Question

A proton of mass mp collides with a heavy particle

Answer 1

Answer:

A proton of mass mp collides with heavy particles After collision proton bounces back with 4/9 of its initial kinetic energy .collision is perfectly elastic . ... Here in this given link fraction of kinetic energy lost is found and you need to find here the mass of the heavy particle . Hope this helps you .

Answer 2

Given that,

A proton of mass mp collides with a heavy particle at rest. After collision proton bounces back with $\dfrac{9}{4}$ of its initial kinetic energy, Collision is perfectly elastic.Find mass of heavy particle.

Let the mass of heavy particle and velocity after collision is M and V.

Velocity of proton before and after collision is u.

So, from question

We need to calculate the velocity after collision

The  kinetic energy after and before collision is

$\dfrac{1}{2}m_{p}v^2=\dfrac{4}{9}\times\dfrac{1}{2}m_{p}u^2$

$v^2=\dfrac{4}{9}u^2$

$v=\dfrac{2}{3}u$.....(I)

Collision is perfectly elastic so the kinetic energy is conserved

So, $\dfrac{1}{2}MV^2+\dfrac{1}{2}m_{p}v^2=\dfrac{1}{2}m_{p}u^2$

Put the value of v

$\dfrac{1}{2}MV^2+\dfrac{1}{2}m_{p}\times\dfrac{4}{9}u^2=\dfrac{1}{2}m_{p}u^2$

$\dfrac{1}{2}MV^2=\dfrac{1}{2}m_{p}u^2(1-\dfrac{4}{9})$

$MV^2=m_{p}u^2\times\dfrac{5}{9}$

$\dfrac{V}{u}=\sqrt{\dfrac{m_{p}5}{M9}}$......(II)

We need to calculate the mass of the heavy particle

Using formula of momentum

$m_{p}u+M\u=m_{p}v+MV$

Put the value into the formula

$m_{p}u+0=m_{p}\times(-\dfrac{2}{3}u)+MV$

$m_{p}u+m_{p}\dfrac{2u}{3}=MV$

$m_{p}u(1+\dfrac{2}{3})=MV$

$m_{p}\dfrac{5}{3}=M\times\dfrac{V}{u}$

$m_{p}\times{5}{3}=M\times\sqrt{\dfrac{m_{p}5}{M9}}$

$M=5 m_{p}$

Put the value of proton mass

$M=5\times1.673\times10^{-27}$

$M=8.365\times10^{-27}\ kg$

Hence, The mass of the heavy particle is $8.365\times10^{-27}\ kg$