Question

A body of mass \(300~\text{g}\) kept at rest breaks into two parts due to internal forces. One part of mass \(200~\text{g}\) is found to move at a velocity of \(12~\text{m s}^{-1}\) towards east. The velocity of the other part is:

Answer 1

Let the mass of body be M.
It splits into two parts.
Let the mass of first part be \(m_1=200~\text{g}=0.2 ~\text{kg}\)
mass of second part be \(m_2\) = \(100~\text{g}=0.1~\text{kg}\)
;final velocity of first part be \(v_1\) =\(12~ \text{m s}^{-1}\)
final velocity of other part be \(v_2\).
Since the body is initially at rest, initial momentum is zero.
According to the law of conservation of linear momentum,
Initial momentum = final momentum (when no external force is applied)
In this case, initial momentum =final momentum = \(0\)
\(\Rightarrow 0 =m_1 v_1+m_2v_2 \Rightarrow 0=0.2 \times 12 +0.1 \times v_2\)
\(\Rightarrow\)\(v_2= \dfrac{-0.2\times 12}{0.1} =-24~ \text{m s}^{-1}\)
The negative sign indicates that it is moving in the opposite direction (west) with a velocity of \(24~\text{m s}^{-1}\)