Question

Amount of energy loss (in J) in the collision is:

Hint:

Energy loss is always $K_{initial} - K_{final}$ in a collision.
Be careful to include all moving parts in the final state; here, both the particle and the rotating disk have energy.
Rotational energy depends on the pivot about which rotation occurs.

Answer 1

Step 1: Understanding the Question:
The energy loss is the difference between the total kinetic energy before the collision and total kinetic energy immediately after the collision.
Initial energy is only in the particle.
Final energy is the sum of the particle's translational kinetic energy and the disk's rotational kinetic energy.

Step 2: Key Formula or Approach:
Energy Loss $\Delta E = K_i - K_f$.
$K_i = \frac{1}{2} m v_i^2$.
$K_f = \frac{1}{2} m v_f^2 + \frac{1}{2} I_C \omega^2$.

Step 3: Detailed Explanation:

• Initial Kinetic Energy ($K_i$):
$m = 0.02\text{ kg}, v_i = 100\text{ ms}^{-1}$.
$K_i = \frac{1}{2} \times 0.02 \times (100)^2 = 0.01 \times 10000 = 100\text{ J}$.

• Final Kinetic Energy of Particle ($K_{pf}$):
$v_f = 90\text{ ms}^{-1}$.
$K_{pf} = \frac{1}{2} \times 0.02 \times (90)^2 = 0.01 \times 8100 = 81\text{ J}$.

• Final Kinetic Energy of Disk ($K_{df}$):
Using $\omega = 15.62\text{ rad/s}$ and $I_C = 0.06\text{ kg m}^2$ from the previous problem.
$K_{df} = \frac{1}{2} \times 0.06 \times (15.62)^2 \approx 0.03 \times 244 = 7.32\text{ J}$.

• Total Energy Loss:
$\Delta E = K_i - (K_{pf} + K_{df}) = 100 - (81 + 7.32) = 100 - 88.32 = 11.68\text{ J}$.

Step 4: Final Answer:
The collision is inelastic, resulting in an energy loss of $11.68\text{ J}$ which is dissipated as heat or internal deformation energy.