, the mass of the unknown atom is \(39.95\;{\rm{u}}\).

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