Chapter 36: Q. 10 (page 1058)

Particle A has half the mass and twice the speed of particle B. Is the momentum $p_{A}$ less than, greater than, or equal to $p_{B}$ ? Explain.

Short Answer

Expert verified

Momentum A is greater than momentum of B.

Step by step solution

Given information

We have given,

Mass of A= $\frac{1}{2}$ Mass of B

Speed of A= $2 \times$ Speed of B

We have to find the momentum of the A and B.

Simplify

Momentum of A, and B is relativistic here,

Then,

$P_{A} = \frac{M_{A} V_{A}}{\sqrt{1 - \frac{V_{A}^{2}}{c^{2}}}} P_{B} = M_{B} V_{B} \times \frac{1}{\sqrt{1 - \frac{V_{B}^{2}}{c^{2}}}} \frac{P_{B}}{P_{A}} = \frac{M_{A} V_{A} \sqrt{1 - \frac{V_{B}^{2}}{c^{2}}}}{M_{B} V_{B} \sqrt{1 - \frac{V_{A}^{2}}{c^{2}}}} \frac{P_{B}}{P_{A}} = \frac{M_{A} (2 V_{B}) \sqrt{1 - \frac{V_{B}^{2}}{c^{2}}}}{(2 M_{A}) V_{B} \sqrt{1 - \frac{4 V_{B}^{2}}{c^{2}}}} {(\frac{P_{B}}{P_{A}})}^{2} = \frac{(1 - \frac{V_{B}^{2}}{c^{2}})}{(1 - \frac{4 V_{B}^{2}}{c^{2}})} {(\frac{P_{B}}{P_{A}})}^{2} = \frac{(c^{2} - V_{B}^{2})}{(c^{2} - 4 V_{B}^{2})}$

Here we can see that the denominator will be greater than numerator then we can say that momentum of A is greater than momentum of B.

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Particle A has half the mass and twice the speed of particle B. Is the momentum $p_{A}$ less than, greater than, or equal to $p_{B}$ ? Explain.

Short Answer

Step by step solution

Given information

Simplify

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Particle A has half the mass and twice the speed of particle B. Is the momentum pA less than, greater than, or equal to pB? Explain.

Short Answer

Step by step solution

Given information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Engineering Physics

Electric Charge Field and Potential

Mechanics and Materials

Physics of Motion

Fundamentals of Physics

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Study anywhere. Anytime. Across all devices.

Particle A has half the mass and twice the speed of particle B. Is the momentum $p_{A}$ less than, greater than, or equal to $p_{B}$ ? Explain.