Let's analyze the problem and solve it systematically.

Extracted Information:

1. Known Values:

   • Cart mass $m = 200 \, \text{g} = 0.2 \, \text{kg}$
   • Initial height $h = 50 \, \text{cm} = 0.5 \, \text{m}$
   • Gravitational acceleration $g = 10 \, \text{m/s}^2$

2. Energy Relations:

   • Initial gravitational potential energy ($\text{GPE}$) is converted to kinetic energy ($\text{KE}$).
   • $\text{GPE} = m \cdot g \cdot h$, and $\text{KE} = \frac{1}{2} m v^2$, where $v$ is the velocity.

   Equating $\text{GPE}$ and $\text{KE}$:
   $m \cdot g \cdot h = \frac{1}{2} m v^2$

   Simplifying:
   $g \cdot h = \frac{1}{2} v^2 \implies v = \sqrt{2 \cdot g \cdot h}$

3. Given Velocity for 200g Cart:

   • From the data, at $h = 0.5 \, \text{m}$, $v = 3.16 \, \text{m/s}$. This confirms the formula works.

Solving the Questions:

1. Final Speed for an 80g Cart Released from $h = 50 \, \text{cm}$:

The mass is irrelevant in the velocity calculation (it cancels out). Using the same formula:
$v = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 10 \cdot 0.5} = \sqrt{10} \approx 3.16 \, \text{m/s}$

Thus, the final speed remains 3.16 m/s.

2. Final Speed for a 600g Cart Released from $h = 85 \, \text{cm}$:

Here, $h = 85 \, \text{cm} = 0.85 \, \text{m}$. Again using the formula:
$v = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 10 \cdot 0.85} = \sqrt{17} \approx 4.12 \, \text{m/s}$

Thus, the final speed is approximately 4.12 m/s.

Final Answers:

1. Final speed of 80g cart (50cm): $\mathbf{3.16 \, \text{m/s}}$
2. Final speed of 600g cart (85cm): $\mathbf{4.12 \, \text{m/s}}$

Would you like a step-by-step explanation or further clarifications?

Related Questions:

1. How does mass influence the gravitational potential energy if height remains constant?
2. Why does the final speed depend only on height and not on mass?
3. Can we apply this formula for heights significantly higher than 1 meter?
4. What happens to the final speed if $g$ changes, e.g., on the Moon?
5. How can energy losses (like air resistance) affect the final speed?

Tip: Always double-check units for consistency before calculating physics problems!