A bomb falling freely bursts after 10 sec. (`g=10ms^2`) into two fragments of masses in the ratio of 2:1. The velocity of heavier fragment immediately after the explosion is `200 ms^(-1)` vertically downwards. The velocity of the lighter fragment immediately after the explosion is

Options

A. `50 ms^(-1)` upward
B. `75 ms^(-1)` downward
C. `100 ms^(-1)` upward
D. `400 ms^(-1)` upward

Answer

To solve the problem step by step, we will use the principles of kinematics and the conservation of momentum.

### Step 1: Determine the velocity of the bomb just before the explosion.
The bomb is falling freely under the influence of gravity. We can use the kinematic equation:

\[ V = U + AT \]

Where:
- \( V \) = final velocity
- \( U \) = initial velocity (0 m/s, since it starts from rest)
- \( A \) = acceleration due to gravity (10 m/s²)
- \( T \) = time of fall (10 s)

Substituting the values:

\[ V = 0 + (10 \, \text{m/s}^2)(10 \, \text{s}) = 100 \, \text{m/s} \]

### Step 2: Calculate the initial momentum of the bomb.
The initial momentum \( P_i \) of the bomb before it bursts can be calculated as:

\[ P_i = M \cdot V \]

Where \( M \) is the mass of the bomb and \( V \) is the velocity just before the explosion.

Substituting the values:

\[ P_i = M \cdot 100 \, \text{m/s} \]

### Step 3: Analyze the explosion and the resulting fragments.
After the explosion, the bomb breaks into two fragments with masses in the ratio of 2:1. Let the mass of the heavier fragment be \( \frac{2}{3}M \) and the mass of the lighter fragment be \( \frac{1}{3}M \).

Given:
- Velocity of the heavier fragment \( V_1 = 200 \, \text{m/s} \) (downwards)
- Velocity of the lighter fragment \( V_2 \) (unknown, assumed to be upwards)

### Step 4: Apply the conservation of momentum.
According to the law of conservation of momentum, the total initial momentum must equal the total final momentum:

\[ P_i = P_f \]

Where \( P_f \) is the final momentum of the two fragments:

\[ P_f = \left(\frac{2}{3}M \cdot V_1\right) + \left(\frac{1}{3}M \cdot V_2\right) \]

Substituting the known values:

\[ M \cdot 100 = \left(\frac{2}{3}M \cdot 200\right) + \left(\frac{1}{3}M \cdot V_2\right) \]

### Step 5: Simplify the equation.
Cancelling \( M \) from both sides (assuming \( M \neq 0 \)):

\[ 100 = \frac{2}{3} \cdot 200 + \frac{1}{3} V_2 \]

Calculating \( \frac{2}{3} \cdot 200 \):

\[ 100 = \frac{400}{3} + \frac{1}{3} V_2 \]

### Step 6: Solve for \( V_2 \).
To isolate \( V_2 \), first multiply through by 3 to eliminate the fraction:

\[ 300 = 400 + V_2 \]

Now, rearranging gives:

\[ V_2 = 300 - 400 = -100 \, \text{m/s} \]

### Step 7: Interpret the result.
The negative sign indicates that the lighter fragment is moving upwards at a speed of \( 100 \, \text{m/s} \).

### Final Answer:
The velocity of the lighter fragment immediately after the explosion is \( 100 \, \text{m/s} \) upwards.

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