JEE PYQ: Motion in a Plane - Question ID 171819735946 (JEE Main 2026)

Core Formula & Concept: This problem deals with the kinematics of a particle under a constant force. When a constant force acts on a body, it produces a constant acceleration according to Newton's Second Law of Motion. The motion can then be described using the standard kinematic equations for constant acceleration. 1. Newton's Second Law: The net force $\vec{F}$ acting on a body of mass $m$ produces an acceleration $\vec{a}$ given by the relation: $\vec{F} = m\vec{a}$ From this, the acceleration can be found as $\vec{a} = \frac{\vec{F}}{m}$. 2. Kinematic Equation for Position (Constant Acceleration): If a body starts with an initial position $\vec{r}_0$ and an initial velocity $\vec{v}_0$, and moves with a constant acceleration $\vec{a}$, its position $\vec{r}$ at time $t$ is given by: $\vec{r} = \vec{r}_0 + \vec{v}_0t + \frac{1}{2}\vec{a}t^2$ In this problem, "starts moving" at $t=0$ implies that the initial velocity $\vec{v}_0 = \vec{0}$. Unless specified otherwise, we assume the body starts from the origin, so $\vec{r}_0 = \vec{0}$. Thus, the equation simplifies to: $\vec{r} = \frac{1}{2}\vec{a}t^2$
Step-by-Step Derivation: 1. Identify Given Parameters and Convert Units: * Mass of the body, $m = 100 \text{ g}$. We must convert this to kilograms for consistency with SI units of force (Newtons). $m = 100 \text{ g} = 0.1 \text{ kg}$. * Force acting on the body, $\vec{F} = (5 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{N}$. * Initial time, $t_0 = 0 \text{ s}$. * Time elapsed, $t = 2 \text{ s}$. * Final position, $\vec{r} = (2 x \hat{\mathrm{i}}+5 y \hat{\mathrm{j}}) \mathrm{m}$. * Initial velocity: "starts moving" implies $\vec{v}_0 = \vec{0}$. * Initial position: Assuming it starts from the origin, $\vec{r}_0 = \vec{0}$. 2. Calculate the Acceleration ($\vec{a}$): Using Newton's Second Law, $\vec{F} = m\vec{a}$. We can find the acceleration vector $\vec{a} = \frac{\vec{F}}{m}$. Substitute the given values: $\vec{a} = \frac{(5 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{N}}{0.1 \text{ kg}}$ $\vec{a} = \left(\frac{5}{0.1}\right) \hat{\mathrm{i}} + \left(\frac{10}{0.1}\right) \hat{\mathrm{j}} \text{ m/s}^2$ $\vec{a} = (50 \hat{\mathrm{i}}+100 \hat{\mathrm{j}}) \text{ m/s}^2$ 3. Apply the Kinematic Equation for Position: Since $\vec{r}_0 = \vec{0}$ and $\vec{v}_0 = \vec{0}$, the position vector at time $t$ is given by: $\vec{r} = \frac{1}{2}\vec{a}t^2$ Substitute the calculated acceleration $\vec{a}$ and the time $t=2 \text{ s}$: $\vec{r} = \frac{1}{2} (50 \hat{\mathrm{i}}+100 \hat{\mathrm{j}}) (2 \text{ s})^2$ $\vec{r} = \frac{1}{2} (50 \hat{\mathrm{i}}+100 \hat{\mathrm{j}}) (4)$ $\vec{r} = (2 \times 50 \hat{\mathrm{i}} + 2 \times 100 \hat{\mathrm{j}}) \text{ m}$ $\vec{r} = (100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}) \text{ m}$ 4. Equate with the Given Final Position and Solve for $x$ and $y$: The problem states that after 2 s, the position is $(2 x \hat{\mathrm{i}}+5 y \hat{\mathrm{j}}) \mathrm{m}$. So, we equate our calculated position vector with the given form: $(100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}) \text{ m} = (2 x \hat{\mathrm{i}}+5 y \hat{\mathrm{j}}) \text{ m}$ Comparing the components: For the $\hat{\mathrm{i}}$ component: $100 = 2x$ $x = \frac{100}{2} = 50$ For the $\hat{\mathrm{j}}$ component: $200 = 5y$ $y = \frac{200}{5} = 40$ 5. Calculate the Ratio $x:y$: Now we find the ratio $x:y$: $\frac{x}{y} = \frac{50}{40}$ $\frac{x}{y} = \frac{5}{4}$ So, the ratio $x:y$ is $5:4$. The final answer corresponds to option D.
Common Traps & Exam Tip: 1. Unit Conversion: A very common mistake is to forget converting the mass from grams to kilograms. If $m=100 \text{ g}$ was used directly, the acceleration would be $500 \hat{\mathrm{i}}+1000 \hat{\mathrm{j}}$, leading to different $x$ and $y$ values and an incorrect ratio. Always ensure all quantities are in a consistent system of units (preferably SI) before performing calculations. 2. Initial Conditions: Incorrectly assuming an initial velocity or initial position other than zero can lead to errors. "Starts moving at $t=0$" conventionally implies starting from rest ($\vec{v}_0 = \vec{0}$) at the origin ($\vec{r}_0 = \vec{0}$), unless explicitly stated otherwise. 3. Vector vs. Scalar: Remember to treat force, acceleration, velocity, and position as vector quantities, performing calculations component-wise. 4. Algebraic Errors: Be careful with the multiplication by $\frac{1}{2}$ and $t^2$ in the kinematic equation, and subsequently when equating components to solve for $x$ and $y$. To avoid such traps, always write down the given information clearly, convert units at the beginning, state any assumptions (like initial conditions), and perform calculations step-by-step, keeping track of vector components. Double-check your final algebra.