Question

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

• A. $24m{s}^{-1}$ towards west
• B. $14m{s}^{-1}$ towards east
• C. $24m{s}^{-1}$ towards east
• D. $54m{s}^{-1}$ towards south

Answer 1

The correct option is A $24m{s}^{-1}$ towards west

Let the mass of body be M.
It splits into two parts.
Let mass of first part be $m_1=200g=0.2kg$ and mass of second part be $m_2$ = $100g=0.1kg$.
Also, let final velocity of first part be $v_1$ =$12m{s}^{-1}$ and 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 = $0$
$\Rightarrow 0=m_1{v}_1+m_2{v}_2\Rightarrow 0=0.2\times 12+0.1\times v_2$
$\Rightarrow$$v_2=\frac{-0.2\times 12}{0.1}=-24m{s}^{-1}$
The negative sign indicates that it is moving in the opposite direction (west) with a velocity of $24m{s}^{-1}$