Two particles A and B of masses 1 kg and 2 kg respectively are kept 1 m apart and are released to move under mutual attraction. Find The speed A when that of B is 3.6 cm/hr. What is the separation between the particles at this instant?

Explanation

To solve the problem step by step, we will follow the principles of conservation of momentum and energy.

### Step 1: Convert the speed of particle B to standard units
The speed of particle B is given as 3.6 cm/hr. We need to convert this to meters per second (m/s).

\[ \text{Speed of B} = 3.6 \, \text{cm/hr} = \frac{3.6}{100} \, \text{m/hr} = \frac{3.6}{100 \times 3600} \, \text{m/s} = 10^{-5} \, \text{m/s} \]
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