Motion in a Plane and Projectile Motion — NEET Physics PYQ MCQs with Solutions

Solution: Distance between the buildings $d=100$ m, height $h=200$ m, speed $v=25$ m/s. As the bullets move towards each other, relative velocity $v_{rel}=25-(-25)=50$ m/s. Time $t=\dfrac{d}{v_{rel}}=\dfrac{100}{50}=2$ s. Vertically the bullet starts from rest ($u=0$) under gravity, so $x=-\dfrac{1}{2}gt^2=-\dfrac{1}{2}\times10\times(2)^2=-20$ m. The negative sign shows they collide 20 m below the top, i.e. at a height $(200-20)=180$ m from the ground after 2 s.

Solution: Using $v^2=u^2-2gh$ with final $v=0$, and for inclined motion effective $g=g\sin\theta$, $h=x$, we get $u^2=2g\sin\theta\,x \Rightarrow x=\dfrac{u^2}{2g\sin\theta}$. So $\dfrac{x_1}{x_2}=\dfrac{2g\sin30^\circ}{2g\sin60^\circ}=\dfrac{1}{2}\times\dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}$ or $1:\sqrt{3}$.

Solution: $v_x=\dfrac{dx}{dt}=\dfrac{d}{dt}(5t-2t^2)=5-4t$, so $a_x=\dfrac{dv_x}{dt}=-4$ m/s$^2$. Also $y=10t \Rightarrow v_y=10$, $a_y=0$. Net acceleration $\vec{a}=a_x\hat{i}+a_y\hat{j}=-4\hat{i}$ m/s$^2$, i.e. $-4$ m/s$^2$ (independent of $t$).

Solution: The shortest distance between the ships is $PQ$. Here $\sin45^\circ=\dfrac{PQ}{OQ} \Rightarrow PQ=100\times\dfrac{1}{\sqrt{2}}=50\sqrt{2}$ km. Relative speed $v_{AB}=\sqrt{v_A^2+v_B^2}=\sqrt{10^2+10^2}=10\sqrt{2}$ km/h. So $t=\dfrac{PQ}{v_{AB}}=\dfrac{50\sqrt{2}}{10\sqrt{2}}=5$ h.

Solution: With $x=4\sin2\pi t$ and $y=4\cos2\pi t$, squaring and adding gives $x^2+y^2=4^2$, so the path is a circle of radius 4 m. Acceleration $\vec{a}=\dfrac{v^2}{R}(-\hat{R})$, so its magnitude is $\dfrac{v^2}{R}$. Velocity components $v_x=4(2\pi)\cos2\pi t$, $v_y=-4(2\pi)\sin2\pi t$ give $v=\sqrt{v_x^2+v_y^2}=8\pi$ m/s (not 8 m/s). Hence statement (c) is wrong.

Solution: The trajectory is $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}$. For identical trajectories at the same angle, $\dfrac{g}{u^2}=$ constant $\Rightarrow \dfrac{9.8}{5^2}=\dfrac{g'}{3^2}$, so $g'=\dfrac{9.8\times9}{25}=3.5$ ms$^{-2}$.

Solution: Position at $t=0$ is $(2\hat{i}+3\hat{j})$ and at $t=5$ s is $(13\hat{i}+14\hat{j})$. $\vec{v}_{av}=\dfrac{\text{Net displacement}}{\text{Time taken}}=\dfrac{(13-2)\hat{i}+(14-3)\hat{j}}{5}=\dfrac{11\hat{i}+11\hat{j}}{5}=\dfrac{11}{5}(\hat{i}+\hat{j})$.

Solution: In projectile motion only the $y$-component of velocity changes, while the $x$-component stays constant. From the figure, at $B$ the $x$-component is unchanged and the $y$-component is reversed, so the velocity at $B$ is $(2\hat{i}-3\hat{j})$ ms$^{-1}$.

Solution: Given range $R=$ maximum height $H$. $R=\dfrac{u^2(2\sin\theta\cos\theta)}{g}$ and $H=\dfrac{u^2\sin^2\theta}{2g}$. Setting $R=H$: $\dfrac{u^2(2\sin\theta\cos\theta)}{g}=\dfrac{u^2\sin^2\theta}{2g} \Rightarrow 2\cos\theta=\dfrac{\sin\theta}{2} \Rightarrow \tan\theta=4 \Rightarrow \theta=\tan^{-1}(4)$.

Solution: For maximum range the projection angle is $45^\circ$, so $R_{max}=\dfrac{u^2}{g}$. With $u=20$ ms$^{-1}$ and $g=10$ ms$^{-2}$, $R_{max}=\dfrac{(20)^2}{10}=\dfrac{400}{10}=40$ m.

