Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m//s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s. When the relative velocity of approach becomes 3m/s, the velocity of the centre of mass is 0.75 m/s.
Text Solution
AI Generated Solution
To solve the problem, we need to analyze the motion of the two particles and the implications of their velocities on the center of mass (CM) of the system.
### Step-by-Step Solution:
1. **Understanding the System**:
We have two particles with masses \( m_1 = 1 \, \text{kg} \) and \( m_2 = 3 \, \text{kg} \). They are moving towards each other due to their mutual gravitational attraction.
2. **Finding the Center of Mass Velocity**:
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Statement-1: Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2 m//s , their centre of mass has a velocity of 0.5 m//s . When the relative velocity of approach becomes 3 m//s the velocity of the centre of mass is 0.75 m//s . Statement-2: The total kinetic energy as seen from ground is 1/2muv_(rel)^(2)+1/2mv_(c)^(2) and in absence of external force, total energy remains conserved.
Two particles of mass m_(1) and m_(2) , approach each other due to their mutual gravitational attraction only. Then
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