To solve the problem, we need to find the velocity of the center of mass (COM) of the two particles after 2 seconds. Let's break down the solution step by step.
### Step 1: Identify the initial conditions
- Particle 1 (mass \( m_1 = 1 \, \text{kg} \)) is projected upwards with an initial velocity \( u_1 = 60 \, \text{m/s} \).
- Particle 2 (mass \( m_2 = 2 \, \text{kg} \)) is dropped from a certain height, so its initial velocity \( u_2 = 0 \, \text{m/s} \).
### Step 2: Calculate the velocity of Particle 1 after 2 seconds
Using the equation of motion for Particle 1:
\[
v_1 = u_1 - g t
\]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity) and \( t = 2 \, \text{s} \):
\[
v_1 = 60 \, \text{m/s} - (10 \, \text{m/s}^2 \times 2 \, \text{s}) = 60 \, \text{m/s} - 20 \, \text{m/s} = 40 \, \text{m/s}
\]
Thus, the velocity of Particle 1 after 2 seconds is \( v_1 = 40 \, \text{m/s} \) upwards.
### Step 3: Calculate the velocity of Particle 2 after 2 seconds
Using the equation of motion for Particle 2:
\[
v_2 = u_2 + g t
\]
Substituting the values:
\[
v_2 = 0 \, \text{m/s} + (10 \, \text{m/s}^2 \times 2 \, \text{s}) = 20 \, \text{m/s}
\]
Thus, the velocity of Particle 2 after 2 seconds is \( v_2 = 20 \, \text{m/s} \) downwards.
### Step 4: Calculate the velocity of the center of mass (COM)
The formula for the velocity of the center of mass \( V_{cm} \) of a two-particle system is given by:
\[
V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}
\]
Substituting the known values:
\[
V_{cm} = \frac{(1 \, \text{kg} \times 40 \, \text{m/s}) + (2 \, \text{kg} \times (-20 \, \text{m/s})}{1 \, \text{kg} + 2 \, \text{kg}}
\]
Note that we take \( v_2 \) as negative since it is downward:
\[
V_{cm} = \frac{40 \, \text{kg m/s} - 40 \, \text{kg m/s}}{3 \, \text{kg}} = \frac{0}{3} = 0 \, \text{m/s}
\]
### Final Answer
The velocity of the center of mass after 2 seconds is \( 0 \, \text{m/s} \).
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