Motion In Two Dimensions JEE Advanced previous year questions with solutions

2 solved JEE Advanced questions on Motion In Two Dimensions, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 18 in the exam simulator.

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.$\vec{F}=m{\alpha }^{2}\left(xx^{∧}+2yy^{∧}\right)$
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   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 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
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   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.