A body of mass `2g`, moving along the positive x-axis in gravity free space with velocity `20cms^-1` explodes at `x=1m`, `t=0` into two pieces of masses `2//3g` and `4/3g`. After `5s`, the lighter piece is at the point `(3m, 2m, -4m)`. Then the position of the heavier piece at this moment, in metres is

Options

A. (a) `(1.5,-1,-2)`
B. (b) `(1.5,-2,-2)`
C. (c) `(1.5,-1,-1)`
D. (d) None of these

Explanation

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

### Step 1: Calculate the initial position of the center of mass (CM)
The initial mass of the body is \(2g\) (or \(0.002 kg\)) moving along the positive x-axis with a velocity of \(20 cm/s\) (or \(0.2 m/s\)). The initial position of the body is at \(x = 1m\) when it explodes.

The position of the center of mass before the explosion can be calculated as:
\[ \text{CM}_{initial} = \text{Position} + \text{Velocity} \times \text{Time} \]
Since the explosion occurs at \(t = 0\), we can consider the time until the explosion as \(0\):
\[ \text{CM}_{initial} = 1m + 0 = 1m \]

### Step 2: Calculate the position of the center of mass after the explosion
After the explosion, the center of mass will remain at the same position since no external forces are acting on the system. The center of mass will still be at:
\[ \text{CM}_{final} = 1m + (0.2 m/s \times 5s) = 1m + 1m = 2m \]

### Step 3: Identify the masses and positions of the pieces after the explosion
The body explodes into two pieces:
- Lighter piece: mass \(m_1 = \frac{2}{3}g\) (or \(0.000667 kg\))
- Heavier piece: mass \(m_2 = \frac{4}{3}g\) (or \(0.001333 kg\))

The position of the lighter piece after 5 seconds is given as:
\[ \text{Position of } m_1 = (3m, 2m, -4m) \]

### Step 4: Use the center of mass formula to find the position of the heavier piece
The center of mass formula is given by:
\[ \text{CM} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2} \]
Where:
- \(r_1\) is the position vector of the lighter piece
- \(r_2\) is the position vector of the heavier piece

Substituting the known values:
\[ 2m = \frac{\left(\frac{2}{3}g \cdot (3, 2, -4)\right) + \left(\frac{4}{3}g \cdot r_2\right)}{2g} \]

### Step 5: Simplifying the equation
Multiply both sides by \(2g\):
\[ 4g = \left(\frac{2}{3}g \cdot (3, 2, -4)\right) + \left(\frac{4}{3}g \cdot r_2\right) \]

Cancelling \(g\) from both sides:
\[ 4 = \frac{2}{3} \cdot (3, 2, -4) + \frac{4}{3} \cdot r_2 \]

Calculating \(\frac{2}{3} \cdot (3, 2, -4)\):
\[ = (2, \frac{4}{3}, -\frac{8}{3}) \]

### Step 6: Rearranging the equation
Now we have:
\[ 4 = (2, \frac{4}{3}, -\frac{8}{3}) + \frac{4}{3} \cdot r_2 \]
Rearranging gives:
\[ 4 - (2, \frac{4}{3}, -\frac{8}{3}) = \frac{4}{3} \cdot r_2 \]
\[ (2, \frac{8}{3}, \frac{8}{3}) = \frac{4}{3} \cdot r_2 \]

### Step 7: Solving for \(r_2\)
Multiply through by \(\frac{3}{4}\):
\[ r_2 = \left(\frac{3}{4} \cdot 2, \frac{3}{4} \cdot \frac{8}{3}, \frac{3}{4} \cdot \frac{8}{3}\right) \]
\[ r_2 = (1.5, 2, 2) \]

### Final Answer
The position of the heavier piece after 5 seconds is:
\[ (1.5 m, 2 m, 2 m) \]