To solve the problem, we will use the principles of kinematics and the conservation of momentum. Here’s a step-by-step breakdown of the solution:
### Step 1: Calculate the velocity of the particle just before the explosion
The particle is thrown upwards with an initial speed \( u = 100 \, \text{m/s} \). After \( t = 5 \, \text{s} \), the velocity \( v \) of the particle can be calculated using the equation of motion:
\[
v = u - gt
\]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
Substituting the values:
\[
v = 100 \, \text{m/s} - (10 \, \text{m/s}^2 \times 5 \, \text{s}) = 100 \, \text{m/s} - 50 \, \text{m/s} = 50 \, \text{m/s}
\]
### Step 2: Determine the masses of the two parts after the explosion
The total mass of the particle is \( 1 \, \text{kg} \). One part has a mass of \( 400 \, \text{g} = 0.4 \, \text{kg} \). Therefore, the mass of the second part is:
\[
m_2 = 1 \, \text{kg} - 0.4 \, \text{kg} = 0.6 \, \text{kg}
\]
### Step 3: Analyze the motion after the explosion
After the explosion, one part (mass \( m_1 = 0.4 \, \text{kg} \)) comes back down with a speed of \( v_1 = -25 \, \text{m/s} \) (negative because it is moving downwards). We need to find the speed of the second part (mass \( m_2 = 0.6 \, \text{kg} \)), which we will denote as \( v_2 \).
### Step 4: Apply the conservation of momentum
According to the law of conservation of momentum:
\[
\text{Initial momentum} = \text{Final momentum}
\]
The initial momentum before the explosion is:
\[
p_{\text{initial}} = (1 \, \text{kg}) \times (50 \, \text{m/s}) = 50 \, \text{kg m/s}
\]
The final momentum after the explosion is:
\[
p_{\text{final}} = m_1 v_1 + m_2 v_2
\]
Substituting the known values:
\[
50 = (0.4)(-25) + (0.6)v_2
\]
Calculating \( (0.4)(-25) \):
\[
50 = -10 + 0.6 v_2
\]
Rearranging the equation to solve for \( v_2 \):
\[
0.6 v_2 = 50 + 10 = 60
\]
\[
v_2 = \frac{60}{0.6} = 100 \, \text{m/s}
\]
### Step 5: Determine the direction of \( v_2 \)
Since the second part is moving upwards, we conclude that:
\[
v_2 = 100 \, \text{m/s} \text{ (upward)}
\]
### Final Answer
The speed of the other part just after the explosion is \( 100 \, \text{m/s} \) upward.
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