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Q48P

Expert-verifiedFound in: Page 170

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**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 \(39.95\;{\rm{u}}\).

** **

**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.

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,

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

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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