A particle of mass 1 kg is subjected to a force which depends on the position as \(\vec{F}=-k(x \hat{i}+y \hat{j}) kg m s^{-2}\) with \(k=1 kg s^{-2}\). 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 \(z=0.5 m\), the value of ( \(x v_{y}-y v_{x}\) ) is \_\_\_\_ \(m^{2} s^{-1}\).

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Rotational Mechanics

A particle of mass 1 kg is subjected to a force which depends on the position as

F = - k (x \hat{i} + y \hat{j}) k g m s^{- 2}

with

k = 1 k g s^{- 2}

. At time

t = 0

, the particle's position

r = (\frac{1}{2} \hat{i} + 2 \hat{j}) m

and its velocity

v = (- 2 \hat{i} + 2 \hat{j} + \frac{2}{π} \hat{k}) 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 = 0.5 m

, the value of (

x v_{y} - y v_{x}

) is ____

m^{2} s^{- 1}

Answer: 3

The force

F = - k (x \hat{i} + y \hat{j})

produces zero torque about the origin since

τ = r \times F = r \times (- k r) = 0

. Therefore, angular momentum about the origin is conserved.

Initial angular momentum:

L_{i} = r \times m v = \hat{i} \frac{1}{2} - 2 \hat{j} 22 \hat{k} 0 \frac{2}{π}

The

z

-component is:

L_{z} = m (\frac{1}{2} \cdot 2 - 2 \cdot (- 2)) = 1 (1 + 2) = 3 kg m^{2} s^{- 1}

At any later time when

z = 0.5

L_{z} = m (x v_{y} - y v_{x}) = 1 \cdot (x v_{y} - y v_{x})

By conservation:

x v_{y} - y v_{x} = 3

m²s⁻¹

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer