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:

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

Answer 1

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

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 = $0$
$\Rightarrow 0=m_1{v}_1+m_2{v}_2\Rightarrow 0=0.2\times 12+0.1\times v_2$
$\Rightarrow$$v_2=\frac{-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 $24{\text{m s}}^{-1}$