这是一份ucl伦敦大学学院 math0054作业代写的成功案
The expression for the lab opening angle $\theta\left(=\theta_{1}+\theta_{2}\right)$ is therefore given by
$$
\tan \theta=\tan \left(\theta_{1}+\theta_{2}\right)=\frac{\tan \theta_{1}+\tan \theta_{2}}{1+\tan \theta_{1} \tan \theta_{2}}=\left(\frac{m_{1}+m_{2}}{m_{1}-m_{2}}\right) \cot \left(\frac{1}{2} \psi\right)
$$
after some simplification.
To find the final energies, we observe that
$$
u_{2}=2 V \sin \left(\frac{1}{2} \psi\right)
$$
so that $E_{2}$, the final lab energy of the mass $m_{2}$ is given by
$$
\frac{E_{2}}{E_{0}}=\frac{\frac{1}{2} m_{2}\left(2 V \sin \left(\frac{1}{2} \psi\right)\right)^{2}}{\frac{1}{2} m_{1} u^{2}}=\frac{4 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)^{2}} \sin ^{2}\left(\frac{1}{2} \psi\right)
$$
MATH0054 COURSE NOTES ：
$$
\frac{1}{2} m_{1} V^{2}-\frac{m_{1} m_{2} G}{c} \geq 0 .
$$
Hence, when $P_{2}$ is not fixed, $P_{1}$ will escape if
$$
\frac{1}{2} m_{1} V^{2}-\frac{m_{1} m_{2} G^{\prime}}{c} \geq 0
$$
that is, if
$$
V^{2} \geq \frac{2\left(m_{1}+m_{2}\right) G}{c}
$$