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 $k=1 \, kgs ^{-2}$ . At time $t=0$, the particle's position $\vec{r}=\left(\frac{1}{\sqrt{2}} \hat{i}+\sqrt{2} \hat{j}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$ . Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=05\, m$, the value of (xv_y – yv_x) is _______ m²s^–1.

Answer 1

Correct Answer: 3

The given force is $\vec{F} = -k(x\hat{i} + y\hat{j})$. The units given with the force expression mean that if $k$ is in $\text{kgs}^{-2}$ and $x,y$ are in $\text{m}$, then $\vec{F}$ is in $\text{kg ms}^{-2}$ (Newtons). So, $F_x = -kx$, $F_y = -ky$, and $F_z = 0$ (since gravity is ignored and no other z-component of force is mentioned).

The quantity $(xv_y - yv_x)$ is related to the z-component of the angular momentum $L_z$. The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} \times m\vec{v}$. Its z-component is $L_z = m(xv_y - yv_x)$.

The rate of change of angular momentum is equal to the net torque $\vec{\tau}$ on the particle: $$ \frac{d\vec{L}}{dt} = \vec{\tau} $$ The torque is calculated as $\vec{\tau} = \vec{r} \times \vec{F}$. Given $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{F} = -kx\hat{i} -ky\hat{j} + 0\hat{k}$. $$ \vec{\tau} = (x\hat{i} + y\hat{j} + z\hat{k}) \times (-kx\hat{i} -ky\hat{j}) $$ Expanding the cross product: $$ \vec{\tau} = x(-ky)(\hat{i} \times \hat{j}) + y(-kx)(\hat{j} \times \hat{i}) + z(-kx)(\hat{k} \times \hat{i}) + z(-ky)(\hat{k} \times \hat{j}) $$ Using the vector identities $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{i} = \hat{j}$, $\hat{k} \times \hat{j} = -\hat{i}$: $$ \vec{\tau} = -kxy\hat{k} -kxy(-\hat{k}) -kxz\hat{j} -kyz(-\hat{i}) $$ $$ \vec{\tau} = -kxy\hat{k} + kxy\hat{k} + kyz\hat{i} - kxz\hat{j} $$ $$ \vec{\tau} = (kyz)\hat{i} - (kxz)\hat{j} + 0\hat{k} $$ The z-component of the torque is $\tau_z = 0$. Since $\frac{dL_z}{dt} = \tau_z$, we have $\frac{dL_z}{dt} = 0$. This implies that $L_z$ is a conserved quantity. Thus, $L_z = m(xv_y - yv_x)$ is constant. Given $m=1 \text{ kg}$, the quantity $(xv_y - yv_x)$ is conserved.

We can calculate the value of $(xv_y - yv_x)$ at time $t=0$ using the initial conditions: Initial position: $x_0 = \frac{1}{\sqrt{2}} \text{ m}$, $y_0 = \sqrt{2} \text{ m}$. Initial velocity: $v_{x0} = -\sqrt{2} \text{ ms}^{-1}$, $v_{y0} = \sqrt{2} \text{ ms}^{-1}$. $$ (xv_y - yv_x)_{t=0} = x_0 v_{y0} - y_0 v_{x0} $$ $$ = \left(\frac{1}{\sqrt{2}} \text{ m}\right)(\sqrt{2} \text{ ms}^{-1}) - (\sqrt{2} \text{ m})(-\sqrt{2} \text{ ms}^{-1}) $$ $$ = 1 \text{ m}^2\text{s}^{-1} - (-2 \text{ m}^2\text{s}^{-1}) $$ $$ = 1 + 2 = 3 \text{ m}^2\text{s}^{-1} $$ Since $(xv_y - yv_x)$ is conserved, its value remains $3 \text{ m}^2\text{s}^{-1}$ at all times. The condition "When $z = 0.5 \text{ m}$" refers to a specific instant in time, but it does not alter the value of this conserved quantity.

The value of $(xv_y - yv_x)$ is $\textbf{3}$ $\text{m}^2\text{s}^{-1}$.

Correct Answer: 3