Two particles A and B initially at rest, move towards each other under their mutual force of attraction. at the instant when the speed of A is v and the speed of B is 2v, the spped of the centre of mass of the system is?

Question

Answer

Initially position both A and B are at rest, hence momentum is zero. It is given that no external force is acting on them .So, momentum remains zero. Since mass cannot be zero, therefore velocity is zero.

Answer

Analysis of the Scenario:

In this scenario, two particles A and B are initially at rest and move towards each other under their mutual force of attraction. At a certain instant, the speed of particle A is v and the speed of particle B is 2v.

Calculating the Speed of the Center of Mass:

To find the speed of the center of mass of the system, we can use the formula:
$$ V_{\text{CM}} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} $$

where:
- $ V_{\text{CM}} $ is the speed of the center of mass,
- $ m_1 $ and $ m_2 $ are the masses of particles A and B respectively,
- $ v_1 $ and $ v_2 $ are the speeds of particles A and B respectively.

In this case, both particles A and B have the same mass, so we can simplify the formula to:
$$ V_{\text{CM}} = \frac{m(v + 2v)}{2m} = \frac{3v}{2} $$

Therefore, the speed of the center of mass of the system at the instant when the speed of particle A is v and the speed of particle B is 2v is $ \frac{3v}{2} $.

Conclusion:

The speed of the center of mass of the system is $ \frac{3v}{2} $ at the given instant when the particles are moving towards each other.

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