To determine the energy loss during the collision, we can calculate the initial and final kinetic energy and find the difference.

Step 1: Initial kinetic energy of the cube

The formula for kinetic energy is:

$KE = \frac{1}{2} m v^2$

For the cube: $KE_{\text{initial cube}} = \frac{1}{2} (0.5) (2)^2 = 1 \, \text{J}$

The second object is stationary, so its initial kinetic energy is: $KE_{\text{initial second object}} = 0 \, \text{J}$

Thus, the total initial kinetic energy is: $KE_{\text{initial total}} = 1 \, \text{J}$

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Step 2: Final velocity after collision (using conservation of momentum)

Momentum before collision equals momentum after collision:

$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{final}}$

Substitute the values: $(0.5)(2) + (1)(0) = (0.5 + 1) v_{\text{final}}$

$1 = 1.5 v_{\text{final}}$

$v_{\text{final}} = \frac{1}{1.5} = \frac{2}{3} \, \text{m/s}$

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Step 3: Final kinetic energy of the system

The final kinetic energy is for the combined mass moving at $v_{\text{final}}$:

$KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v_{\text{final}}^2$

Substitute the values: $KE_{\text{final}} = \frac{1}{2} (1.5) \left(\frac{2}{3}\right)^2$

$KE_{\text{final}} = \frac{1}{2} (1.5) \left(\frac{4}{9}\right)$

$KE_{\text{final}} = \frac{1.5 \times 4}{18} = \frac{6}{18} = 0.333 \, \text{J}$

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Step 4: Energy loss

The energy loss during the collision is:

$\text{Energy loss} = KE_{\text{initial total}} - KE_{\text{final}}$

$\text{Energy loss} = 1 - 0.333 = 0.667 \, \text{J}$

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Final Answer:

The energy loss during the collision is:

$\boxed{0.667 \, \text{J}}$

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Would you like further details or have additional questions? Here are 5 related questions to expand on this topic:

1. What would happen to the energy loss if the second object was not stationary initially?
2. How does the coefficient of restitution influence energy conservation in a collision?
3. How is momentum conserved when external forces, like friction, are considered?
4. Can the energy loss be zero? Under what conditions?
5. What if the cube collided elastically with the second object? How would the energy distribution differ?

Tip: For inelastic collisions, kinetic energy is not conserved, but momentum always is!