Motion in a Plane — NEET Physics PYQ MCQs with Solutions

Solution: Let the two vectors be $P$ and $Q$. Given $|P+Q|=|P-Q|$ with angle $\phi$ between them. Then $P^2+Q^2+2PQ\cos\phi=P^2+Q^2-2PQ\cos\phi\Rightarrow 4PQ\cos\phi=0\Rightarrow\cos\phi=0\Rightarrow\phi=\dfrac{\pi}{2}=90^\circ$.

Solution: For perpendicular vectors, $A\cdot B=0$. So $[\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}]\cdot[\cos\dfrac{\omega t}{2}\,\hat{i}+\sin\dfrac{\omega t}{2}\,\hat{j}]=0\Rightarrow\cos\omega t\cos\dfrac{\omega t}{2}+\sin\omega t\sin\dfrac{\omega t}{2}=0$. Using $\cos(A-B)=\cos A\cos B+\sin A\sin B$, $\cos\left(\omega t-\dfrac{\omega t}{2}\right)=0\Rightarrow\cos\dfrac{\omega t}{2}=0\Rightarrow\dfrac{\omega t}{2}=\dfrac{\pi}{2}\Rightarrow t=\dfrac{\pi}{\omega}$.

Solution: If two non-zero vectors are represented by the two adjacent sides of a parallelogram, the resultant is given by the diagonal passing through their point of intersection. Hence $d+e=f$.

Solution: Given $|A\times B|=\sqrt{3}(A\cdot B)\Rightarrow AB\sin\theta=\sqrt{3}\,AB\cos\theta\Rightarrow\tan\theta=\sqrt{3}\Rightarrow\theta=60^\circ$.

Solution: If two vectors are perpendicular, their dot product is zero. Let $a=2\hat{i}+3\hat{j}+8\hat{k}$ and $b=-4\hat{i}+4\hat{j}+\alpha\hat{k}$. Then $a\cdot b=0\Rightarrow(2\hat{i}+3\hat{j}+8\hat{k})\cdot(-4\hat{i}+4\hat{j}+\alpha\hat{k})=0\Rightarrow-8+12+8\alpha=0\Rightarrow 8\alpha=-4\Rightarrow\alpha=-\dfrac{4}{8}=-\dfrac{1}{2}$.

Solution: Given $|A\times B|=\sqrt{3}\,A\cdot B$. Since $|A\times B|=AB\sin\theta$ and $A\cdot B=AB\cos\theta$, we get $AB\sin\theta=\sqrt{3}\,AB\cos\theta\Rightarrow\tan\theta=\sqrt{3}\Rightarrow\theta=60^\circ$. By the parallelogram law, $|A+B|=\sqrt{A^2+B^2+2AB\cos60^\circ}=\sqrt{A^2+B^2+2AB\times\dfrac{1}{2}}=(A^2+B^2+AB)^{1/2}$.

Solution: Let $A$ and $B$ be the two forces. Sum $F_1=A+B$ and difference $F_2=A-B$. Since the sum is perpendicular to the difference, $F_1\cdot F_2=0\Rightarrow(A+B)\cdot(A-B)=0\Rightarrow A^2-B^2=0\Rightarrow A^2=B^2\Rightarrow|A|=|B|$. Thus the forces are equal in magnitude.

Solution: For a unit vector, $|\hat{n}|=1$. So $\sqrt{(0.5)^2+(0.8)^2+c^2}=1\Rightarrow 0.25+0.64+c^2=1\Rightarrow 0.89+c^2=1\Rightarrow c^2=1-0.89=0.11\Rightarrow c=\sqrt{0.11}$.

Solution: Speed is a scalar quantity as it gives no information about the direction of motion. Velocity, displacement and torque each possess both magnitude and direction, so they are vector quantities. Hence speed is not a vector quantity.

Solution: $\cos\theta=\dfrac{A\cdot B}{AB}$. Here $A=\sqrt{3^2+4^2+5^2}=\sqrt{50}$, $B=\sqrt{3^2+4^2+(-5)^2}=\sqrt{50}$ and $A\cdot B=9+16-25=0$. So $\cos\theta=\dfrac{0}{\sqrt{50}\cdot\sqrt{50}}=0\Rightarrow\theta=90^\circ$.

Solution: From the properties of the vector product, the cross product of any vector with the zero vector is a null (zero) vector.

Solution: In a scalar triple product the positions of dot and cross can be interchanged: $A\cdot(B\times A)=(A\times B)\cdot A=(A\times A)\cdot B$. But $A\times A=0$, so $A\cdot(B\times A)=0$. Alternatively, if $A\times B=C$ then $C$ is perpendicular to $A$, so $A\cdot C=0$.

Solution: Given $|\vec{A}|=3$, $|\vec{B}|=4$, $|\vec{C}|=5$ and $\vec{A}+\vec{B}=\vec{C}$. So $5^2=3^2+4^2+2\cdot4\cdot3\cos\theta \Rightarrow \cos\theta=0 \Rightarrow \theta=\dfrac{\pi}{2}$. Thus $\vec{A}$ is perpendicular to $\vec{B}$.

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.