Center of Mass

Step 1: Find the final velocities after the collisionThe first particle has mass $8\,\mathrm{\mu g}$ and moves with speed $u_1$. The second particle has mass $4\,\mathrm{\mu g}$ and is at rest, so $u_2=0$.The formula for the final speed of the first particle after an elastic, one-dimensional collision is: $v_1=\frac{(m_1-m_2) u_1+2 m_2 u_2}{m_1+m_2}$ Plug in the numbers:$v_1=\frac{(8-4) u_1+2 \times 4 \times 0}{8+4}=\frac{4 u_1}{12}=\frac{u_1}{3}$The formula for the final speed of the second particle is: $v_2=\frac{2 m_1 u_1+(m_2-m_1) u_2}{m_1+m_2}$ Plug in the numbers:$v_2=\frac{2 \times 8 u_1+(4-8) \times 0}{8+4}=\frac{16 u_1}{12}=\frac{4}{3} u_1$Step 2: Find the de-Broglie wavelength for each particleThe de-Broglie wavelength formula is: $\lambda=\frac{h}{m v}$ where $h$ is Planck's constant, $m$ is mass, and $v$ is velocity.To compare their wavelengths, find the ratio: $\frac{\lambda_2}{\lambda_1}=\frac{m_1 v_1}{m_2 v_2}$Plug in the values found earlier: $\frac{\lambda_2}{\lambda_1}=\frac{8 \times \frac{u_1}{3}}{4 \times \frac{4}{3} u_1}$Simplify the expression: $=\frac{8 \times \frac{u_1}{3}}{4 \times \frac{4 u_1}{3}}=\frac{\frac{8 u_1}{3}}{\frac{16 u_1}{3}}=\frac{8 u_1}{16 u_1}=\frac{1}{2}$ So, $\frac{\lambda_1}{\lambda_2}=2$ which means the ratio of their wavelengths after the collision is $2: 1$