A bomb of `1kg` is thrown vertically up with speed `100m//s` After 5 seconds it explodes into two parts. One part of mass `400gm` goes down with speed `25m//s` What will happen to the other part just after explosion .

Options

A. `100 ms^(-1)` upward
B. `600 ms^(-1)` upward
C. `100 ms^(-1)` downward
D. `300 ms^(-1)` upward

Explanation

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

### Step 1: Determine the initial velocity of the bomb just before the explosion.
The bomb is thrown vertically upwards with an initial speed of \(100 \, \text{m/s}\). After \(5 \, \text{s}\), we need to calculate its velocity just before the explosion using the equation of motion:

\[ v = u + at \]

Where:
- \(u = 100 \, \text{m/s}\) (initial velocity)
- \(a = -g = -9.8 \, \text{m/s}^2\) (acceleration due to gravity, acting downwards)
- \(t = 5 \, \text{s}\)

Substituting the values:

\[ v = 100 - 9.8 \times 5 = 100 - 49 = 51 \, \text{m/s} \]

### Step 2: Apply the conservation of momentum principle.
Before the explosion, the total momentum of the system is given by:

\[ \text{Total momentum before explosion} = m \cdot v \]

Where:
- \(m = 1 \, \text{kg}\) (mass of the bomb)
- \(v = 51 \, \text{m/s}\) (velocity just before the explosion)

Thus, the total momentum before the explosion is:

\[ \text{Total momentum before} = 1 \cdot 51 = 51 \, \text{kg m/s} \]

### Step 3: Calculate the momentum after the explosion.
After the explosion, the bomb splits into two parts:
- Part 1: Mass \(m_1 = 400 \, \text{g} = 0.4 \, \text{kg}\) going down with speed \(v_1 = -25 \, \text{m/s}\) (downward direction is negative).
- Part 2: Mass \(m_2 = 1 - 0.4 = 0.6 \, \text{kg}\) (the remaining mass).

Let \(v_2\) be the velocity of the second part (mass \(m_2\)) after the explosion.

Using conservation of momentum:

\[ \text{Total momentum after explosion} = m_1 \cdot v_1 + m_2 \cdot v_2 \]

Setting the total momentum before and after equal:

\[ 51 = (0.4)(-25) + (0.6)v_2 \]

### Step 4: Solve for \(v_2\).
Calculating the left side:

\[ 51 = -10 + 0.6v_2 \]

Rearranging gives:

\[ 51 + 10 = 0.6v_2 \]
\[ 61 = 0.6v_2 \]
\[ v_2 = \frac{61}{0.6} = 101.67 \, \text{m/s} \]

### Step 5: Determine the direction of the second part.
Since \(v_2\) is positive, it indicates that the second part moves upwards.

### Final Answer:
The second part of the bomb (mass \(600 \, \text{g}\)) moves upwards with a speed of approximately \(101.67 \, \text{m/s}\).

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