To solve the problem, we will use the principle of conservation of momentum. The total momentum before and after the interaction must remain constant since there are no external forces acting on the system.
### Step 1: Identify the initial conditions
- Both spacemen A and B are initially at rest, so their initial velocities are:
- \( V_{A_i} = 0 \, \text{m/s} \)
- \( V_{B_i} = 0 \, \text{m/s} \)
- Mass of A, \( m_A = 120 \, \text{kg} \)
- Mass of B, \( m_B = 90 \, \text{kg} \) (which includes a 5 kg radio box)
- Mass of the radio box, \( m_{box} = 5 \, \text{kg} \)
- Mass of B without the box, \( m_{B_{effective}} = m_B - m_{box} = 90 \, \text{kg} - 5 \, \text{kg} = 85 \, \text{kg} \)
### Step 2: Calculate the momentum after the box is thrown
- When B throws the box towards A with a velocity of \( V_{box} = 2 \, \text{m/s} \), the momentum of the box is:
\[
p_{box} = m_{box} \cdot V_{box} = 5 \, \text{kg} \cdot 2 \, \text{m/s} = 10 \, \text{kg m/s}
\]
- By conservation of momentum, the momentum of B must change in the opposite direction. Let \( V_{B_f} \) be the final velocity of B after throwing the box. The momentum of B is:
\[
p_{B} = -m_{B_{effective}} \cdot V_{B_f}
\]
### Step 3: Set up the conservation of momentum equation
- The total initial momentum is zero, and the total final momentum must also be zero:
\[
0 = p_{box} + p_{B}
\]
\[
0 = 10 \, \text{kg m/s} - 85 \, \text{kg} \cdot V_{B_f}
\]
### Step 4: Solve for \( V_{B_f} \)
\[
85 \, \text{kg} \cdot V_{B_f} = 10 \, \text{kg m/s}
\]
\[
V_{B_f} = \frac{10}{85} \approx 0.1176 \, \text{m/s} \approx 11.76 \, \text{cm/s}
\]
### Step 5: Calculate the change in velocity of B
- The change in velocity of B is:
\[
\Delta V_B = V_{B_f} - V_{B_i} = 0.1176 \, \text{m/s} - 0 = 0.1176 \, \text{m/s}
\]
### Step 6: Calculate the change in velocity of A after catching the box
- After A catches the box, the total mass of A and the box is:
\[
m_{total} = m_A + m_{box} = 120 \, \text{kg} + 5 \, \text{kg} = 125 \, \text{kg}
\]
- Let \( V_{A_f} \) be the final velocity of A after catching the box. By conservation of momentum:
\[
0 = p_{box} + p_{A}
\]
\[
0 = 10 \, \text{kg m/s} - 125 \, \text{kg} \cdot V_{A_f}
\]
\[
125 \, \text{kg} \cdot V_{A_f} = 10 \, \text{kg m/s}
\]
\[
V_{A_f} = \frac{10}{125} = 0.08 \, \text{m/s} = 8 \, \text{cm/s}
\]
### Step 7: Calculate the change in velocity of A
- The change in velocity of A is:
\[
\Delta V_A = V_{A_f} - V_{A_i} = 0.08 \, \text{m/s} - 0 = 0.08 \, \text{m/s}
\]
### Final Results
- Change in velocity of B: \( \Delta V_B \approx 11.76 \, \text{cm/s} \)
- Change in velocity of A: \( \Delta V_A = 8 \, \text{cm/s} \)
