The solution to this problem involves using conservation of momentum and conservation of energy principles.

First, we can determine the total system (two carts) momentum before the collision:

P₁ = m₁v₁ + m₂v₂, where m₁ and m₂ are the masses of the carts and v₁ and v₂ are the velocities.

Plugging in the values, we get:

P₁ = (0.18 kg)(0.1 m/s) + (0.09 kg)(0.15 m/s) = 0.03 kg·m/s

After the collision, the carts stick together and move with a common velocity, v_f. To find v_f, we use the conservation of momentum equation:

P₁ = P₂ = (m₁+m₂)v_f

Plugging in the values again, we get:

0.03 kg·m/s = (0.18 kg + 0.09 kg)v_f

Solving for v_f, we get v_f = 0.03 kg·m/s / (0.18 kg + 0.09 kg) = 0.1 m/s

Therefore, the speed of the carts after the collision is 0.1 m/s.

To calculate the amount of kinetic energy lost during the collision, we can use the conservation of energy equation:

KE_i = KE_f + ΔKE_internal, where KE_i is the initial kinetic energy, KE_f is the final kinetic energy, and ΔKE_internal is the change in internal kinetic energy (i.e. the energy lost during the collision).

Since there is no external work done on the system, the initial kinetic energy is equal to the final kinetic energy. Therefore, we can rewrite the equation as:

KE_i = KE_f + ΔKE_internal = 0

Solving for ΔKE_internal, we get:

ΔKE_internal = KE_i - KE_f = (½)mv_i² - (½)mv_f² = (½)(0.18 kg)(0.1 m/s)² - (½)(0.18 kg + 0.09 kg)(0.1 m/s)² = -0.6 mJ

Therefore, after the collision, 0.6 mJ of kinetic energy is converted into internal energy in the system. This is why the carts stick together after the collision instead of bouncing off each other.

1. Use the conservation of energy law to solve this problem. According to this law, the energy of a system remains constant, meaning that the initial energy of the stone must be equal to its energy at maximum height.
2. Start by finding the potential energy of the stone at the beginning of its ascent. This can be calculated using the formula P = mgh, where m is the mass of the stone (200 g), g is the acceleration due to gravity (9.8 m/s²), and h is the initial height of the stone (in this case, the height of the Earth's surface).
3. The stone also has kinetic energy at the start of its ascent, which can be calculated using the formula K = 0.5mv², where v is the initial velocity of the stone (6 m/s).
4. At maximum height, the stone has no kinetic energy (since it stops moving) and all of its energy is in the form of potential energy. Therefore, we can set the potential energy at maximum height equal to the initial potential energy.
5. To find the maximum height, set the potential energies at the beginning and end of the stone's ascent equal to each other and solve for h. This gives you the equation P = mgh = K = 0.5mv². Substitute in the values we know and solve for h. You should get a maximum height of approximately 2.45 meters.
6. Finally, to find the kinetic energy at the midpoint of the stone's flight, you can use the formula K = 0.5mv² again, this time plugging in the midpoint velocity (which is half of the initial velocity) for v. This should give you a kinetic energy of approximately 0.45 joules.

Remember, the conservation of energy law applies to all types of energy, not just potential and kinetic. So even if the stone's height changes during its flight, the total energy of the system remains the same!

The work done by the force of gravity can be calculated using the formula:

W = mgh

where m is the mass of the object, g is the acceleration due to gravity, and h is the height or distance the object falls.

In this case, the mass (m) of the object is 4 kg, the acceleration due to gravity (g) is 9.8 m/s², and the change in velocity (v) is 8 m/s - 2 m/s = 6 m/s.

Therefore, h can be calculated using the formula:

h = v² / 2g

Substituting the values, we get:

h = (6 m/s)² / 2(9.8 m/s²) = 18.37 m

Thus, the work done by the force of gravity on the object is 18.37 joules along the entire path.

This increase in the object's velocity can be explained by the conservation of energy principle, where the potential energy of the object at the top of the path is transformed into kinetic energy as it falls.

P.S. Did the object do some exercises while falling? It seems to have gained some speed 😜