Question

A body of mass $1kg$ initially at rest, explodes and breaks into three fragments of masses in the ratio $1:1:3$. The two pieces of equal mass fly-off perpendicular to each other with a speed of $30m/s$ each. What is the velocity of the heavier fragment?

Answer 1

Given:

$m_1:m_2:m_3=1:1:3$

$m_1+m_2+m_3=1kg$

Solving,

$m_1=m_2=0.2kg,m_3=0.6kg$

Applying conservation of momentum,

$m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+m_3{\overrightarrow{v}}_3=0$

$0.2m\times v\hat{j}+0.2m\times v\hat{i}+0.6m\overrightarrow{v^{\prime }}=0$

$\overrightarrow{v^{\prime }}=-v/3(\hat{i}+\hat{j})$

$v^{\prime }=v\sqrt{2}/3=30\times \sqrt{2}/3\approx 14.14m/s$

Hence, the velocity of heavier fragment is $14.14m/s$ at an angle of ${135}^o$ to the direction of other two fragments and in the plane of directions of other two fragments.