QJEEM20Ph19bb: JEE Main 2020 Physics - Class 11 JEEM20Ph19bb Solution: A particle moving in the xy plane experiences a velocity…

Question

Verified Solution • 2025–26 Session moderate

A particle moving in the xy plane experiences a velocity dependent force

\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

, where v_x and v_y are the
x and y components of its velocity

\overrightarrow v

. If

\overrightarrow a

is the
acceleration of the particle, then
which of the following statements is true for the particle?

Accepted Answer

Given

\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

\Rightarrow

m

\overrightarrow a

=

k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

\Rightarrow

\overrightarrow a = {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

Also

{{d{v_x}} \over {dt}} = {k \over m}{v_x}

and

{{d{v_y}} \over {dt}} = {k \over m}{v_y}

{{d{v_x}} \over {d{v_y}}} = {{{v_y}} \over {{v_x}}}

\Rightarrow

\int {{v_x}d{v_x}} = \int {{v_y}} d{v_y}

\Rightarrow

v_y^2 = v_x^2 + C

\Rightarrow

v_y^2 - v_x^2 = C

= Constant

From Option (B),

\overrightarrow v \times \overrightarrow a

=

\left( {{v_x}\widehat i + {v_y}\widehat j} \right) \times {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} \right)

=

\left( {v_x^2\widehat k - v_y^2\widehat k} \right) \times {k \over m}

=

\left( {v_x^2 - v_y^2} \right) \times {k \over m}\widehat k

= Constant

Step-by-Step Verified Solution