A particle moving in the xy plane experiences a velocity dependent force \overrightarrow F = k( v_y\widehat i + v_x\widehat j ) , 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?

quantity \overrightarrow v \times \overrightarrow a is constant in time

kinetic energy of particle is constant in time

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

\overrightarrow F arises due to a magnetic field

Laws of Motion PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q48 with Solution

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Correct answer: B

Given force law

The force is

\vec F = k\,(v_y\hat i + v_x\hat j)

So the acceleration is

\vec a = \frac{\vec F}{m} = \frac{k}{m}(v_y\hat i + v_x\hat j)

Hence,

a_x = \frac{k}{m}v_y, \qquad a_y = \frac{k}{m}v_x

Check option A: Is kinetic energy constant?

Rate of change of kinetic energy is

\frac{d}{dt}\left(\frac12 mv^2\right)= \vec F\cdot \vec v

Now,

\vec F\cdot \vec v = k(v_y\hat i + v_x\hat j)\cdot (v_x\hat i + v_y\hat j)

= k(v_xv_y + v_yv_x)=2kv_xv_y

This is not zero in general, so kinetic energy is not constant.

So, A is false.

Check option C: Is $\vec v\cdot \vec a$ constant?

We have

\vec v\cdot \vec a = (v_x\hat i+v_y\hat j)\cdot \frac{k}{m}(v_y\hat i+v_x\hat j)

= \frac{k}{m}(v_xv_y+v_yv_x)=\frac{2k}{m}v_xv_y

This is generally not constant because $v_x$ and $v_y$ change with time.

So, C is false.

Check option B: Is $\vec v\times \vec a$ constant?

Since motion is in the $xy$ -plane, $\vec v\times \vec a$ is along $\hat k$ :

\vec v\times \vec a = \begin{vmatrix} \hat i & \hat j & \hat k\\ v_x & v_y & 0\\ a_x & a_y & 0 \end{vmatrix} = (v_xa_y-v_ya_x)\hat k

Substitute

a_x=\frac{k}{m}v_y,\qquad a_y=\frac{k}{m}v_x

Then,

v_xa_y-v_ya_x = v_x\left(\frac{k}{m}v_x\right)-v_y\left(\frac{k}{m}v_y\right)

=\frac{k}{m}(v_x^2-v_y^2)

So,

\vec v\times \vec a = \frac{k}{m}(v_x^2-v_y^2)\hat k

Now check whether $v_x^2-v_y^2$ is constant:

\frac{d}{dt}(v_x^2-v_y^2)=2v_x\dot v_x-2v_y\dot v_y

Using

\dot v_x=a_x=\frac{k}{m}v_y,\qquad \dot v_y=a_y=\frac{k}{m}v_x

we get

\frac{d}{dt}(v_x^2-v_y^2)=2v_x\frac{k}{m}v_y-2v_y\frac{k}{m}v_x=0

Hence $v_x^2-v_y^2$ is constant, so

\vec v\times \vec a

is constant in time.

So, B is true.

Check option D: Does this force arise due to a magnetic field?

Magnetic force is always of the form

\vec F = q(\vec v\times \vec B)

which is always perpendicular to $\vec v$ , so

\vec F\cdot \vec v = 0

But here,

\vec F\cdot \vec v = 2kv_xv_y

which is not zero in general. Therefore this force cannot be purely magnetic.

So, D is false.

Final answer

The correct option is:

\boxed{\text{B}}

More from Laws of Motion

View written solutionFree

Correct answer: B

Given force law

The force is

\vec F = k\,(v_y\hat i + v_x\hat j)

So the acceleration is

\vec a = \frac{\vec F}{m} = \frac{k}{m}(v_y\hat i + v_x\hat j)

Hence,

a_x = \frac{k}{m}v_y, \qquad a_y = \frac{k}{m}v_x

Check option A: Is kinetic energy constant?

Rate of change of kinetic energy is

\frac{d}{dt}\left(\frac12 mv^2\right)= \vec F\cdot \vec v

Now,

\vec F\cdot \vec v = k(v_y\hat i + v_x\hat j)\cdot (v_x\hat i + v_y\hat j)

= k(v_xv_y + v_yv_x)=2kv_xv_y

This is not zero in general, so kinetic energy is not constant.

So, A is false.

Check option C: Is $\vec v\cdot \vec a$ constant?

We have

\vec v\cdot \vec a = (v_x\hat i+v_y\hat j)\cdot \frac{k}{m}(v_y\hat i+v_x\hat j)

= \frac{k}{m}(v_xv_y+v_yv_x)=\frac{2k}{m}v_xv_y

This is generally not constant because $v_x$ and $v_y$ change with time.

So, C is false.

Check option B: Is $\vec v\times \vec a$ constant?

Since motion is in the $xy$ -plane, $\vec v\times \vec a$ is along $\hat k$ :

\vec v\times \vec a = \begin{vmatrix} \hat i & \hat j & \hat k\\ v_x & v_y & 0\\ a_x & a_y & 0 \end{vmatrix} = (v_xa_y-v_ya_x)\hat k

Substitute

a_x=\frac{k}{m}v_y,\qquad a_y=\frac{k}{m}v_x

Then,

v_xa_y-v_ya_x = v_x\left(\frac{k}{m}v_x\right)-v_y\left(\frac{k}{m}v_y\right)

=\frac{k}{m}(v_x^2-v_y^2)

So,

\vec v\times \vec a = \frac{k}{m}(v_x^2-v_y^2)\hat k

Now check whether $v_x^2-v_y^2$ is constant:

\frac{d}{dt}(v_x^2-v_y^2)=2v_x\dot v_x-2v_y\dot v_y

Using

\dot v_x=a_x=\frac{k}{m}v_y,\qquad \dot v_y=a_y=\frac{k}{m}v_x

we get

\frac{d}{dt}(v_x^2-v_y^2)=2v_x\frac{k}{m}v_y-2v_y\frac{k}{m}v_x=0

Hence $v_x^2-v_y^2$ is constant, so

\vec v\times \vec a

is constant in time.

So, B is true.

Check option D: Does this force arise due to a magnetic field?

Magnetic force is always of the form

\vec F = q(\vec v\times \vec B)

which is always perpendicular to $\vec v$ , so

\vec F\cdot \vec v = 0

But here,

\vec F\cdot \vec v = 2kv_xv_y

which is not zero in general. Therefore this force cannot be purely magnetic.

So, D is false.

Final answer

The correct option is:

\boxed{\text{B}}

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