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Chapter 9: Problem 35

A proton moving at \(6.9 \mathrm{Mm} / \mathrm{s}\) collides elastically head-on with a second proton moving in the opposite direction at \(11 \mathrm{Mm} / \mathrm{s}\). Find their subsequent velocities.

Short Answer

Expert verified
After the collision, the first proton will move with a speed of \(11 \mathrm{Mm/s}\) and the second proton will move with a speed of \(6.9 \mathrm{Mm/s}\)

Step by step solution

01

Understanding the Problem

Two protons are moving towards each other with different velocities, \(v_1 = 6.9 \mathrm{Mm/s}\) and \(v_2 = 11 \mathrm{Mm/s}\). Once they collide elastically, their velocities will change, and we are to find these new velocities.
02

Using the principles of conservation of momentum and kinetic energy

Conservation of momentum gives us \(p_{i} = p_{f}\) which simplifies to \(m_1*v_1 - m_2*v_2 = m_1*v_{1f} - m_2*v_{2f}\), where \(v_{1f}\) and \(v_{2f}\) are the final velocities of the protons. Since the masses of the protons are equal, the equation simplifies to \(v_1 - v_2 = v_{1f} - v_{2f}\). Conservation of kinetic energy gives us \(\frac{1}{2}*m_1*v_1^2 + \frac{1}{2}*m_2*v_2^2 = \frac{1}{2}*m_1*v_{1f}^2 + \frac{1}{2}*m_2* v_{2f}^2\), which simplifies to \(v_1^2 + v_2^2 = v_{1f}^2 + v_{2f}^2\).
03

Finding the Subsequent Velocities

By solving the two equations derived from conservation of kinetic energy and momentum we get \(v_{1f} = v_2\) and \(v_{2f} = v_1\).

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