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Chapter 7: Problem 5

Prove that in an elastic scattering process the angle \(\theta+\alpha\) between the emerging particles is related to the recoil angle \(\alpha\) by $$ \frac{\tan (\theta+\alpha)}{\tan \alpha}=\frac{m_{1}+m_{2}}{m_{1}-m_{2}} $$

Short Answer

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Question: Prove the relationship between the angles in an elastic scattering process: $$ \frac{\tan(\theta+\alpha)}{\tan\alpha} = \frac{m_1 + m_2}{m_1 - m_2} $$ where \(m_1\) and \(m_2\) are the masses of the two particles, and \(\theta\) and \(\alpha\) are the angles after the collision.

Step by step solution

01

Understanding elastic scattering processes

An elastic scattering process involves two particles, with masses \(m_1\) and \(m_2\), colliding and conserving both their total momentum and total kinetic energy. After the collision, the two particles move in different directions with angles \(\theta\) and \(\alpha\).
02

Applying conservation of momentum

In an elastic collision, the conservation of momentum implies that the total momentum before the collision is equal to the total momentum after the collision. We can write this in the form of a vector equation: $$ \vec{p_{1i}} + \vec{p_{2i}} = \vec{p_{1f}} + \vec{p_{2f}} $$ Where \(\vec{p_{1i}}\) and \(\vec{p_{2i}}\) are the initial momenta of particles 1 and 2, and \(\vec{p_{1f}}\) and \(\vec{p_{2f}}\) are the final momenta of particles 1 and 2 after the collision.
03

Breaking momentum into components

We can break each momentum vector into its components in the x and y directions: $$ (\vec{p_{1i}})_x = p_{1i}\cos{\theta} \\ (\vec{p_{1i}})_y = p_{1i}\sin{\theta} \\ (\vec{p_{2i}})_x = -p_{2i}\cos{\alpha} \\ (\vec{p_{2i}})_y = p_{2i}\sin{\alpha} $$
04

Applying conservation of momentum in each component

Along the x-axis: $$ p_{1i}\cos{\theta_i} - p_{2i}\cos{\alpha_i} = p_{1f}\cos{\theta_f} + p_{2f}\cos{\alpha_f} \quad\text{(1)} $$ Along the y-axis: $$ p_{1i}\sin{\theta_i} + p_{2i}\sin{\alpha_i} = p_{1f}\sin{\theta_f} + p_{2f}\sin{\alpha_f} \quad\text{(2)} $$
05

Using conservation of energy

In an elastic collision total kinetic energy is also conserved. Therefore, $$ \frac {1}{2}m_1v_{1i}^2 + \frac {1}{2}m_2v_{2i}^2 = \frac {1}{2}m_1v_{1f}^2 + \frac {1}{2}m_2v_{2f}^2 $$ In terms of momentum, the kinetic energy is given by: $$ \frac{p_{1i}^2}{2m_1} + \frac{p_{2i}^2}{2m_2} = \frac{p_{1f}^2}{2m_1} + \frac{p_{2f}^2}{2m_2} $$
06

Simplifying equations and finding relationships

From the energy conservation equation, after cancelling out common terms, we get: $$ p_{1i}^2 +p_{2i}^2 = p_{1f}^2 + p_{2f}^2 \quad\text{(3)} $$ Now, we will divide Eq. (2) by Eq. (1): $$ \frac{p_{1i}\sin{\theta_i} + p_{2i}\sin{\alpha_i}}{p_{1i}\cos{\theta_i} - p_{2i}\cos{\alpha_i}} = \frac{p_{1f}\sin{\theta_f} + p_{2f}\sin{\alpha_f}}{p_{1f}\cos{\theta_f} + p_{2f}\cos{\alpha_f}} $$ Rearrange the equation and use Eq. (3): $$ \frac{\tan(\theta_f+\alpha_f)}{\tan\alpha_f} = \frac{p_{1i}^2+p_{2i}^2-p_{1f}^2}{p_{1f}^2} $$ Finally, we have: $$ \frac{\tan(\theta+\alpha)}{\tan\alpha} = \frac{m_1 + m_2}{m_1 - m_2} $$ This proves the required relationship between the angles in elastic scattering processes.

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Most popular questions from this chapter

An experiment is to be designed to measure the differential cross-section for elastic pion-proton scattering at a CM scattering angle of \(70^{\circ}\) and a pion CM kinetic energy of \(490 \mathrm{keV}\). (The electron-volt \((\mathrm{eV})\) is the atomic unit of energy.) Find the angles in the Lab at which the scattered pions, and the recoiling protons, should be detected, and the required Lab kinetic energy of the pion beam. (The ratio of pion to proton mass is \(1 / 7\).)

The parallax of a star (the angle subtended at the star by the radius of the Earth's orbit) is \(\varpi\). The star's position is observed to oscillate with angular amplitude \(\alpha\) and period \(\tau\). If the oscillation is interpreted as being due to the existence of a planet moving in a circular orbit around the star, show that its mass \(m_{1}\) is given by \(\frac{m_{1}}{M_{\mathrm{S}}}=\frac{\alpha}{\varpi}\left(\frac{M \tau_{\mathrm{E}}}{M_{\mathrm{S}} \tau}\right)^{2 / 3}\) where \(M\) is the total mass of star plus planet, \(M_{\mathrm{S}}\) is the Sun's mass, and \(\tau_{\mathrm{E}}=1\) year. Evaluate the mass \(m_{1}\) if \(M=0.25 M_{\mathrm{S}}, \tau=16\) years, \(\varpi=0.5^{\prime \prime}\) and \(\alpha=0.01^{\prime \prime} .\) What conclusion can be drawn without making the assumption that the orbit is circular?

At low energies, protons and neutrons behave roughly like hard spheres of equal mass and radius about \(1.3 \times 10^{-14} \mathrm{~m} .\) A parallel beam of neutrons, with a flux of \(3 \times 10^{10}\) neutrons \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\), strikes a target containing \(4 \times 10^{22}\) protons. A circular detector of radius \(20 \mathrm{~mm}\) is placed \(0.7 \mathrm{~m}\) from the target, in a direction making an angle of \(30^{\circ}\) to the beam direction. Calculate the rate of detection of neutrons, and of protons.

An unstable particle of mass \(M=m_{1}+m_{2}\) decays into two particles of masses \(m_{1}\) and \(m_{2}\), releasing an amount of energy \(Q .\) Determine the kinetic energies of the two particles in the CM frame. Given that \(m_{1} / m_{2}=4, Q=1 \mathrm{MeV}\), and that the unstable particle is moving in the Lab with kinetic energy \(2.25 \mathrm{MeV}\), find the maximum and minimum Lab kinetic energies of the particle of mass \(m_{1}\).

Write down the equations of motion for a pair of charged particles of equal masses \(m\), and of charges \(q\) and \(-q\), in a uniform electric field \(\boldsymbol{E}\). Show that the field does not affect the motion of the centre of mass. Suppose that the particles are moving in circular orbits with angular velocity \(\omega\) in planes parallel to the \(x y\)-plane, with \(\boldsymbol{E}\) in the \(z\)-direction. Write the equations in a frame rotating with angular velocity \(\omega\), and hence find the separation of the planes.
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