91Ó°ÊÓ

Kinetic energy, a form of mechanical energy, can change in a collision.
To understand this concept, think about an object's motion.
The formula for kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] For our exercise, two blocks collide and their speeds alter after impact.
Initially, both blocks have kinetic energies calculated based on their masses and velocities.
When they collide, the speeds change, leading to new kinetic energy values.

• Initial Kinetic Energy: \[ KE_i = \frac{1}{2} \times 5.0 \times (3.0)^2 + \frac{1}{2} \times 10.0 \times (2.0)^2 \]
• Final Kinetic Energy for the first scenario: \[ KE_f = \frac{1}{2} \times 5.0 \times (2.0)^2 + \frac{1}{2} \times 10.0 \times (2.5)^2 \]
• Final Kinetic Energy for the second scenario (new condition): \[ KE_f' = \frac{1}{2} \times 5.0 \times (2.0)^2 + \frac{1}{2} \times 10.0 \times (4.0)^2 \]

The change in kinetic energy (\( \Delta KE \) and \( \Delta KE' \)) reflects the energy lost or gained from the original value after collision.

In an inelastic collision, some kinetic energy turns into other energy forms like heat or sound.
However, total momentum is still conserved.
For the exercise problem, the collision between the two blocks is assumed to be inelastic.A key characteristic of inelastic collisions is that objects might move together post-collision, though not required in our situation.
The main takeaway is:

• Kinetic energy is not conserved.
• Some energy is lost as heat, sound, or deformation of objects.

In our exercise, the initial total kinetic energy is higher than the final, signifying that energy has been "lost" to other forms.This loss causes the phenomenon where \( \Delta KE = -1.25 \, \text{J} \) in the first scenario.

Momentum conservation is an essential rule in physics.
It states that the total momentum of a closed system remains constant, regardless of the interactions within it.
For our exercise:Initial momentum is calculated for both blocks before collision and is expressed as:\[ p_{initial} = m_1 \times v_1 + m_2 \times v_2 \]Where \( m_1 \) and \( m_2 \) are the masses and \( v_1 \) and \( v_2 \) are the initial velocities.The final momentum after collision:\[ p_{final} = m_1 \times v'_1 + m_2 \times v'_2 \]Momentum conservation simplifies to:\[ 5.0 \times 3.0 + 10.0 \times 2.0 = 5.0 \times v_1' + 10.0 \times 2.5 \]Solving this yields the unknown velocity \( v_1' \) of the 5.0 kg block after collision.
Even with kinetic energy loss, momentum stays steady in magnitude, emphasizing its inherent stability.