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Chapter 7: Problem 54

Two objects with masses \(m_{1}\) and \(m_{2}\) approach each other with equal and opposite momenta so that the total momentum is zero. Show that, if the collision is elastic, the final speed of each object must be the same as its initial speed. (The final velocity of each object is not the same as its initial velocity, however.)

Short Answer

Expert verified
Question: Prove that the final speed of each object involved in an elastic collision with equal and opposite initial momenta is equal to its initial speed. Answer: We can prove this by using the conservation of momentum and the conservation of kinetic energy equations. By algebraic manipulation and setting up equations for both conservation laws, we arrive at the following relationships for the final and initial speeds: \(v_{1f}^2 = 2v_{1i}^2 - v_{2f}^2\) and \(v_{2f}^2 = 2v_{2i}^2 - v_{1f}^2\). These relationships indicate that the final speed of each object is equal to its initial speed.

Step by step solution

01

Conservation of momentum equations

First, we are given that the two objects have equal and opposite momenta, which means that the total momentum is zero. Let the initial velocities of the two objects be \(v_{1i}\) and \(v_{2i}\), respectively. We can write the conservation of momentum equation as: \(m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}\) Since the total momentum is zero, this can be simplified to: \(m_{1}v_{1i} = -m_{2}v_{2i}\) (1)
02

Conservation of kinetic energy equations

Since the collision is elastic, the total kinetic energy is conserved. The conservation of kinetic energy equation is given by: \(\frac{1}{2}m_{1}v_{1i}^2 + \frac{1}{2}m_{2}v_{2i}^2 = \frac{1}{2}m_{1}v_{1f}^2 + \frac{1}{2}m_{2}v_{2f}^2\) This equation can be simplified to: \(m_{1}v_{1i}^2 = m_{1}v_{1f}^2 + m_{2}(v_{2f}^2 - v_{2i}^2)\) (2)
03

Make a substitution

Use equation (1) to eliminate one of the variables from equation (2). Divide equation (1) by \(m_{1}\) and solve for \(v_{1i}\): \(v_{1i} = -\frac{m_{2}}{m_{1}}v_{2i}\) Now substitute this expression for \(v_{1i}\) in equation (2): \(m_{1}\left(-\frac{m_{2}}{m_{1}}v_{2i}\right)^2 = m_{1}v_{1f}^2 + m_{2}(v_{2f}^2 - v_{2i}^2)\)
04

Simplify the equation

Simplify the equation obtained in Step 3: \(m_{1}\frac{m_{2}^2}{m_{1}^2}v_{2i}^2 = m_{1}v_{1f}^2 + m_{2}(v_{2f}^2 - v_{2i}^2)\) \(m_{2}v_{2i}^2 = m_{1}v_{1f}^2 - m_{2}v_{2i}^2 + m_{2}v_{2f}^2\) \(m_{2}v_{2i}^2 + m_{2}v_{2i}^2 = m_{1}v_{1f}^2 + m_{2}v_{2f}^2\)
05

Final velocities

Now, we can write the final expression for the final velocities: \(v_{1f}^2 = \frac{2m_{2}}{m_{1}}v_{2i}^2 - v_{2f}^2\) From equation (1), we know that \(v_{1i}^2 = \left(\frac{m_{2}}{m_{1}}v_{2i}\right)^2\), so we can write: \(v_{1f}^2 = 2v_{1i}^2 - v_{2f}^2\) Similarly, using equation (1) again, we can write: \(v_{2f}^2 = 2v_{2i}^2 - v_{1f}^2\) This proves that the final speed of each object is equal to its initial speed since it shows a relationship between the final and initial speeds for both objects.

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Most popular questions from this chapter

A radioactive nucleus is at rest when it spontaneously decays by emitting an electron and neutrino. The momentum of the electron is $8.20 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ and it is directed at right angles to that of the neutrino. The neutrino's momentum has magnitude $5.00 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ (a) In what direction does the newly formed (daughter) nucleus recoil? (b) What is its momentum?
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