Question

Paragraph for question nos. (16-17) Meson is composed of two sub-atomic particles and the interaction between the sub-atomic particles is complicated. Research of meson can be done by studying the collisions between the meson and high energy electron. As the collision is quite complicated, scientists invented a simplified model called "parton model" to grasp the main content during collision. In the model, the electron first collides with pan of meson (e.g. one of the sub-atomic particles) elastically. Then the energy and momentum are transferred to the other sub-atomic particles and thus the whole meson during subsequent interaction. This simplified model is described by the following: [figure] An electron of mass $m$ and energy $E_0$ collides with a neutral sub-atomic particles of mass $m_1$ in a meson. The other neutral sub-atomic particles in the meson has mass $m_2$. The sub-atomic particles are connected by a massless spring at natural length $L$ which is at equilibrium before collision. All movements are on a straight line and neglect the effect of relativity. Find, after collision. (given $m_1=5m$ & $m_2=2m_1$)

16. The energy gain by the sub-atomic particle of mass $m_1$ due to collision is $\frac{E_0}{x}$. Find $x$

17. The minimum kinetic energy of the meson as a whole system is $\frac{E_0}{y}$. Find the value of $y$.

• A. x = 9/5
• B. x = 5/9
• C. y = 27/5
• D. y = 5/27

Answer 1

16. $x = \frac{9}{5}$, 17. $y = \frac{27}{5}$

Answer 2

Question 16: The collision is elastic between the electron (mass $m$, initial velocity $v_e$) and particle $m_1$ (mass $m_1$, initially at rest). Initial energy of electron: $E_0 = \frac{1}{2} m v_e^2$. Final velocity of $m_1$ after elastic collision: $v_1' = \frac{2m}{m + m_1} v_e$. Given $m_1 = 5m$, so $v_1' = \frac{2m}{m + 5m} v_e = \frac{2m}{6m} v_e = \frac{1}{3} v_e$. Energy gain by $m_1$: $\Delta KE_1 = \frac{1}{2} m_1 (v_1')^2 = \frac{1}{2} (5m) \left(\frac{1}{3} v_e\right)^2 = \frac{5}{9} \left(\frac{1}{2} m v_e^2\right) = \frac{5}{9} E_0$. Given $\Delta KE_1 = \frac{E_0}{x}$, so $\frac{5}{9} E_0 = \frac{E_0}{x} \implies x = \frac{9}{5}$.

Question 17: The kinetic energy of the meson's center of mass ($KE_{CM}$). $KE_{CM} = \frac{1}{2} (m_1 + m_2) V_{CM}^2$. Momentum of meson system after collision: $P_{meson} = m_1 v_1'$. $V_{CM} = \frac{P_{meson}}{m_1 + m_2} = \frac{m_1 v_1'}{m_1 + m_2}$. $KE_{CM} = \frac{(m_1 v_1')^2}{2(m_1 + m_2)}$. Given $m_1 = 5m$ and $m_2 = 2m_1 = 10m$, so $m_1 + m_2 = 15m$. $v_1' = \frac{1}{3} v_e$, and $v_e^2 = \frac{2E_0}{m}$. $KE_{CM} = \frac{(5m)^2 \left(\frac{1}{3} v_e\right)^2}{2(15m)} = \frac{25m^2 \cdot \frac{1}{9} \frac{2E_0}{m}}{30m} = \frac{25m \cdot \frac{2E_0}{9}}{30m} = \frac{50}{270} E_0 = \frac{5}{27} E_0$. Since $V_{CM}$ is constant, $KE_{CM}$ is the minimum kinetic energy. Given $KE_{CM} = \frac{E_0}{y}$, so $\frac{5}{27} E_0 = \frac{E_0}{y} \implies y = \frac{27}{5}$.