JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 13)
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 vx and vy 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?
$$\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$$ , where vx and vy 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?
kinetic energy of particle is constant in time
quantity $$\overrightarrow v \times \overrightarrow a $$
is constant in time
quantity $$\overrightarrow v .\overrightarrow a $$
is constant in time
$$\overrightarrow F $$ arises due to a magnetic field
Explanation
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
$$ \Rightarrow $$ 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
Comments (0)