Solution: Using $\vec{v}=\vec{u}+\vec{a}t$ with $\vec{u}=3\hat{i}+4\hat{j}$, $\vec{a}=0.4\hat{i}+0.3\hat{j}$, $t=10$ s: $\vec{v}=3\hat{i}+4\hat{j}+10(0.4\hat{i}+0.3\hat{j})=7\hat{i}+7\hat{j}$. So $|\vec{v}|=\sqrt{7^2+7^2}=7\sqrt{2}$.

Solution: $\Delta\vec{p}=m(\vec{v}-\vec{u})$. Only the vertical component reverses, so $|\Delta\vec{p}|=|mv(\cos45^\circ\hat{i}-\sin45^\circ\hat{j})-mv(\cos45^\circ\hat{i}+\sin45^\circ\hat{j})|=2mv\sin45^\circ=\sqrt{2}\,mv$.

Solution: The slope of the path $OA$ gives the angle with the $x$-axis. $\tan\theta=\dfrac{3-0}{\sqrt{3}-0}=\sqrt{3}$, so $\theta=60^\circ$.

Solution: The angles $(45^\circ-\theta)$ and $(45^\circ+\theta)$ are complementary (they add up to $90^\circ$). Horizontal ranges for complementary angles of projection are equal, so the required ratio is $1:1$.

Solution: Range $R=\dfrac{u^2\sin2\theta}{g}$, so $\dfrac{R_1}{R_2}=\dfrac{\sin(2\times30^\circ)}{\sin(2\times60^\circ)}=\dfrac{\sin60^\circ}{\sin120^\circ}=1$, i.e. their ranges are equal ($30^\circ$ and $60^\circ$ are complementary). Heights ratio $=\dfrac{\sin^2 30^\circ}{\sin^2 60^\circ}=\dfrac{1}{3}$ and time-of-flight ratio $=\dfrac{\sin30^\circ}{\sin60^\circ}=\dfrac{1}{\sqrt{3}}$ are unequal.

Solution: Let velocities along the $x$ and $y$ axes be $v_x$ and $v_y$. From the figure, $\tan\alpha=\dfrac{y}{x}\Rightarrow y=x\tan\alpha$. Differentiating w.r.t. $t$, $\dfrac{dy}{dt}=\dfrac{dx}{dt}\tan\alpha\Rightarrow v_y=v_x\tan\alpha$. With $v_x=10$ m/s and $\alpha=60^\circ$, $v_y=10\tan60^\circ=10\sqrt{3}=17.3$ m/s.

Solution: Time taken by the bullet to cover the horizontal distance, $t=\dfrac{100}{1000}=\dfrac{1}{10}$ s. During this time the bullet falls vertically by $h=ut+\dfrac{1}{2}gt^2=0+\dfrac{1}{2}\times10\times(0.1)^2=0.05$ m $=5$ cm.

Solution: Velocity is the rate of change of the position vector, $\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=(-a\omega\sin\omega t)\hat{\mathbf{i}}+(a\omega\cos\omega t)\hat{\mathbf{j}}=\omega[(-a\sin\omega t)\hat{\mathbf{i}}+(a\cos\omega t)\hat{\mathbf{j}}]$. Slope of position vector $=\dfrac{a\sin\omega t}{a\cos\omega t}=\tan\omega t$ and slope of velocity vector $=\dfrac{-a\cos\omega t}{a\sin\omega t}=-\dfrac{1}{\tan\omega t}$, so velocity is perpendicular to the displacement. Alternatively $\mathbf{v}\cdot\mathbf{r}=-a^2\omega\sin\omega t\cos\omega t+a^2\omega\sin\omega t\cos\omega t=0$, hence they are perpendicular.

Solution: Horizontal range $R_1=\dfrac{u^2\sin2\theta}{g}$. For the complementary angle $(90^\circ-\theta)$, $R_2=\dfrac{u^2\sin2(90^\circ-\theta)}{g}=\dfrac{u^2}{g}\sin(180^\circ-2\theta)=\dfrac{u^2}{g}\sin2\theta$. Since $30^\circ$ and $60^\circ$ are complementary, $R_1=R_2$, so $\dfrac{R_1}{R_2}=1$.

Solution: Range $R=\dfrac{u^2\sin2\theta}{g}$; it is maximum at $\theta=45^\circ$, giving $R_{max}=\dfrac{u^2}{g}=16\,\text{km}=16000\,\text{m}$. Hence $u=\sqrt{16000g}=\sqrt{16000\times10}=400\,\text{ms}^{-1}$.