A particle of mass 1 kg is thrown vertically upwards with speed 100 m/s. After 5 s, it explodes into two parts. One part of mass 300 g comes back with speed 24 m/s, What is the speed of other part just after explosion?
(a) 100 m/s upwards (b) 600 m/s upwards
(c) 600 m/s upwards (d)81.7 m/s upwards

Options

A. 100 m/s upwards
B. 600 m/s upwards
C. 100 m/s downwards
D. 81.7 m/s upwards

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the velocity of the particle after 5 seconds
The particle is thrown upwards with an initial velocity \( u = 100 \, \text{m/s} \). The acceleration due to gravity \( g \) acts downwards, which we take as \( -10 \, \text{m/s}^2 \).

Using the first equation of motion:
\[ v = u + at \]
where:
- \( v \) = final velocity after time \( t \)
- \( u = 100 \, \text{m/s} \)
- \( a = -g = -10 \, \text{m/s}^2 \)
- \( t = 5 \, \text{s} \)

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

### Step 2: Apply the conservation of momentum
Before the explosion, the total momentum of the system is:
\[ \text{Initial Momentum} = \text{mass} \times \text{velocity} = 1 \, \text{kg} \times 50 \, \text{m/s} = 50 \, \text{kg m/s} \]

After the explosion, the particle splits into two parts:
- Mass \( m_1 = 300 \, \text{g} = 0.3 \, \text{kg} \) comes back with a speed of \( v_1 = -24 \, \text{m/s} \) (downward).
- The other mass \( m_2 = 700 \, \text{g} = 0.7 \, \text{kg} \) has an unknown speed \( v_2 \) (upward).

Using the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ 50 = (0.3 \times -24) + (0.7 \times v_2) \]

### Step 3: Solve for \( v_2 \)
Calculating the momentum of the first part:
\[ 0.3 \times -24 = -7.2 \, \text{kg m/s} \]
Substituting this into the momentum equation:
\[ 50 = -7.2 + 0.7 v_2 \]
Adding \( 7.2 \) to both sides:
\[ 50 + 7.2 = 0.7 v_2 \]
\[ 57.2 = 0.7 v_2 \]
Now, divide both sides by \( 0.7 \):
\[ v_2 = \frac{57.2}{0.7} = 81.7142857 \approx 81.7 \, \text{m/s} \]

### Final Answer
The speed of the other part just after the explosion is approximately \( 81.7 \, \text{m/s} \) upwards.