To solve for the kinetic energy in this perfectly inelastic collision, let's start by understanding what "perfectly inelastic" means. In a perfectly inelastic collision, the two objects stick together after colliding. This type of collision typically results in a loss of some kinetic energy, which we'll calculate below.

Steps to Find the Kinetic Energy

1. Calculate Initial Kinetic Energy of Both Objects
   The kinetic energy $KE$ of an object is given by the formula: $KE = \frac{1}{2}mv^2$

   For object A: $KE_A = \frac{1}{2} \cdot 3.2 \, \text{kg} \cdot (4.5 \, \text{m/s})^2$ $KE_A = \frac{1}{2} \cdot 3.2 \cdot 20.25 = 32.4 \, \text{J}$

   For object B: $KE_B = \frac{1}{2} \cdot 5.6 \, \text{kg} \cdot (7.2 \, \text{m/s})^2$ $KE_B = \frac{1}{2} \cdot 5.6 \cdot 51.84 = 145.15 \, \text{J}$

   Total Initial Kinetic Energy: $KE_{\text{initial}} = KE_A + KE_B = 32.4 \, \text{J} + 145.15 \, \text{J} = 177.55 \, \text{J}$

2. Calculate Final Velocity of the Combined Mass (Using Conservation of Momentum)
   Since it's a perfectly inelastic collision, the momentum is conserved. The final velocity $v_f$ of the combined mass after collision can be calculated by: $m_A \cdot v_A + m_B \cdot v_B = (m_A + m_B) \cdot v_f$

   Solving for $v_f$: $v_f = \frac{m_A \cdot v_A + m_B \cdot v_B}{m_A + m_B}$

   Substitute the values: $v_f = \frac{3.2 \cdot 4.5 + 5.6 \cdot 7.2}{3.2 + 5.6}$ $v_f = \frac{14.4 + 40.32}{8.8} = \frac{54.72}{8.8} \approx 6.22 \, \text{m/s}$

3. Calculate Final Kinetic Energy of the Combined Mass
   The final kinetic energy $KE_{\text{final}}$ of the combined object (mass $m_A + m_B$) moving at $v_f$ is: $KE_{\text{final}} = \frac{1}{2} \cdot (m_A + m_B) \cdot v_f^2$ $KE_{\text{final}} = \frac{1}{2} \cdot 8.8 \, \text{kg} \cdot (6.22 \, \text{m/s})^2$ $KE_{\text{final}} = \frac{1}{2} \cdot 8.8 \cdot 38.7 \approx 170.28 \, \text{J}$

4. Calculate the Kinetic Energy Lost in the Collision
   The energy lost $\Delta KE$ in the collision is the difference between the initial and final kinetic energies: $\Delta KE = KE_{\text{initial}} - KE_{\text{final}}$ $\Delta KE = 177.55 \, \text{J} - 170.28 \, \text{J} \approx 7.27 \, \text{J}$

Summary

• Initial Kinetic Energy: $177.55 \, \text{J}$
• Final Kinetic Energy: $170.28 \, \text{J}$
• Kinetic Energy Lost: $7.27 \, \text{J}$

Would you like more details or have any questions?

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Related Questions

1. How would the energy lost change if the masses were different?
2. What if the collision were elastic; how would that impact the energy calculation?
3. How is momentum conserved in inelastic collisions compared to elastic ones?
4. How does this result help understand the concept of energy dissipation?
5. What if one of the objects was initially at rest? How would that affect the calculations?

Tip:

In inelastic collisions, kinetic energy is not conserved, but total momentum is always conserved.