Question

A bomb of mass $12kg$ at rest explodes into two fragments of masses in the ratio $1:3$. The K.E of the smaller fragment is $216J$. The momentum of heavier is (in $kg.m/s$):

• A. $36$
• B. $72$
• C. $108$
• D. Insufficient data

Answer 1

The correct option is A $36$

$\frac{K_1}{K_2}=\frac{\frac{p_1^2}{2m_1}}{\frac{p_2^2}{2m_2}}$

Since $p_1=p_2$

$\Rightarrow \frac{K_1}{K_2}=\frac{m_2}{m_1}\Rightarrow \frac{216}{K_2}=\frac{3}{1}\Rightarrow K_2=\frac{216}{3}=72J$

Momentum of heavier fragment

$p_2=\sqrt{2m_2{K}_2}$

$m_2=\frac{3}{4}\times 12=9kg$

$\therefore p_2=\sqrt{2\times 9\times 72}=36kgm/s$