A particle of mass 1 kg is subjected to a force which depends on the position as with . At time , the particle's position and its velocity . Let and denote the and the components of the particle's velocity, respectively. Ignore gravity. When , the v

Question

A particle of mass 1 kg is subjected to a force which depends on the position as $\vec{F} = - k (x \hat{i} + y \hat{j}) {kg ms}^{- 2}$ with $k = 1 {kg s}^{- 2}$ . At time $t = 0$ , the particle's position $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) m$ and its velocity $\vec{v} = (- \sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{π} \hat{k}) {ms}^{- 1}$ . Let $v_{x}$ and $v_{y}$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5 m$ , the value of $(x v_{y} - y v_{x})$ is ______ $m^{2} s^{- 1}$ . [2022]

Smart Achievers · Accepted Answer

(3)

Here, $\vec{F} = - k \vec{r}$ . So force passes through origin. So, $τ_{origin} = 0 \Rightarrow$ angular momentum about origin will be conserved

So, $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{\sqrt{2}} & \sqrt{2} & 0 \\ - \sqrt{2} & \sqrt{2} & \frac{2}{π} \end{matrix} |$ $=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & 0.5 \\ v_{x} & v_{y} & \frac{2}{π} \end{matrix} |$

$\Rightarrow \hat{k} [\frac{1}{\sqrt{2}} \times \sqrt{2} - (- \sqrt{2}) \times \sqrt{2}] = \hat{k} (x v_{y} - y v_{x})$

$\Rightarrow x v_{y} - y v_{x} = 3$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) Here, F→=-kr→. So force passes through origin. So, τorigin=0⇒ angular momentum about origin will be conserved So, |i^j^k^1220-222π|=|i^j^k^xy0.5vxvy2π| ⇒k^[12×2-(-2)×2]=k^(xvy-yvx) ⇒xvy-yvx=3

Want Us to Email you About Special Offers & Updates?