Laws of Motion JEE 2023: Velocity-Field Force Solution

What is the correct option for the Laws of Motion JEE 2023 velocity-field question?

Option A is correct for this Laws of Motion JEE 2023 question from JEE Advanced 2023, Paper 2, Physics, a single-correct MCQ, not JEE Main. A particle moves in the xy-plane with constant mass and position-dependent velocity:

mass = m, \vec{v} = α (y \hat{x} + 2 x \hat{y}), α \neq 0 .

Here, alpha is a constant. Determine the net force at the particle’s current position: $(x, y) .$

The four options are:

\vec{F} = 2 m α^{2} (x \hat{x} + y \hat{y})

\vec{F} = m α^{2} (y \hat{x} + 2 x \hat{y})

\vec{F} = 2 m α^{2} (y \hat{x} + x \hat{y})

\vec{F} = m α^{2} (x \hat{x} + 2 y \hat{y})

The question bank tags this question medium, with an expected solving time of 90 seconds. These are question-bank metadata, not official JEE ratings or measured student performance. The deciding step is to differentiate each scalar coefficient without changing its unit vector.

How do you identify the two coordinate rates?

Read each component from its unit vector, not from the coordinate inside its coefficient. The coefficient multiplying the horizontal unit vector gives the horizontal velocity, even though it depends on the vertical coordinate.

v_{x} = \frac{d x}{d t} = α y

v_{y} = \frac{d y}{d t} = 2 α x

Along the particle’s motion, both coordinates depend on time:

x = x (t), y = y (t) .

The absence of an explicit time symbol in the field does not mean that the particle’s velocity is constant. As the particle changes position, it samples different velocity values. During time differentiation, alpha and both Cartesian unit vectors remain constant; only the coordinates change.

How do you differentiate the velocity with respect to time?

Differentiate the coordinate in each coefficient, keeping its original unit vector attached. Then replace each coordinate rate using the identities above. Start the official component-by-component method with the definition of acceleration:

\vec{a} = \frac{d \vec{v}}{d t} = \frac{d}{d t} [α (y \hat{x} + 2 x \hat{y})] .

Because alpha and the unit vectors are constant:

\vec{a} = α \frac{d y}{d t} \hat{x} + 2 α \frac{d x}{d t} \hat{y} .

For the horizontal component, differentiate its dependence on the vertical coordinate. Substitute the vertical velocity, not the horizontal velocity:

a_{x} = α \frac{d y}{d t} = α (2 α x) = 2 α^{2} x .

For the vertical component, differentiate its dependence on the horizontal coordinate. This time, substitute the horizontal velocity:

a_{y} = 2 α \frac{d x}{d t} = 2 α (α y) = 2 α^{2} y .

Combining the two components gives:

\vec{a} = 2 α^{2} x \hat{x} + 2 α^{2} y \hat{y} = 2 α^{2} (x \hat{x} + y \hat{y}) .

The substitutions cross: the derivative of the horizontal velocity requires the vertical velocity, and the derivative of the vertical velocity requires the horizontal velocity. Write those substitutions separately rather than doing them mentally. You do not need the trajectory to calculate this instantaneous acceleration.

How does Newton’s second law confirm option A?

Multiplying the acceleration by the particle’s mass gives option A. The force is a positive multiple of the position vector, so at every non-zero position it points away from the origin.

\vec{F} = m \vec{a} = 2 m α^{2} (x \hat{x} + y \hat{y}) .

The velocity field gives the units of alpha: velocity divided by length. Therefore, either force component has the required dimensions:

[α] = \frac{L T^{- 1}}{L} = T^{- 1} .

[m α^{2} x] = M (T^{- 1})^{2} L = M L T^{- 2} .

Changing the sign of alpha reverses the velocity at a specified position, but it does not change the force there: the result contains alpha squared. All four options have force dimensions, so dimensional consistency alone cannot select A.

How does changing a component’s direction incorrectly produce option C?

