A mass of 1 kg is thrown up with a velocity of 100 m/s. After 5 seconds. It explodes into two parts. One parts of mass 400 g comes down with a velocity 25 m/s Calaculate the velocity of other parts:

Options

A. 40 m/s upward
B. 40 m/s downward
C. 100 m/s upward
D. 60 m/s downward

Explanation

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

### Step 1: Determine the velocity of the mass just before the explosion
The mass is thrown upwards with an initial velocity (u) of 100 m/s. After 5 seconds, we need to find the velocity (v) just before the explosion. We can use the equation of motion:
\[ v = u + at \]
where:
- \( u = 100 \, \text{m/s} \) (initial velocity)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it acts downwards)
- \( t = 5 \, \text{s} \)

Substituting the values:
\[ v = 100 - 10 \times 5 \]
\[ v = 100 - 50 \]
\[ v = 50 \, \text{m/s} \]

### Step 2: Set up the conservation of momentum equation
Before the explosion, the total momentum (P_initial) of the system is given by:
\[ P_{\text{initial}} = m \cdot v \]
where:
- \( m = 1 \, \text{kg} \)
- \( v = 50 \, \text{m/s} \)

Calculating the initial momentum:
\[ P_{\text{initial}} = 1 \times 50 = 50 \, \text{kg m/s} \]

After the explosion, the mass splits into two parts:
- Part 1: mass \( m_1 = 400 \, \text{g} = 0.4 \, \text{kg} \) moving down with velocity \( v_1 = -25 \, \text{m/s} \) (downward direction is negative)
- Part 2: mass \( m_2 = 600 \, \text{g} = 0.6 \, \text{kg} \) with an unknown velocity \( v_2 \)

The conservation of momentum states:
\[ P_{\text{initial}} = P_{\text{final}} \]
\[ 50 = (0.4 \times -25) + (0.6 \times v_2) \]

### Step 3: Solve for the unknown velocity \( v_2 \)
Substituting the values into the equation:
\[ 50 = -10 + 0.6 v_2 \]

Rearranging the equation to solve for \( v_2 \):
\[ 50 + 10 = 0.6 v_2 \]
\[ 60 = 0.6 v_2 \]
\[ v_2 = \frac{60}{0.6} \]
\[ v_2 = 100 \, \text{m/s} \]

### Conclusion
The velocity of the other part after the explosion is \( 100 \, \text{m/s} \) in the upward direction.

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