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A body is moving with a velocity of $10\,m{s^{ - 1}}$ burst into part $A$ and $B$ of masses $6\,kg$ and $1\,kg$.If part A continues to move in the original direction with $12.5\,m{s^{ - 1}}$ ,the velocity of part $B$ is-
A. $ + 5\,m{s^{ - 1}}$
B. $ - 5\,m{s^{ - 1}}$
C. $ + 2.5\,m{s^{ - 1}}$
D. $ - 2.5\,m{s^{ - 1}}$

Last updated date: 17th Jul 2024
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Hint:The relationship between speed, mass, and direction is defined by momentum in physics. It also applies to the force that is used to stop and keep moving objects. If an object has sufficient energy, it may exert considerable force.

Complete step by step answer:
The linear momentum of a particle is defined as the product of its mass and velocity. It's a vector number, and its direction is the same as the particle's velocity. The symbol for linear momentum is \[p\]. If \[m\] is the mass of a particle moving at $v$ , then \[p = mv\] is the linear momentum of the particle.

So, according to the question,
Initial velocity of the object was $u=10\,m{s^{ - 1}}$.
Mass of $A$ was ${m_A}=6\,kg$
Mass of $B$ was ${m_B}=1\,kg$
Final velocity of $A$ was ${v_A}=12.5\,m{s^{ - 1}}$
Now, we know that
$({m_A}+{m_B})u = {m_A}\,{v_A} + {m_B}\,{v_B}$
$ \Rightarrow \left( {6 + 1} \right)10 = 6 \times 12.5 + 1 \times {v_B}$
$ \Rightarrow 70 = 75 + {v_B}$
$\therefore {v_B} = - 5\,m/s$
So, the velocity of part $B$ is $ - 5\,m/s$ as the direction is opposite to the original.

Hence the correct option is B.

Note: The product of the total mass \[M\] of the system and the velocity of the centre of mass gives the linear momentum of a system of particles. This expression shows that the linear momentum of a system of particles is conserved when the net external force acting on it is zero.