A stone is dropped at t = 0. A second stone, with 4 times the mass of the first, is dropped from the same point at t = 110 ms. (a) How far below the release point is the center of mass of the two stones at t = 410 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

To find the distance below the release point that the center of mass of the two stones is at t = 410 ms, we need to use the concept of center of mass and the conservation of momentum.

(a) Finding the distance:

1. First, let's assign variables to the important parameters:
- Let m1 be the mass of the first stone.
- Let m2 be the mass of the second stone (which is 4 times the mass of the first stone).
- Let d1 be the distance of the first stone from the release point at t = 410 ms.
- Let d2 be the distance of the second stone from the release point at t = 410 ms.
- Let d_cm be the distance of the center of mass from the release point at t = 410 ms.

2. We know that the center of mass of a system is given by:
d_cm = (m1 * d1 + m2 * d2) / (m1 + m2)

3. Since the stones are dropped from the same point, both d1 and d2 will be negative values (representing below the release point).

4. We need to find the relation between d1 and d2. At t = 410 ms, the second stone has been falling for 300 ms longer than the first stone, so the distance it has fallen will be greater. Therefore, we can write:
d2 = d1 + 300 ms

5. We can substitute this relation into the center of mass equation:
d_cm = (m1 * d1 + m2 * (d1 + 300 ms)) / (m1 + m2)

6. Substitute the given values:
m2 = 4 * m1
d_cm = (m1 * d1 + (4 * m1) * (d1 + 300 ms)) / (m1 + 4 * m1)

7. Simplify the equation:
d_cm = (m1 * d1 + 4 * m1 * d1 + 1200 ms * m1) / (5 * m1)
d_cm = (5 * d1 + 1200 ms) / 5

Now we have the equation for the distance of the center of mass (d_cm) in terms of d1. We can solve for d1.

8. Substitute t = 410 ms (or 0.41 s) into the equation for d_cm:
d_cm = (5 * d1 + 1200 ms) / 5
d_cm = (5 * d1 + 1200 ms) / 5 = d_cm = (5 * d1 + 1200 ms) / 5

9. Cross-multiply and solve for d1:
5 * d_cm = 5 * d1 + 1200 ms
5 * d_cm - 1200 ms = 5 * d1
d1 = (5 * d_cm - 1200 ms) / 5

(b) Finding the speed of the center of mass:

The speed of the center of mass is the rate at which it changes its position.

1. We know that velocity is the rate of change of displacement over time. So, the speed of the center of mass (v_cm) can be found by taking the derivative of the equation for d_cm with respect to time.

2. Differentiating d_cm with respect to time:
v_cm = d(d_cm)/dt

3. Differentiate the equation derived for d_cm in part (a) with respect to time:
v_cm = d(5 * d_cm)/dt = d(5 * ((5 * d1 + 1200 ms) / 5))/dt

4. Simplify the equation:
v_cm = d1/5

Therefore, the speed of the center of mass of the two-stone system at t = 410 ms is given by d1/5, where d1 is the distance calculated in part (a), as explained above.

To answer these questions, we can use the concept of the center of mass and equations of motion.

(a) To find the distance below the release point of the center of mass of the two stones at t = 410 ms, we can use the equation:

\(d = \frac{1}{2}gt^2\),

where d is the distance below the release point, g is the acceleration due to gravity, and t is the time.

Since the stone is dropped, the initial velocity is zero. Therefore, we can use the equation:

\(t = \sqrt{\frac{2d}{g}}\).

Using this equation, we can calculate the time it takes for the stone to reach a certain distance below the release point. Let's calculate it for t = 410 ms:

\(t = \sqrt{\frac{2d}{g}}\)

\(0.41 = \sqrt{\frac{2d}{9.8}}\)

Squaring both sides:

\(0.1681 = \frac{2d}{9.8}\)

Now solve for d:

\(d = 0.1681 \times 9.8/2\)

\(d = 0.826\ m\)

So, the center of mass of the two stones is 0.826 meters below the release point at t = 410 ms.

(b) To find the velocity of the center of mass of the two-stone system at t = 410 ms, we can use the equation:

\(v = gt\),

where v is the velocity and g is the acceleration due to gravity.

Let's calculate it for t = 410 ms:

\(v = 9.8 \times 0.41\)

\(v = 4.018\ m/s\)

So, the center of mass of the two-stone system is moving at a velocity of 4.018 m/s at t = 410 ms.