Rotational Motion Question 105

Question: Two particles $A$ and $B$ start moving due to their mutual interaction only. If at any time t, $\vec{a}_A$ and $\vec{a}_B$ are their respective accelerations, $\vec{v}_A$ and $\vec{v}_B$ are their respective velocities, and up to that time $W_A$ and $W_B$ are the work done on $A$ and $B$ respectively by the mutual force, $m_A$ and $m_B$ are their masses respectively, then which of the following is always correct?

Options:

A) $ \vec{v}_A+\vec{v}_B=0 $

B) $ m_A \vec{v}_A+m_B \vec{v}_B=0 $

C) $ W_A + W_B = 0 $

D) $ \vec{a}_A+\vec{a}_B=0 $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Since $ \Sigma \vec{F} _{e x t}=\overrightarrow{0} \Rightarrow m_A \vec{a}_A+m_B \vec{a}_B=0 $

$ \therefore $ Momentum of system will remain conserved, equal to zero.

$ m_A \vec{v}_A+m_B \vec{v}_B=0$

But $W_A+W_B \neq 0 $



NCERT Chapter Video Solution

Dual Pane