A particle of mass $100\ \text{g}$ is projected at time $t=0$ with a speed $20\ \text{m s}^{-1}$ at an angle $45^\circ$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2\ \text{s}$ is found to be $\sqrt{K}\ \text{kg m}^2\ \text{s}^{-1}$. The value of $K$ is ______. (Take $g=10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 800
The angular momentum about the launch point is $L=m\,|\vec r\times\vec v|$. With $\vec r=(u_xt,\ u_yt-\tfrac12gt^2)$ and $\vec v=(u_x,\ u_y-gt)$:
$$L=m\left[u_xt(u_y-gt)-u_x\left(u_yt-\tfrac12gt^2\right)\right]=-\tfrac12mgu_xt^2$$
Magnitude: $L=\tfrac12mg\,u\cos45^\circ\,t^2=\tfrac12\times0.1\times10\times\dfrac{20}{\sqrt2}\times4=20\sqrt2=\sqrt{800}$.
So $K=800$.
Solution by Sreeraj P, M.Sc Physics