The initial kinetic energy of the bullet is:
\[ K_i = \frac{1}{2} m (100)^2 \]
The final kinetic energy of the bullet is:
\[ K_f = \frac{1}{2} m (40)^2 \]
The percentage loss in kinetic energy is given by:
\[ \% \text{loss} = \frac{|K_f - K_i|}{K_i} \times 100 \]
Substituting the expressions for \( K_i \) and \( K_f \):
\[ \% \text{loss} = \frac{\left| \frac{1}{2} m (40)^2 - \frac{1}{2} m (100)^2 \right|}{\frac{1}{2} m (100)^2} \times 100 \]
Simplify:
\[ \% \text{loss} = \frac{|1600 - 10000|}{10000} \times 100 \]
\[ \% \text{loss} = \frac{8400}{10000} \times 100 = 84\% \]
Final Answer: 84% (Option 4)
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
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: