### Related Formula The magnitude of the angular momentum L$L$ of a particle of mass m$m$ moving with velocity v$v$ is: L = m v d$L = m v d$ where: * d$d$ is the perpendicular distance from the axis of rotation (origin) to the line of motion of the particle. ### Core Logic Given parameters: * Mass, m = 5text kg$m = 5\text{ kg}$ * Velocity, v = 3sqrt2text ms^-1$v = 3\sqrt{2}\text{ ms}^{-1}$ * Line of motion: y = x + 4 implies x - y + 4 = 0$y = x + 4 \implies x - y + 4 = 0$ ### Step 1: Calculate Perpendicular Distance The perpendicular distance d$d$ from the origin (0,0)$(0,0)$ to the line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|A(0) + B(0) + C|sqrtA^2 + B^2$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$ For the line x - y + 4 = 0$x - y + 4 = 0$: d = frac|4|sqrt1^2 + (-1)^2 = frac4sqrt2 = 2sqrt2text m$d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ m}$ ### Step 2: Calculate Angular Momentum Substitute the values into the angular momentum formula: L = m v d$L = m v d$ L = 5text kg times (3sqrt2text ms^-1) times (2sqrt2text m)$L = 5\text{ kg} \times (3\sqrt{2}\text{ ms}^{-1}) \times (2\sqrt{2}\text{ m})$ L = 5 times 3 times 4 = 60text kg m^2texts^-1$L = 5 \times 3 \times 4 = 60\text{ kg m}^2\text{s}^{-1}$ Thus, the angular momentum of the particle about the origin is 60text kg m^2texts^-1$60\text{ kg m}^2\text{s}^{-1}$. ### Pattern Recognition Instead of complicated vector cross products, find the perpendicular distance of the straight line from the origin using standard coordinate geometry. L = mvd$L = mvd$ is extremely fast and reliable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion