To solve the problem, we will use the principle of conservation of momentum. We will analyze the situation step by step:
### Step 1: Initial Momentum Calculation
The initial momentum of the system (cart + boys) can be calculated before any boy jumps off.
- Mass of the cart, \( m_c = 100 \, \text{kg} \)
- Mass of each boy, \( m_b = 20 \, \text{kg} \)
- Total mass of the boys, \( m_{total\_boys} = 5 \times m_b = 100 \, \text{kg} \)
- Total mass of the system, \( m_{total} = m_c + m_{total\_boys} = 100 \, \text{kg} + 100 \, \text{kg} = 200 \, \text{kg} \)
- Initial speed of the cart, \( v_{initial} = 10 \, \text{m/s} \)
The initial momentum \( p_{initial} \) is given by:
\[
p_{initial} = m_{total} \cdot v_{initial} = 200 \, \text{kg} \cdot 10 \, \text{m/s} = 2000 \, \text{kg m/s}
\]
### Step 2: First Boy Jumps Off
When the first boy jumps off in the opposite direction (backward) with a speed of \( 2 \, \text{m/s} \) relative to the cart, we need to calculate the new velocities.
- Speed of the first boy relative to the ground:
\[
v_{boy1} = v_{cart} - 2 = 10 \, \text{m/s} - 2 \, \text{m/s} = 8 \, \text{m/s} \, \text{(backward)}
\]
- Momentum of the first boy:
\[
p_{boy1} = m_b \cdot v_{boy1} = 20 \, \text{kg} \cdot (-8 \, \text{m/s}) = -160 \, \text{kg m/s}
\]
- Momentum of the cart after the first boy jumps off:
Let \( v_{cart1} \) be the new speed of the cart:
\[
p_{cart1} = m_c \cdot v_{cart1} + p_{boy1} = 2000 \, \text{kg m/s}
\]
\[
100 \, \text{kg} \cdot v_{cart1} - 160 \, \text{kg m/s} = 2000 \, \text{kg m/s}
\]
\[
100 \, \text{kg} \cdot v_{cart1} = 2000 + 160 = 2160 \, \text{kg m/s}
\]
\[
v_{cart1} = \frac{2160}{100} = 21.6 \, \text{m/s}
\]
### Step 3: Second Boy Jumps Off
Now, the second boy jumps off perpendicular to the motion of the cart with the same speed of \( 2 \, \text{m/s} \) relative to the cart.
- Speed of the second boy relative to the ground:
Since he jumps perpendicular, we only need to consider the y-direction:
\[
v_{boy2} = 2 \, \text{m/s} \, \text{(perpendicular)}
\]
- Momentum of the second boy:
\[
p_{boy2} = m_b \cdot v_{boy2} = 20 \, \text{kg} \cdot 2 \, \text{m/s} = 40 \, \text{kg m/s}
\]
### Step 4: Conservation of Momentum in the Y-Direction
Since there is no external force acting in the y-direction, the total momentum in the y-direction must be conserved. Initially, the total momentum in the y-direction is zero.
Let \( v_{cart2} \) be the new speed of the cart after the second boy jumps off. The momentum equation in the y-direction is:
\[
0 = m_c \cdot v_{cart2} + p_{boy2}
\]
\[
m_c \cdot v_{cart2} + 40 = 0
\]
\[
100 \cdot v_{cart2} = -40
\]
\[
v_{cart2} = -0.4 \, \text{m/s}
\]
### Step 5: Final Speed of the Cart
The final speed of the cart after both boys have jumped off is a combination of its x and y components:
- \( v_{cart1} = 21.6 \, \text{m/s} \) (in the x-direction)
- \( v_{cart2} = -0.4 \, \text{m/s} \) (in the y-direction)
Using Pythagorean theorem to find the resultant speed:
\[
v_{final} = \sqrt{(v_{cart1})^2 + (v_{cart2})^2} = \sqrt{(21.6)^2 + (-0.4)^2} = \sqrt{466.56 + 0.16} = \sqrt{466.72} \approx 21.6 \, \text{m/s}
\]
### Conclusion
The speed of the cart after the second boy jumps will be approximately \( 21.6 \, \text{m/s} \).
---
