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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

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

Most popular questions for Physics Textbooks

Billiard balls A and B, of equal mass, move at right angles and meet at the origin of an xy coordinate system as shown in Fig. 7–36. Initially ball A is moving along the y axis at \({\bf{ + 2}}{\bf{.0}}\;{\bf{m/s}}\), and ball B is moving to the right along the x axis with speed \({\bf{ + 3}}{\bf{.7}}\;{\bf{m/s}}\). After the collision (assumed elastic), ball B is moving along the positive y axis (Fig. 7–36) with velocity What is the final direction of ball A, and what are the speeds of the two balls?

FIGURE 7-36

Problem 46. (Ball A after the Collison is not shown)

A ball is dropped from a height of 1.60 m and rebounds to a height of 1.20 m. Approximately how many rebounds will the ball make before losing 90% of its energy?

A child in a boat throws a 5.30-kg package out horizontally with a speed of 10.0 m/s Fig. 7–31. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 24.0 kg and the mass of the boat is 35.0 kg.

FIGURE 7-31

Problem 7.

When a high jumper is in a position such that his arms and lower legs are hanging vertically, and his thighs, trunk, and head are horizontal just above the bar, estimate how far below the torso’s median line the CM will be. Will this CM be outside the body? Use Table 7–1.

A meteor whose mass was about \(1.5 \times {10^8}\;{\rm{kg}}\) struck the Earth \(\left( {{m_{\rm{E}}} = 6.0 \times {{10}^{24}}\;{\rm{kg}}} \right)\) with a speed of about 25 km/s and came to rest in the Earth.

(a) What was the Earth’s recoil speed (relative to Earth at rest before the collision)?

(b) What fraction of the meteor’s kinetic energy was transformed to kinetic energy of the Earth?

(c) By how much did the Earth’s kinetic energy change as a result of this collision?
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