Question 142: Laws of Motion

Question

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 $$\overrightarrow F  = k\left( {{v_y}\widehat i + {v_x}\widehat j} ight)$$
$$ \Rightarrow $$ m$$\overrightarrow a $$ = $$k\left( {{v_y}\widehat i + {v_x}\widehat j} ight)$$
$$ \Rightarrow $$ $$\overrightarrow a  = {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} ight)$$
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),
$$\overrightarrow v  	imes \overrightarrow a $$
= $$\left( {{v_x}\widehat i + {v_y}\widehat j} ight) 	imes {k \over m}\left( {{v_y}\widehat i + {v_x}\widehat j} ight)$$
= $$\left( {v_x^2\widehat k - v_y^2\widehat k} ight) 	imes {k \over m}$$
= $$\left( {v_x^2 - v_y^2} ight) 	imes {k \over m}\widehat k$$
= Constant

Answer

kinetic energy of particle is constant in time

Answer

quantity $$\overrightarrow v  	imes \overrightarrow a $$
 is constant in time

Answer

quantity $$\overrightarrow v .\overrightarrow a $$
 is constant in time

Answer

$$\overrightarrow F $$ arises due to a magnetic field

Laws of Motion

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