Motion In Two Dimensions JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Motion In Two Dimensions, free to read — no sign-in needed. The full chapter has 17 questions; sign in to attempt the remaining 14 in the exam simulator.

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1. Q1JEE Advanced Adv 2023 (Paper 2)
   A particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x,y)$ is given as $\vec{v}=\alpha \left(yx^{∧}+2xy^{∧}\right)$ where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?
   1. A.$\vec{F}=2m{\alpha }^{2}\left(xx^{∧}+yy^{∧}\right)$
   2. B.$\vec{F}=m{\alpha }^{2}\left(yx^{∧}+2xy^{∧}\right)$
   3. C.$\vec{F}=2m{\alpha }^{2}\left(yx^{∧}+xy^{∧}\right)$
   4. D.$\vec{F}=m{\alpha }^{2}\left(xx^{∧}+2yy^{∧}\right)$
   Show answer & solution

   Answer: (A)

   As force is given by, $\vec{F}=m\dfrac{d\vec{v}}{dt}$. Given: $\vec{v}=\alpha \left(yx^{∧}+2xy^{∧}\right)$, where velocity along x-axis is ${v}_{x}=\alpha y$ and velocity along y-axis is ${v}_{y}=2x\alpha$. Therefore, $\dfrac{d\vec{v}}{dt}=\alpha \left(\dfrac{dy}{dt}x^{∧}+2\dfrac{dx}{dt}y^{∧}\right)$ $=\alpha \left({v}_{y}x^{∧}+2{v}_{x}y^{∧}\right)$ $=\alpha \left[2x\alpha x^{∧}+2\alpha yy^{∧}\right]$ $=2{\alpha }^{2}\left[xx^{∧}+yy^{∧}\right]$ Therefore, required value of $\vec{F}=2m{\alpha }^{2}\left[xx^{∧}+yy^{∧}\right]$.
2. Q2JEE Advanced Adv 2016 (Paper 1)
   The positon vector $\vec{r}$ of a particle of mass m is given by the following equation $\vec{r}\left(t\right)=\alpha {t}^{3}\hat{i}+\beta {t}^{2}\hat{j},where\alpha =\dfrac{10}{3}m{s}^{-3},\beta =5m{s}^{-2}andm=0.1kg.Att=1s,$ which of the following statements (s) is (are) true about the particle?
   1. A.The velocity $\vec{v}$ is given by $\vec{v}=\left(10\hat{i}+10\hat{j}\right)m{s}^{-1}$
   2. B.The angular momentum $\vec{L}$ with respect to the origin is given by $\vec{L}=-\left(\dfrac{5}{3}\right)\hat{k}Nms$
   3. C.The force $\vec{F}$ is given by $\vec{F}=\left(\hat{i}+2\hat{j}\right)N$
   4. D.The torque $\vec{\tau }$ with respect to the origin is given by $\vec{\tau }=-\left(\dfrac{20}{3}\right)\hat{k}Nm$
   Show answer & solution

   Answer: A,B,D

   $\vec{r}=\alpha {t}^{3}\hat{i}+\beta {t}^{2}\hat{j}$ $\vec{v}=\dfrac{d\vec{r}}{dt}=3\alpha {t}^{2}\hat{i}+2\beta t\hat{j}$ $\vec{a}=\dfrac{{d}^{2}\vec{r}}{d{t}^{2}}=6\alpha t\hat{i}+2\beta \hat{j}$ At t = 1 (i) $\vec{v}=3\times \dfrac{10}{3}\times 1\hat{i}+2\times J\times 1\hat{j}$ $=10\hat{i}+10\hat{j}$ (ii) $\vec{L}=\vec{r}\times \vec{p}$ $=\left(\dfrac{10}{3}\times 1\hat{i}+5\times 1\hat{j}\right)\times 0.1\left(10\hat{i}+10\hat{j}\right)$ $=-\dfrac{5}{3}\hat{k}$ (iii) $\vec{F}=m\times \left(6\times \dfrac{10}{3}\times 1\hat{i}+2\times 5\hat{j}\right)=2\hat{i}+\hat{j}$ (iv) $\vec{\tau }=r\times \vec{F}$ $=\left(\dfrac{10}{3}\hat{i}+5\hat{j}\right)\times \left(2\hat{i}+\hat{j}\right)$ $=+\dfrac{10}{3}\hat{k}+10\left(-\hat{k}\right)$ $=-\dfrac{20}{3}\hat{k}$
3. Q3JEE Advanced Adv 2011 (Paper 2)
   A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of $10 \mathrm{~m} / \mathrm{s}$, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 \mathrm{~m}$ inside the train to catch the ball back at the initial height. The acceleration of the train in $\mathrm{m} / \mathrm{s}^2$, is
   Show answer & solution

   Answer: 5

   $\begin{aligned} t & =T=\frac{2 u \sin \theta}{g} \\ & =\frac{2 \times 10 \times \sin 60^{\circ}}{10}=\sqrt{3} \mathrm{~s}\end{aligned}$ Displacement of train in time $t=\frac{1}{2} a t^2$ Displacement of boy with respect to train $$=1.15 \mathrm{~m}$$ $\therefore$ Displacement of boy with respect to ground $=\left(1.15+\frac{1}{2} a t^2\right)$ Displacement of ball with respect to ground $=\left(u \cos 60^{\circ}\right) t$ To catch the ball back at initial height, $$\begin{aligned} 1.15+\frac{1}{2} a t^2 & =\left(u \cos 60^{\circ}\right) t \\ \therefore 1.15+\frac{1}{2} a(\sqrt{3})^2 & =10 \times \frac{1}{2} \times \sqrt{3} \end{aligned}$$ Solving this equation, we get $$a=5 \mathrm{~ms}^{-2}$$ $\therefore$ Answer is 5 . Analysis of Question (i) Question is moderately tough. (ii) Velocity of ball given in the question is with respect to ground.