Kinematics 2D – JEE Main Previous Year Questions with Solutions

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Q. A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$ where K is a constant. The general equation for its path is : (1) $\mathrm{y}^{2}=\mathrm{x}^{2}+$ constant (2) $\mathrm{y}=\mathrm{x}^{2}+$ constant (3) $\mathrm{y}^{2}=\mathrm{x}+$ constant (4) xy = constant [AIEEE – 2010]

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Sol. (1) $\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$ $\overrightarrow{\mathrm{v}}=\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$ $\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} \hat{\mathrm{i}}+\frac{\mathrm{dy}}{\mathrm{dt}} \hat{\mathrm{j}}$ $\frac{d x}{d t}=k y \& \frac{d y}{d t}=k x$ $\Rightarrow \frac{d x}{d y}=\frac{y}{x}$ $\Rightarrow \mathrm{y}^{2}=\mathrm{x}^{2}+$ constant

Q. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :- (1) $\frac{\pi}{2} \frac{v^{4}}{g^{2}}$ (2) $\pi \frac{\mathrm{v}^{2}}{\mathrm{g}^{2}}$ ( 3)$\pi \frac{\mathrm{v}^{2}}{\mathrm{g}}$ ( 4)$\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$ [AIEEE – 2011]

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Sol. (4) $\mathrm{r}=\mathrm{R}_{\max }=\frac{\mathrm{v}^{2}}{\mathrm{g}}$ area $=\pi \mathrm{r}^{2}$ $=\pi\left(\frac{\mathrm{v}^{2}}{\mathrm{g}}\right)^{2}$ $=\pi \frac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$

Q. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_{0} \mathrm{e}^{-\mathrm{bt}}$ in the x direction. Its speed v(t) is depicted by which of the following curves ? [AIEEE – 2012]

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Sol. (3) $\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}}$ $\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\frac{\mathrm{F}_{0}}{\mathrm{m}} \int_{0}^{\mathrm{t}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$ $\mathrm{v}=\frac{-\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[\mathrm{e}^{-\mathrm{bt}}-1\right]$ $=\frac{\mathrm{F}_{0}}{\mathrm{m} \mathrm{b}}\left[1-\mathrm{e}^{-\mathrm{bt}}\right]$

Q. A projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) m / s$ where $\hat{\mathbf{i}}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is : (1) $\mathrm{y}=\mathrm{x}-5 \mathrm{x}^{2}$ (2) $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ (3) $4 y=2 x-5 x^{2}$ (4) $4 y=2 x-25 x^{2}$ [AIEEE – 2013]

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Sol. (2) $\mathrm{u}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}$ $\mathrm{u}_{\mathrm{x}}=1$ $\mathrm{u}_{\mathrm{y}}=2$ $\tan \theta=\frac{2}{1}$ $\mathrm{y}=\mathrm{x} \tan \theta-\frac{1}{2} \mathrm{g} \frac{\mathrm{x}^{2}}{\mathrm{u}^{2} \cos ^{2} \theta}$ $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$

Q. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 $\mathrm{M} / \mathrm{S}^{2}$) (The figure are schematic and not drawn to scale) [JEE Mains – 2015]

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Sol. (1) $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ For particle 2 $-240=40 \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$ $5 t^{2}-40 t-240=0$ $\mathrm{t}_{2}=12 \mathrm{sec}$ for $0<\mathrm{t}<8$ sec $\rightarrow \mathrm{a}_{\text {rel }}=0$ straight line x-t graph for $8<\mathrm{t}<12$ sec $\rightarrow \mathrm{a}_{\mathrm{rel}}=-\mathrm{g}$ downward parabola for $\mathrm{t}>12 \mathrm{sec} \rightarrow$ Both particles comes to rest