Question

A body of mass $300\text{g}$ kept at rest breaks into two parts due to internal force,One part of mass $200\text{g}$ is found to move at a speed of $12\text{m/s}$ towards the east. What will be the velocity of the other part ?

• A. $12\text{m/s, Westwards}$
• B. $12\text{m/s, Eastwards}$
• C. $24\text{m/s, Westwards}$
• D. $23\text{m/s, Eastwards}$

Answer 1

The correct option is C $24\text{m/s, Westwards}$


Before it broke, the body was at rest.The linear momentum of the body was thus $p=mv=0$.
The body breaks due to internal forces. As the external force acting on it is zero,its linear momentum will remain constant, i.e.,zero.

The linear momentum of the first part is
$P_1=m_1{v}_1=\left(200g\right)\times \left(12\text{m/s}\right)$, towards the east.
For the total momentum to remain zero, the linear momentum of the other part must have the same magnitude and should be opposite in direction. It therefore moves towards the west. If its speed is $v_2$, its linear momentum is
$P_2=m_2{v}_2=\left(100g\right)\times v_2$
Thus,
$\left(200g\right)\times \left(12\text{m/s}\right)=\left(100g\right)\times v_2$
$v_2=24\text{m/s}$
The velocity of the other part is $24\text{m/s}$ towards the west.