AP Physics 1 Unit 4B: Conservation of Linear Momentum & Collisions — practice questions — page 4

Explore 92 free Unit 4B: Conservation of Linear Momentum & Collisions questions in AP Physics 1 Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 4B: Conservation of Linear Momentum & Collisions questions and exercises with answers

Showing questions 91–92 of 92.

The **rms speed** of gas molecules at absolute temperature $T$ is:Read the answer and worked solution →Average translational kinetic energy of a gas molecule at temperature $T$ is:Read the answer and worked solution →

Worked practice questions

The rms speed of gas molecules at absolute temperature $T$ is:

$\sqrt{3RT/M_0}$ (also $\sqrt{3 k_B T/m}$)
$\sqrt{2RT/M_0}$
$RT/M_0$
$3RT/M_0$

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Correct answer: A. $\sqrt{3RT/M_0}$ (also $\sqrt{3 k_B T/m}$)

$v_{rms} = \sqrt{\overline{v^2}} = \sqrt{3RT/M_0} = \sqrt{3k_B T/m}$, where $M_0$ is molar mass, $m$ is molecular mass.

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Average translational kinetic energy of a gas molecule at temperature $T$ is:

$k_B T$
$\frac{1}{2} k_B T$
$\frac{3}{2} k_B T$ (= $\frac{3}{2} RT$ per mole)
$3 k_B T$

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Correct answer: C. $\frac{3}{2} k_B T$ (= $\frac{3}{2} RT$ per mole)

Average KE per molecule = $\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T$. **Per mole:** $\frac{3}{2} RT$. Temperature is a direct measure of average kinetic energy.

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