91Ó°ÊÓ

Mass of part A is$m_A$

Mass of part B is$m_B$

Velocity of part A is$v_A$

Velocity of part B is$v_B$

The phenomenon of breaking up of a large nucleus into multiple small nuclei, releasing a large amount of energy in the process, is known as Nuclear Fission.

Answer is not given in the Drive.

Let the mass of the large nucleus before fission bem and it is stationary.

So the momentum of the nucleus before the burst will be-

$$\begin{gathered} p=m×0 \\ =0 \end{gathered}$$

After the burst, it broke into two parts- A and B. Part A having mass$m_A$moves to the left with velocity$V_A$and Part B having mass$m_B$moves to the right with velocity $v_B$.

So the momentum of the nucleus after the burst will be-

$p^{\text{'}}=m_A{v}_A+\left(-m_B{v}_B\right)$

By the law of conversation of momentum-

$$\begin{gathered} P=p^{\text{'}} \\ 0=m_A{v}_A-m_B{v}_B \\ {m}_A{v}_A=m_B{v}_B \\ {v}_B=\frac{m_A{v}_A}{m_B} \end{gathered}$$

Hence, the value of is $\frac{m_A{v}_A}{m_B}$.

The kinetic energy of part A is:

$K_A=\frac{1}{2}{m}_A{v}_A^2$

The kinetic energy of part B is:

role="math" localid="1665128046268" $K_B=\frac{1}{2}{m}_B{v}_B^2$

Dividing kinetic energy for part A by that of part B, we get:

$$\begin{gathered} \frac{K_A}{K_B}=\frac{{\displaystyle \frac{1}{2}}{m}_A{v}_A^2}{{\displaystyle \frac{1}{2}}{m}_B{v}_B^2} \\ =\frac{m_A{v}_A^2}{m_B{\left({\displaystyle \frac{m_A{v}_A}{m_B}}\right)}^2} \\ =\frac{m_A}{m_B}×\frac{m_B^2}{m_A^2} \\ =\frac{m_B}{m_A} \end{gathered}$$

Hence, it is proved.