Question

A particle makes a head-on-collision with an unknown particle at rest.The proton rebounds back with 4/9 of its initial kinetic energy.Find the ratio of mass of unknown particle to mass of proton assuming the collision to be elastic.

Answer 1

$$\frac{m_{1}V_{1}^2}{2} =\frac{m_{1}V_{1}'^2}{2} +\frac{m_{2}V_{2}^2}{2}$$

Where m1 - mass proton, m2- mass unknown particle, V1- initial speed of proton, V1' - speed of proton after collision, V2 - speed of unknown particle after collision.

$$\frac{m_{1}V_{1}'^2}{2}= 4/9 \frac{m_{1}V_{1}^2}{2}$$

$$V_{1}' = 2/3 V_{1}$$

$$m_{1}V_{1}=m_{1}V_{1}'+ m_{2}V_{2}$$

$$m_{1}V_{1}= 2/3 m_{1}V_{1}+ m_{2}V_{2}$$

$$1/3m_{1}V_{1}=m_{2}V_{2}$$

$$V_{2}= 1/3 \frac{m_{1}V_{1}}{m_{2}}$$

$$\frac{m_{1}V_{1}^2}{2} =4/9 \frac{m_{1}V_{1}^2}{2} +\frac{m_{2}V_{2}^2}{2}$$

$$5/9\frac{m_{1}V_{1}^2}{2} =\frac{m_{2}(1/3 \frac{m_{1}V_{1}}{m_{2}})^2}{2}$$

$$5/9=1/9 \frac{m_{1}}{m_{2}}$$ $$\frac{m_{1}}{m_{2}}= 5$$

Or

$$\frac{m_{2}}{m_{1}}=1/5$$