Question 18

Initial Kinetic Energy: $K_i = \frac{1}{2} m v_1^2 = 0.5 \times 0.02 \times (100)^2 = 100 \text{ J}$.

Final Kinetic Energy: $K_{f, particle} = \frac{1}{2} m v_2^2 = 0.5 \times 0.02 \times (90)^2 = 81 \text{ J}$. $K_{f, disk} = \frac{1}{2} I_C \omega^2$ (from previous question calculation) $\approx 1.528 \text{ J}$. Total $K_f = 81 + 1.528 = 82.528 \text{ J}$.

Energy Loss: Loss $= K_i - K_f = 100 - 82.528 = 17.472 \text{ J}$. Rounding to one decimal place, we get $17.5 \text{ J}$.