Option C follows if you differentiate the scalar coefficients correctly but wrongly swap their unit vectors. Depending on the vertical coordinate does not make a term a vertical component. Its unit vector determines its direction.

The error is to assign the derivative of the first term to the vertical direction merely because its coefficient contains the vertical coordinate. The same mistake assigns the derivative of the second term to the horizontal direction.

Wrong intermediate expression:

{\vec{a}}_{w r o n g} = α \frac{d y}{d t} \hat{y} + 2 α \frac{d x}{d t} \hat{x} .

Even substituting the correct coordinate rates now gives the wrong vector:

{\vec{a}}_{w r o n g} = 2 α^{2} x \hat{y} + 2 α^{2} y \hat{x} .

Multiplying by mass produces option C:

{\vec{F}}_{w r o n g} = 2 m α^{2} (y \hat{x} + x \hat{y}) .

The repair is to keep the fixed Cartesian direction unchanged:

\frac{d}{d t} (y \hat{x}) = \frac{d y}{d t} \hat{x} .

Differentiating a scalar coefficient changes its value, not its component direction. Mark the unit vectors first if you mix up the coordinates.

How do you solve two related force questions?

Use the same sequence: identify the coordinate rates, differentiate each velocity component, then substitute. Both questions below are original related practice, not verified JEE PYQs, using the same particle mass as above. Attempt each before reading its calculation.

Question 1: A particle moves along the x-axis with the following velocity, where beta is constant. Find its net force at its current horizontal coordinate.

v_{x} = β x, β \neq 0 .

Differentiate and substitute the horizontal coordinate rate:

a_{x} = \frac{d (β x)}{d t} = β \frac{d x}{d t} = β (β x) = β^{2} x .

Hence: $\vec{F} = m β^{2} x \hat{x} .$

For a worked numerical example, choose:

m = 3 k g, β = 2 s^{- 1}, x = 1 m .

Then:

a_{x} = (2)^{2} (1) = 4 m s^{- 2}, \vec{F} = 3 (4) \hat{x} = 12 \hat{x} N .

Question 2: A particle has the velocity below, with constant non-zero beta. Determine its net force. $\vec{v} = β (y \hat{x} - x \hat{y}) .$

First record the coordinate rates:

\frac{d x}{d t} = β y, \frac{d y}{d t} = - β x .

Now differentiate each component:

a_{x} = β \frac{d y}{d t} = - β^{2} x, a_{y} = - β \frac{d x}{d t} = - β^{2} y .

Therefore:

\vec{F} = - m β^{2} (x \hat{x} + y \hat{y}) .

At a non-zero position, this force points towards the origin, whereas the original PYQ’s force points away from it. Redo this second drill with the solution covered, keeping the minus sign attached through both substitutions.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Keep going with How to Study Differential Equations JEE: Methods and Checks.

Frequently asked questions

Is the JEE 2023 velocity-field question from Main or Advanced?

This question is from JEE Advanced 2023, Paper 2, Physics, not JEE Main. It is a single-correct MCQ asking for the net force on a particle with a position-dependent velocity.

What is the correct answer to the Laws of Motion JEE 2023 velocity-field question?

Option A is correct: the force components are F_x = 2mα²x and F_y = 2mα²y. For the given velocity components v_x = αy and v_y = 2αx, time differentiation gives a_x = 2α²x and a_y = 2α²y. Multiplying by the constant mass m gives the net force, which points away from the origin at every non-zero position.

Why is acceleration not zero when the velocity field has no time term?

A position-dependent velocity can change as the particle moves, even without explicit time dependence in the field. Here, x and y depend on time, so differentiating v_x = αy gives a_x = α(dy/dt), while differentiating v_y = 2αx gives a_y = 2α(dx/dt). Substitute the coordinate rates to find the acceleration.

Why is option C wrong in the JEE 2023 velocity-field question?

Option C results from incorrectly swapping the Cartesian unit vectors while differentiating the velocity components. A coefficient containing y does not make a term a vertical component; its unit vector determines its direction. Differentiate each scalar coefficient while keeping its original unit vector attached.