A particle moving in the xy plane experiences a velocity dependent force F-kvy-vx- where vx and vy are x and y components of its velocity v- If a is the acceleration of the particle- then which of the following statements is true for the particle-

Given, F=k(v_yî+v_xĵ) ∴ F_x=kv_yî, F_y=kv_xĵ mdv_xdt=kv_y⇒dv_xdt=kmv_y Similarly, nbsp;dv_y dt=kmv_x dv_ydv_x=v_xv_y⇒∫ v_ydv_y=∫ v_xdv_x v_y^2=v_x^2+C v_ ...

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Byju's Answer

Standard XII

Physics

Vector Components

A particle mo...

Question

A particle moving in the $x y$ plane experiences a velocity dependent force $\to F = k (v_{y}^i + v_{x}^j),$ where $v_{x}$ and $v_{y}$ are $x$ and $y$ components of its velocity $\to v .$ If $\to a$ is the acceleration of the particle, then which of the following statements is true for the particle?

Quantity

\to v \times \to a

is constant in time

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\to F

arises due to a magnetic field

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Kinetic energy of particle is constant in time

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Quantity

\to v . \to a

is constant in time

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Solution

The correct option is A Quantity $\to v \times \to a$ is constant in time
Given, $\to F = k (v_{y}^i + v_{x}^j)$

$∴ F_{x} = k v_{y}^i, F_{y} = k v_{x}^j$

$\frac{m d v_{x}}{d t} = k v_{y} \Rightarrow \frac{d v_{x}}{d t} = \frac{k}{m} v_{y}$

Similarly, $\frac{d v_{y}}{d t} = \frac{k}{m} v_{x}$

$\frac{d v_{y}}{d v_{x}} = \frac{v_{x}}{v_{y}} \Rightarrow \int v_{y} d v_{y} = \int v_{x} d v_{x}$

$v_{y}^{2} = v_{x}^{2} + C$

$v_{y}^{2} - v_{x}^{2} = constant$

$\to v \times \to a = (v_{x}^i + v_{y}^j) \times \frac{k}{m} (v_{y}^i + v_{x}^j)$

$= (v_{x}^{2}^k - v_{y}^{2}^k) \frac{k}{m} = (v_{x}^{2} - v_{y}^{2}) \frac{k}{m}^k = constant$

Hence, option $(A)$ is correct.

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A particle moving in the xy plane experiences a velocity dependent force →F=k(vy^i+vx^j), where vx and vy are x and y components of its velocity →v. If →a is the acceleration of the particle, then which of the following statements is true for the particle?