Center Of Mass Momentum And Collision Question 2
Question 2 - 25 January - Shift 2
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$. The ratio of their nuclear sizes will be $(\frac{x}{3})^{\frac{1}{3}}$. The value of ’ $x$ ’ is :
Show Answer
Answer: (2)
Solution:
Formula: Conservation of Momentum
$m_1$
$m_2$
$\frac{v_1}{v_2}=\frac{3}{2}$
$m_1 v_1=m_2 v_2 \Rightarrow \frac{m_1}{m_2}=\frac{2}{3}$
Since, Nuclear mass density is constant
$\frac{m_1}{\frac{4}{3} \pi r_1^{3}}=\frac{m_2}{\frac{4}{3} \pi r_2^{3}}$
$(\frac{r_1}{r_2})^{3}=\frac{m_1}{m_2}$
$\frac{r_1}{r_2}=(\frac{2}{3})^{\frac{1}{3}}$
So, $x=2$
