Chapter 4: Problem 452
Two bodies of masses \(m_{1}\) and \(m_{2}\) have equal kinetic energies. If \(P_{1}\) and \(P_{2}\) are their respective momentum, what is ratio of \(\mathrm{P}_{2}: \mathrm{P}_{1}\) ? (A) \(\mathrm{m}_{1}: \mathrm{m}_{2}\) (B) \(\sqrt{\mathrm{m}}_{2} / \sqrt{\mathrm{m}_{1}}\) (C) \(\sqrt{m_{1}}: \sqrt{m_{2}}\) (D) \(\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}\)
Short Answer
Step by step solution
Kinetic Energy formula:
Momentum formula:
Express the equal kinetic energies:
Simplify the equation and solve for ratio of velocities:
Express the momenta:
Find the ratio P2 : P1:
Use the ratio of velocities squared from step 2:
Simplify the expression:
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy
- \(K = \frac{1}{2}mv^2\)
- Doubling the mass will double the kinetic energy, assuming velocity stays constant.
- Doubling the velocity will fourfold increase the kinetic energy because velocity is squared in the formula.
Mass
- It influences the momentum of an object (\(P = m \cdot v\)). More mass means more momentum if velocity remains constant.
- Contributes directly to kinetic energy (\(K = \frac{1}{2}mv^2\)). A larger mass leads to greater kinetic energy at the same speed.
Velocity
- Velocity is squared in the kinetic energy formula \(K = \frac{1}{2}mv^2\), which shows its exponential influence on the energy of a moving object.
- Velocity is used directly in the momentum equation \(P = m \cdot v\), impacting the momentum linearly. Thus, an increase in velocity results in a proportional increase in momentum if mass is steady.
Formula
- Kinetic energy formula \(K = \frac{1}{2}mv^2\) highlights the relationship between movement and energy. It shows how mass and speed of an object contribute to its energy.
- Momentum formula \(P = m \cdot v\) links the mass and velocity of an object to describe its motion's impact.
- The problems involving these formulas require a clear understanding of their derivations and implications—for instance, knowing how to isolate variables or substitute values useful in solving real-world problems.