Step 1: Relation between kinetic energy and momentum - Kinetic energy \[ K = \frac{P^2}{2m}, \] where \(P\) is momentum and \(m\) is mass.
Step 2: Calculate new kinetic energy - Initial momentum \(P_1\), kinetic energy \[ K_1 = \frac{P_1^2}{2m}. \] New momentum \[ P_2 = P_1 + 0.5P_1 = 1.5P_1. \] New kinetic energy \[ K_2 = \frac{P_2^2}{2m} = \frac{(1.5P_1)^2}{2m} = \frac{2.25P_1^2}{2m} = 2.25K_1. \]
Step 3: Calculate percentage increase in kinetic energy - Percentage increase is given by: \[ \frac{K_2 - K_1}{K_1} \times 100 = \frac{2.25K_1 - K_1}{K_1} \times 100 = 125\%. \]
Final Answer: The percentage increase in kinetic energy is 125%.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: