Mix Examples-Motion in Plane Questions in English

(A) Given: Initial speed $u = 50 \, m/s$,angle $\theta = 53^{\circ}$,time $t = 2 \, s$.
The horizontal and vertical components of initial velocity are:
$u_x = u \cos 53^{\circ} = 50 \times \frac{3}{5} = 30 \, m/s$
$u_y = u \sin 53^{\circ} = 50 \times \frac{4}{5} = 40 \, m/s$
At any time $t$,the position coordinates $(x, y)$ of the projectile are given by:
$x = u_x t = 30 \times 2 = 60 \, m$
$y = u_y t - \frac{1}{2} g t^2 = 40 \times 2 - \frac{1}{2} \times 10 \times (2)^2 = 80 - 20 = 60 \, m$
The distance $d$ of the particle from the origin $(0, 0)$ at $t = 2 \, s$ is:
$d = \sqrt{x^2 + y^2} = \sqrt{60^2 + 60^2} = \sqrt{2 \times 60^2} = 60\sqrt{2} \, m$.

(D) Let the particles be at vertices $A, B,$ and $C$ of an equilateral triangle with side $L$. Each particle moves towards the next one. The velocity of particle $A$ is directed towards $B$,and the velocity of particle $B$ is directed towards $C$. The component of the velocity of $A$ along the line $AB$ is $v$,and the component of the velocity of $B$ along $AB$ is $-v \cos(60^{\circ}) = -v/2$. The relative velocity of approach between $A$ and $B$ is $v_{rel} = v - (-v/2) = 3v/2$. The time taken for the particles to meet at the centroid is $t = L / v_{rel} = L / (3v/2) = 2L / (3v)$. Since each particle moves with a constant speed $v$,the distance covered by each particle is $s = v \times t = v \times (2L / 3v) = 2L/3$.

(A) Let the two wires intersect at point $O$ with an angle $\alpha$ between them. Each wire moves perpendicular to its length with a speed $v$.
Consider the bisector of the angle $\alpha$. Due to symmetry,the intersection point $O$ must move along this bisector.
Let the speed of the intersection point be $V$.
Consider one wire. The velocity of the intersection point $V$ makes an angle $\alpha /2$ with the direction of the wire's velocity $v$.
The component of the velocity of the intersection point $V$ along the direction of the wire's motion must be equal to the speed of the wire,$v$.
Therefore,$V \sin(\alpha /2) = v$.
Solving for $V$,we get $V = v / \sin(\alpha /2) = v \csc(\alpha /2)$.

(D) For a particle moving along a curve,the acceleration vector $\vec{a}$ is given by $\vec{a} = \frac{dv}{dt} \hat{t} + \frac{v^2}{R} \hat{n}$,where $v$ is the speed,$R$ is the radius of curvature,$\hat{t}$ is the unit tangent vector,and $\hat{n}$ is the unit normal vector.
If the speed is constant,$\frac{dv}{dt} = 0$,so $\vec{a} = \frac{v^2}{R} \hat{n}$. This acceleration is directed towards the center of curvature (normal direction),not along the tangent. Thus,option $(B)$ is correct.
Since $\vec{a} = \frac{v^2}{R} \hat{n}$,the magnitude of acceleration is $|\vec{a}| = \frac{v^2}{R} = v^2 \kappa$,where $\kappa = \frac{1}{R}$ is the curvature. Thus,the magnitude of acceleration is proportional to the curvature $\kappa$. Thus,option $(C)$ is correct.
Therefore,both $(B)$ and $(C)$ are correct.

(D) The angular velocity is $\omega = v / R$. The angle covered in time $t = \pi R / 3v$ is $\Delta \theta = \omega t = (v / R) \times (\pi R / 3v) = \pi / 3$ radians $(60^\circ)$.
Average velocity is given by $|\vec{v}_{avg}| = |\Delta \vec{r}| / t$. The magnitude of displacement is $|\Delta \vec{r}| = 2R \sin(\Delta \theta / 2) = 2R \sin(30^\circ) = 2R \times (1 / 2) = R$.
Thus,$|\vec{v}_{avg}| = R / (\pi R / 3v) = 3v / \pi$.
Average acceleration is given by $|\vec{a}_{avg}| = |\Delta \vec{v}| / t$. The magnitude of change in velocity is $|\Delta \vec{v}| = 2v \sin(\Delta \theta / 2) = 2v \sin(30^\circ) = 2v \times (1 / 2) = v$.
Thus,$|\vec{a}_{avg}| = v / (\pi R / 3v) = 3v^2 / \pi R$.
Since both $(A)$ and $(B)$ are correct,the answer is $(D)$.

(D) Given: $m = 1 \, kg$,$u = \sqrt{20} \, m/s$,$R = \sqrt{3} \, m$,$g = 10 \, m/s^2$.
Using range formula: $R = \frac{u^2 \sin 2\theta}{g} \Rightarrow \sqrt{3} = \frac{20 \sin 2\theta}{10} \Rightarrow \sin 2\theta = \frac{\sqrt{3}}{2}$.
Thus,$2\theta = 60^\circ$ or $120^\circ$,so $\theta = 30^\circ$ or $60^\circ$.
For $\theta = 30^\circ$: $H = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{20 \times (1/4)}{20} = 0.25 \, m$.
For $\theta = 60^\circ$: $H = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{20 \times (3/4)}{20} = 0.75 \, m$.
Minimum velocity $v_{\min} = u \cos \theta$. For $\theta = 30^\circ$,$v_{\min} = \sqrt{20} \times \frac{\sqrt{3}}{2} = \sqrt{15} \, m/s$.
Time of flight $T = \frac{2u \sin \theta}{g}$. For $\theta = 60^\circ$,$T = \frac{2 \sqrt{20} \times \frac{\sqrt{3}}{2}}{10} = \sqrt{\frac{3}{5}} \, s$.
Maximum potential energy $U_{\max} = mgH_{\max} = 1 \times 10 \times 0.75 = 7.5 \, J$.
Comparing with options: $(a)$ is correct ($0.25 \, m$ is possible),$(b)$ is correct ($\sqrt{15} \, m/s$ is possible),$(c)$ is correct ($\sqrt{3/5} \, s$ is possible),$(d)$ is incorrect $(7.5 \, J \neq 6 \, J)$. Thus,option $(d)$ is the incorrect statement.

(D) $1$. For option $A$: Maximum height $H = u^2 / (2g)$ when $\theta = 90^\circ$. Maximum range $R_{max} = u^2 / g$ when $\theta = 45^\circ$. Thus,$R_{max} = 2H$. Option $A$ is correct.
$2$. For option $B$: $R = (u^2 \sin 2\theta) / g$ and $H = (u^2 \sin^2 \theta) / (2g)$. Given $R = nH$,we have $(2u^2 \sin \theta \cos \theta) / g = n \cdot (u^2 \sin^2 \theta) / (2g)$. This simplifies to $4 \cos \theta = n \sin \theta$,so $\tan \theta = 4/n$ or $\theta = \tan^{-1}(4/n)$. Option $B$ is correct.
$3$. For option $C$: $T = (2u \sin \theta) / g$,so $T^2 = (4u^2 \sin^2 \theta) / g^2$. Also $R = (u^2 \sin 2\theta) / g = (2u^2 \sin \theta \cos \theta) / g$. Then $2R \tan \theta = 2 \cdot [(2u^2 \sin \theta \cos \theta) / g] \cdot [\sin \theta / \cos \theta] = (4u^2 \sin^2 \theta) / g = gT^2$. Option $C$ is correct.
Since all statements are correct,the answer is $D$.

(D) For a projectile launched with initial velocity $u$ at an angle $\theta$ with the horizontal:
$1$. The horizontal displacement is $x = (u \cos \theta)t$. Since $R = (u \cos \theta)T$,we have $x/R = t/T$.
$2$. The vertical displacement is $y = (u \sin \theta)t - \frac{1}{2}gt^2$. Since $h = \frac{u^2 \sin^2 \theta}{2g}$ and $T = \frac{2u \sin \theta}{g}$,we can write $u \sin \theta = \frac{gT}{2}$.
$3$. Substituting $u \sin \theta$ and $g = \frac{2u \sin \theta}{T}$ into the equation for $y$:
$y = \left( \frac{gT}{2} \right)t - \frac{1}{2}g t^2 = \frac{gT^2}{2} \left( \frac{t}{T} \right) - \frac{gT^2}{2} \left( \frac{t}{T} \right)^2 = \frac{gT^2}{2} \left( \frac{t}{T} \right) \left( 1 - \frac{t}{T} \right)$.
$4$. Since $h = \frac{(u \sin \theta)^2}{2g} = \frac{(gT/2)^2}{2g} = \frac{gT^2}{8}$,we have $\frac{gT^2}{2} = 4h$.
$5$. Thus,$y = 4h \left( \frac{t}{T} \right) \left( 1 - \frac{t}{T} \right)$.
$6$. Substituting $t/T = x/R$,we get $y = 4h \left( \frac{x}{R} \right) \left( 1 - \frac{x}{R} \right)$.
Therefore,both $(A)$ and $(B)$ are correct.

(D) The vertical displacement is given by $y = u_y t + \frac{1}{2} a_y t^2$. Since the ball is rolled horizontally,the initial vertical velocity $u_y = 0$. Taking downward as positive,$h = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (0.4)^2 = 0.8 \, m$. So,option $A$ is correct.
The vertical velocity at time $t$ is $v_y = u_y + g t = 0 + 10 \times 0.4 = 4 \, m/s$. So,option $B$ is correct.
The horizontal distance covered is $x = u_x t = 4 \times 0.4 = 1.6 \, m$. So,option $C$ is correct.
Since all statements $A, B,$ and $C$ are correct,the correct answer is $D$.

(B) Given the formulas: $R = \frac{u^2 \sin 2\theta}{g}$,$H = \frac{u^2 \sin^2 \theta}{2g}$,and $T = \frac{2u \sin \theta}{g}$.
Since $u$ is constant,we analyze the trigonometric functions as $\theta$ varies from $30^o$ to $60^o$.
For $R$: The term $\sin 2\theta$ varies from $\sin 60^o = \frac{\sqrt{3}}{2} \approx 0.866$ to $\sin 120^o = \frac{\sqrt{3}}{2} \approx 0.866$,passing through $\sin 90^o = 1$ at $\theta = 45^o$. Thus,$R$ increases until $45^o$ and then decreases.
For $H$: The term $\sin^2 \theta$ increases monotonically as $\theta$ increases from $30^o$ to $60^o$ (since $\sin 30^o = 0.5$ and $\sin 60^o \approx 0.866$). Thus,$H$ increases.
For $T$: The term $\sin \theta$ increases monotonically as $\theta$ increases from $30^o$ to $60^o$. Thus,$T$ increases.
Therefore,$R$ first increases then decreases,while $H$ and $T$ both increase.

(A) For a conical pendulum,the bob moves in a horizontal circle of radius $r = l \sin \theta$,where $l$ is the length of the string and $\theta$ is the angle with the vertical.
The forces acting on the bob are tension $T$ along the string and weight $mg$ downwards.
The vertical component of tension balances the weight: $T \cos \theta = mg \implies T = \frac{mg}{\cos \theta}$.
The horizontal component provides the centripetal force: $T \sin \theta = m \omega^2 r = m \omega^2 (l \sin \theta)$.
From the horizontal component,$T = ml \omega^2$. This implies that for a fixed length $l$,$T$ is directly proportional to $\omega^2$.
However,as $\omega$ increases,the angle $\theta$ increases. At $\omega = 0$,the bob is at rest,and $T = mg$. As $\omega$ increases,the bob starts moving in a circle,and the tension increases as $T = ml \omega^2$ (for small angles,$T \approx mg + \frac{1}{2}ml\omega^2$ is a common approximation,but generally $T$ increases with $\omega^2$).
Looking at the options,option $A$ represents the behavior where $T$ starts at $mg$ (at $\omega=0$) and increases with $\omega^2$ as the pendulum gains angular velocity.

(A) In circular motion,the acceleration of a particle has two components: centripetal acceleration $(a_c)$ directed towards the center and tangential acceleration $(a_t)$ directed along the tangent.
The resultant force $F = ma$ is the vector sum of the centripetal force $(F_c = ma_c)$ and the tangential force $(F_t = ma_t)$.
If the motion is uniform circular motion,$a_t = 0$,so the resultant force is purely centripetal (towards the center).
If the motion is non-uniform circular motion,$a_t \neq 0$,so the resultant force is not directed towards the center.
Therefore,the resultant force may be towards the center (in uniform circular motion) or at an angle to the radius (in non-uniform circular motion).
Option $A$ is correct because it states the force 'may' be towards the center,which covers the uniform case.

(C) Let the angular positions at $t=0$ be $\theta_P(0) = \pi$, $\theta_Q(0) = \pi/2$, and $\theta_R(0) = 0$.
The angular positions at time $t$ are:
$\theta_P(t) = 5\pi t + \pi$
$\theta_Q(t) = 2\pi t + \pi/2$
$\theta_R(t) = 3\pi t + 0$
For particles $P$ and $Q$ to be at the same position: $\theta_P(t) - \theta_Q(t) = 2n\pi$ (where $n$ is an integer).
$3\pi t + \pi/2 = 2n\pi \implies 3t = 2n - 1/2 \implies t = (4n-1)/6$.
For $n=1, t = 3/6 = 1/2 \, s$. For $n=2, t = 7/6 \, s$. For $n=3, t = 11/6 \, s$.
For particles $Q$ and $R$ to be at the same position: $\theta_Q(t) - \theta_R(t) = 2m\pi$.
$-\pi t + \pi/2 = 2m\pi \implies t = 1/2 - 2m$.
For $m=0, t = 1/2 \, s$. For $m=-1, t = 2.5 \, s$.
Comparing the times, they all meet at $t = 1/2 \, s$.

(B) Let the initial angular positions be $\theta_P(0) = \pi$,$\theta_Q(0) = \pi/2$,and $\theta_R(0) = 0$ (taking $R$ at the positive $x$-axis).
The angular positions at time $t$ are given by $\theta(t) = \theta(0) + \omega t$.
$\theta_P(t) = \pi + 5\pi t$
$\theta_Q(t) = \pi/2 + 2\pi t$
Particles $P$ and $Q$ meet when $\theta_P(t) = \theta_Q(t) + 2n\pi$,where $n$ is an integer.
$\pi + 5\pi t = \pi/2 + 2\pi t + 2n\pi$
$3\pi t = 2n\pi - \pi/2$
$3t = 2n - 0.5$
$t = (2n - 0.5) / 3$
For $0 \le t \le 1$:
$0 \le (2n - 0.5) / 3 \le 1$
$0 \le 2n - 0.5 \le 3$
$0.5 \le 2n \le 3.5$
$0.25 \le n \le 1.75$
Since $n$ must be an integer,$n = 1$.
Thus,they meet only once at $t = (2(1) - 0.5) / 3 = 1.5 / 3 = 0.5 \text{ s}$.

(B) Given $K.E. = \frac{1}{2}mv^2 = 2t^2$. Since $m = 1 \, kg$,we have $\frac{1}{2}(1)v^2 = 2t^2$,which implies $v^2 = 4t^2$,so $v = 2t \, m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(2t) = 2 \, m/s^2$. Thus,option $A$ is incorrect.
Power $P = \frac{d(K.E.)}{dt} = \frac{d}{dt}(2t^2) = 4t$. At $t = 2 \, s$,$P = 4(2) = 8 \, W$. Thus,option $B$ is correct.
Tangential force $F_t = m \cdot a_t = 1 \cdot 2 = 2 \, N$. Thus,option $D$ is incorrect.
For the first round,distance $s = 2\pi r = 2\pi(8) = 16\pi \, m$. Since $v = 2t$,$s = \int v \, dt = \int_0^T 2t \, dt = T^2$. Setting $T^2 = 16\pi$,$T = 4\sqrt{\pi} \, s \neq 2 \, s$. Thus,option $C$ is incorrect.

(A) At the highest point,the velocity of the projectile just before the explosion is $v_x = u \cos 53^{\circ} = 50 \times 0.6 = 30 \, m/s$.
Let the mass of the projectile be $2m$. At the highest point,the momentum is $P_i = (2m)v_x = 60m$.
After the explosion,one part of mass $m$ comes to rest $(v_1 = 0)$,and the other part of mass $m$ moves with velocity $v_2$.
By conservation of linear momentum: $P_i = P_f \implies 60m = m(0) + m(v_2) \implies v_2 = 60 \, m/s$.
The radius of curvature is given by $R = \frac{v^2}{a_n}$,where $a_n$ is the normal acceleration.
At the highest point,the only acceleration acting is gravity $(g = 10 \, m/s^2)$,which is perpendicular to the velocity.
Radius of curvature just before explosion: $R_1 = \frac{v_x^2}{g} = \frac{30^2}{10} = 90 \, m$.
Radius of curvature just after explosion for the moving particle: $R_2 = \frac{v_2^2}{g} = \frac{60^2}{10} = 360 \, m$.
The ratio is $\frac{R_1}{R_2} = \frac{90}{360} = \frac{1}{4}$.

(D) Case $1$: When both move in opposite directions.
Let the angular positions be $\theta_1 = \omega t$ and $\theta_2 = 5\omega t$.
They meet when the sum of their angular displacements equals $2\pi$ radians: $\theta_2 + \theta_1 = 2\pi$.
$5\omega t + \omega t = 2\pi \implies 6\omega t = 2\pi \implies \omega t = \frac{\pi}{3} = 60^{\circ}$.
Thus,they cross at points subtending $60^{\circ}$ at the center.
The time interval between crossings is $T = \frac{2\pi}{\omega_{rel}} = \frac{2\pi}{5\omega + \omega} = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega}$.
Case $2$: When both move in the same direction.
They meet when the difference of their angular displacements equals $2\pi$ radians: $\theta_2 - \theta_1 = 2\pi$.
$5\omega t - \omega t = 2\pi \implies 4\omega t = 2\pi \implies \omega t = \frac{\pi}{2} = 90^{\circ}$.
Thus,they cross at points subtending $90^{\circ}$ at the center.
Since all statements are correct,the answer is $D$.

(D) When the particle is thrown vertically upward from the moving cart,it possesses the same horizontal velocity as the cart at the instant of projection. This horizontal velocity is directed tangentially to the circular path.
Since the cart continues to move along the circular path while the particle is in the air,the particle's horizontal motion is linear (tangential),whereas the cart's motion is curved. Consequently,the particle will land outside the circular path.
Additionally,because the particle has both a constant horizontal velocity (tangential) and a vertical velocity influenced by gravity,its trajectory relative to the ground is a parabola. Therefore,both $(B)$ and $(C)$ are correct.

(C) Given the equations of motion:
$x = a \sin(\omega t) \Rightarrow \frac{x}{a} = \sin(\omega t)$
$y = a(1 - \cos(\omega t)) \Rightarrow \frac{y}{a} = 1 - \cos(\omega t) \Rightarrow \frac{y}{a} - 1 = -\cos(\omega t) \Rightarrow 1 - \frac{y}{a} = \cos(\omega t)$
Squaring and adding both equations:
$(\frac{x}{a})^2 + (1 - \frac{y}{a})^2 = \sin^2(\omega t) + \cos^2(\omega t)$
$\frac{x^2}{a^2} + \frac{(a - y)^2}{a^2} = 1$
$x^2 + (y - a)^2 = a^2$
This is the equation of a circle with center $(0, a)$ and radius $a$.

(D) Given: Initial velocity $\vec{u} = (3\hat{i} + 4\hat{j}) \; ms^{-1}$,acceleration $\vec{a} = (0.4\hat{i} + 0.3\hat{j}) \; ms^{-2}$,and time $t = 10 \; s$.
Using the first equation of motion for vectors: $\vec{v} = \vec{u} + \vec{a}t$.
Substituting the values: $\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \times 10$.
$\vec{v} = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) = 7\hat{i} + 7\hat{j}$.
The speed is the magnitude of the velocity vector $\vec{v}$.
$|\vec{v}| = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \; ms^{-1}$.

(A) Given the velocity vector $\vec v = K(y\hat i + x\hat j)$.
We know that $\vec v = v_x \hat i + v_y \hat j = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j$.
Comparing the components,we get $\frac{dx}{dt} = Ky$ and $\frac{dy}{dt} = Kx$.
To find the path of the particle,we divide the two equations: $\frac{dy/dt}{dx/dt} = \frac{Kx}{Ky}$.
This simplifies to $\frac{dy}{dx} = \frac{x}{y}$.
Rearranging the terms,we get $y \, dy = x \, dx$.
Integrating both sides,we obtain $\int y \, dy = \int x \, dx$.
This results in $\frac{y^2}{2} = \frac{x^2}{2} + C'$,where $C'$ is a constant.
Multiplying by $2$,we get $y^2 = x^2 + C$,where $C = 2C'$ is another constant.
Thus,the equation of the path is $y^2 = x^2 + \text{constant}$.

(C) Let the particle move on a circle of radius $R$ with constant angular speed $\omega$. At $t = 0$,the particle is at a position $P_0$. At time $t$,the particle is at position $P$,such that the angle $\angle P_0OP = \theta = \omega t$.
The displacement $S$ is the chord length $P_0P$. In the isosceles triangle $\triangle P_0OP$,the length of the chord is given by $S = 2R \sin(\frac{\theta}{2}) = 2R \sin(\frac{\omega t}{2})$.
This function represents a sine curve with a period $T = \frac{2\pi}{\omega}$.
At $t = 0$,$S = 0$. At $t = \frac{\pi}{\omega}$,$S = 2R \sin(\frac{\pi}{2}) = 2R$ (maximum displacement). At $t = \frac{2\pi}{\omega}$,$S = 2R \sin(\pi) = 0$.
Thus,the graph is a sine curve starting from $0$,reaching a maximum at $t = \frac{\pi}{\omega}$,and returning to $0$ at $t = \frac{2\pi}{\omega}$. This corresponds to the graph shown in option $C$.

(D) The speed $v$ at which the tape moves towards the reel is given by the relation $v = \omega R$,where $\omega$ is the angular velocity of the reel and $R$ is its current radius.
Given that the reel rotates with a constant angular velocity,$\omega = \text{constant}$.
The radius $R$ of the reel is increasing at a steady rate,which means $\frac{dR}{dt} = \text{constant}$.
To find the variation of speed $v$ with time $t$,we differentiate the expression for $v$ with respect to time:
$\frac{dv}{dt} = \frac{d}{dt}(\omega R) = \omega \cdot \frac{dR}{dt}$.
Since both $\omega$ and $\frac{dR}{dt}$ are constants,their product $\frac{dv}{dt}$ is also a constant.
$A$ constant rate of change of speed with respect to time implies that the $v-t$ graph is a straight line with a constant positive slope.

(C) For particle $A$,the path is given by $y = x$.
Differentiating with respect to time $t$,we get $\frac{dy}{dt} = \frac{dx}{dt}$,which implies $v_{y_A} = v_{x_A}$.
Since particle $B$ moves along the $X$-axis with a uniform speed of $3 \ m/s$,its velocity is $\vec{v}_B = 3 \hat{i}$.
Given that the $X$-coordinates of $A$ and $B$ are always equal,the $X$-component of the velocity of $A$ must be equal to the velocity of $B$,so $v_{x_A} = 3 \ m/s$.
Since $v_{y_A} = v_{x_A}$,we have $v_{y_A} = 3 \ m/s$.
Thus,the velocity vector of particle $A$ is $\vec{v}_A = 3 \hat{i} + 3 \hat{j}$.
The speed of particle $A$ is the magnitude of its velocity vector: $|\vec{v}_A| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \ m/s$.

(D) Let the initial velocity of the projectile be $u$ at an angle $\theta$ with the horizontal. The horizontal component of velocity remains constant throughout the motion,$v_x = u \cos \theta$.
At any point,the velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j}$.
For point $P$,$\vec{v}_P = v_x \hat{i} + v_{yP} \hat{j}$. The angle $\alpha$ with the horizontal is given by $\tan \alpha = \frac{v_{yP}}{v_x}$,so $v_{yP} = v_x \tan \alpha$.
For point $Q$,$\vec{v}_Q = v_x \hat{i} + v_{yQ} \hat{j}$.
Since $\vec{v}_P \perp \vec{v}_Q$,their dot product is zero: $\vec{v}_P \cdot \vec{v}_Q = 0$.
$(v_x \hat{i} + v_{yP} \hat{j}) \cdot (v_x \hat{i} + v_{yQ} \hat{j}) = 0 \implies v_x^2 + v_{yP} v_{yQ} = 0$.
Thus,$v_{yQ} = -\frac{v_x^2}{v_{yP}} = -\frac{v_x^2}{v_x \tan \alpha} = -v_x \cot \alpha$.
The magnitudes are $v_P = \sqrt{v_x^2 + v_{yP}^2} = \sqrt{v_x^2 + v_x^2 \tan^2 \alpha} = v_x \sec \alpha$.
And $v_Q = \sqrt{v_x^2 + v_{yQ}^2} = \sqrt{v_x^2 + v_x^2 \cot^2 \alpha} = v_x \csc \alpha$.
Therefore,$\frac{v_Q}{v_P} = \frac{v_x \csc \alpha}{v_x \sec \alpha} = \frac{1/\sin \alpha}{1/\cos \alpha} = \cot \alpha$.

(B) Initial velocity $\vec{v}_1 = 5\hat{i}\ m/s$.
Final velocity $\vec{v}_2 = 5\hat{j}\ m/s$.
Change in velocity $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1 = 5\hat{j} - 5\hat{i}$.
The magnitude of change in velocity is $|\Delta\vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\ m/s$.
The direction of $\Delta\vec{v}$ is North-West.
Average acceleration $\vec{a}_{avg} = \frac{\Delta\vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\ m/s^2$.
Thus,the average acceleration is $\frac{1}{\sqrt{2}}\ m/s^2$ in the North-West direction.

(B) Since the tension force is always perpendicular to the velocity vector of the disc,the work done by the tension force is zero.
According to the work-energy theorem,the kinetic energy and thus the speed of the disc remain constant throughout its motion.
Therefore,the time taken $t$ is given by $t = \frac{s}{v_0}$,where $s$ is the total distance traversed by the disc.
At an arbitrary position,the infinitesimal distance $ds$ covered by the disc is $ds = (l_0 - R\theta) d\theta$.
The total distance $s$ is obtained by integrating from $\theta = 0$ to $\theta = \frac{l_0}{R}$:
$s = \int_{0}^{l_0/R} (l_0 - R\theta) d\theta = [l_0\theta - \frac{1}{2}R\theta^2]_{0}^{l_0/R} = l_0(\frac{l_0}{R}) - \frac{1}{2}R(\frac{l_0}{R})^2 = \frac{l_0^2}{R} - \frac{l_0^2}{2R} = \frac{l_0^2}{2R}$.
Substituting the values $l_0 = 2 \ m$,$R = 1 \ m$,and $v_0 = 1 \ m/s$:
$t = \frac{l_0^2}{2Rv_0} = \frac{2^2}{2 \times 1 \times 1} = \frac{4}{2} = 2 \ s$.

(A) $1$. The path consists of two chords $AB$ and $DA$, and a circular arc $BCD$.
$2$. Given the radius $R = 42\ cm$ and the central angles $\angle BAC = 60^\circ$ and $\angle DAC = 60^\circ$, the total angle subtended by the arc $BCD$ at the center is $240^\circ$ (or $\frac{4\pi}{3}\ radians$).
$3$. Length of chord $AB = 2R \sin(60^\circ/2) = 2R \sin(30^\circ) = R = 42\ cm$. Similarly, $DA = 42\ cm$.
$4$. Length of arc $BCD = R \theta = 42 \times \frac{4\pi}{3} = 56\pi \approx 56 \times 3.14159 = 175.93\ cm$.
$5$. Total distance $D = AB + \text{arc } BCD + DA = 42 + 175.93 + 42 = 259.93\ cm$.
$6$. Time $t = \frac{D}{v} = \frac{259.93}{1.3} \approx 199.95\ s$.
$7$. The value is closest to $200\ s$.

(B) Let the boy's velocity be $\vec{v}_B = v \hat{i}$.
Let the position of the point mass on the circle of radius $R$ be given by $\theta = \omega t$,where $\omega = \frac{v}{R}$.
The velocity of the point mass is $\vec{v}_P = -v \sin(\omega t) \hat{i} + v \cos(\omega t) \hat{j}$.
The relative velocity is $\vec{v}_{rel} = \vec{v}_P - \vec{v}_B = (-v \sin(\omega t) - v) \hat{i} + v \cos(\omega t) \hat{j}$.
The magnitude of relative velocity is $|\vec{v}_{rel}| = \sqrt{(-v \sin(\omega t) - v)^2 + (v \cos(\omega t))^2}$.
$|\vec{v}_{rel}| = \sqrt{v^2 \sin^2(\omega t) + v^2 + 2v^2 \sin(\omega t) + v^2 \cos^2(\omega t)} = \sqrt{2v^2 + 2v^2 \sin(\omega t)} = v \sqrt{2(1 + \sin(\omega t))}$.
The magnitude varies periodically with the angular frequency $\omega = \frac{v}{R}$.
The time period of this variation is $T = \frac{2\pi}{\omega} = \frac{2\pi R}{v}$.
The frequency of change is $f = \frac{1}{T} = \frac{v}{2\pi R}$.
Thus,the correct option is $B$.

(A) The motion of the sphere is analogous to projectile motion where the acceleration is constant in the $x$-direction.
Given: Mass $m = 500 \ g = 0.5 \ kg$,Force $F_x = 0.9 \ N$,Initial velocity $v = 3 \ m/s$ at an angle $\theta = 30^{\circ}$ with the $y$-axis.
The acceleration in the $x$-direction is $a_x = \frac{F_x}{m} = \frac{0.9 \ N}{0.5 \ kg} = 1.8 \ m/s^2$.
The initial velocity component in the $x$-direction is $u_x = v \sin(30^{\circ}) = 3 \times 0.5 = 1.5 \ m/s$.
The sphere crosses the $y$-axis again when its displacement in the $x$-direction becomes zero,i.e.,$s_x = 0$.
Using the equation of motion $s_x = u_x t + \frac{1}{2} a_x t^2$:
$0 = (1.5)t + \frac{1}{2} (-1.8) t^2$ (Note: The force acts in the $+x$ direction,but the initial velocity component is in the $-x$ direction,so acceleration is opposite to initial velocity).
$0 = 1.5 t - 0.9 t^2$
$0.9 t^2 = 1.5 t$
$t = \frac{1.5}{0.9} = \frac{15}{9} = \frac{5}{3} \approx 1.66 \ \sec$.

(A) The horizontal acceleration due to the electric field is $a_x = \frac{qE}{m} = \frac{10^{-6} \times 2 \times 10^7}{2} = 10 \ ms^{-2}$.
The time of flight $T$ is determined by the vertical motion: $T = \frac{2u \sin \theta}{g} = \frac{2 \times 10 \times \sin 45^{\circ}}{10} = \frac{2 \times 10 \times \frac{1}{\sqrt{2}}}{10} = \sqrt{2} \ s$.
The horizontal range $R$ is given by the kinematic equation $R = u_x T + \frac{1}{2} a_x T^2$.
Here,$u_x = u \cos 45^{\circ} = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \ ms^{-1}$.
Substituting the values: $R = (\frac{10}{\sqrt{2}}) \times \sqrt{2} + \frac{1}{2} \times 10 \times (\sqrt{2})^2$.
$R = 10 + \frac{1}{2} \times 10 \times 2 = 10 + 10 = 20 \ m$.

(B) Let the distance between the hinges $A$ and $B$ be $l$. The ring moves in a circular path of radius $R = 1 \ m$ centered at $A$.
When the angle $\phi$ is at its maximum value,the rod $Q$ is tangent to the circular path of the ring.
In this configuration,the triangle formed by the hinges $A, B$ and the position of the ring is a right-angled triangle with the right angle at the ring's position.
Given that the angle at $A$ is $\theta = 60^{\circ}$ at the maximum value of $\phi$,we have:
$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{R}{l}$
$\cos 60^{\circ} = \frac{1}{l}$
$\frac{1}{2} = \frac{1}{l}$
$l = 2 \ m$
Thus,the distance between the hinges $A$ and $B$ is $2 \ m$.

(D) Given velocity: $\vec{v} = 2t\hat{i} + t^2\hat{j}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 2t\hat{j}$.
At $t = 1 \ s$,$\vec{a} = 2\hat{i} + 2\hat{j}$,so magnitude $a = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \ m/s^2$.
Speed $v = |\vec{v}| = \sqrt{(2t)^2 + (t^2)^2} = \sqrt{4t^2 + t^4}$.
At $t = 1 \ s$,$v = \sqrt{4+1} = \sqrt{5} \ m/s$.
Tangential acceleration $a_t = \frac{dv}{dt} = \frac{1}{2\sqrt{4t^2 + t^4}} \cdot (8t + 4t^3)$.
At $t = 1 \ s$,$a_t = \frac{8+4}{2\sqrt{5}} = \frac{12}{2\sqrt{5}} = \frac{6}{\sqrt{5}} \ m/s^2$.
Radial (centripetal) acceleration $a_c = \sqrt{a^2 - a_t^2} = \sqrt{8 - \frac{36}{5}} = \sqrt{\frac{40-36}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \ m/s^2$.
Radius of curvature $R = \frac{v^2}{a_c} = \frac{(\sqrt{5})^2}{2/\sqrt{5}} = \frac{5 \cdot \sqrt{5}}{2} = \frac{5\sqrt{5}}{2} \ m$.
Thus,option $D$ is correct.

(B) For a projectile,the range $R$ is the same for two complementary angles of projection,$\theta$ and $(90^\circ - \theta)$.
The time of flight for angle $\theta$ is $T_1 = \frac{2u \sin \theta}{g}$.
The time of flight for angle $(90^\circ - \theta)$ is $T_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g}$.
Multiplying $T_1$ and $T_2$,we get:
$T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right) = \frac{4u^2 \sin \theta \cos \theta}{g^2} = \frac{2u^2 (2 \sin \theta \cos \theta)}{g^2} = \frac{2u^2 \sin 2\theta}{g^2}$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we can substitute this into the equation:
$T_1 T_2 = \frac{2R}{g}$.
Therefore,$R = \frac{T_1 T_2 g}{2}$.

(D) The angular velocity of $B$ with respect to $A$ is given by $\omega_{BA} = \frac{v_{\perp}}{r}$,where $v_{\perp}$ is the relative velocity of $B$ with respect to $A$ perpendicular to the line joining $A$ and $B$.
Let the line $AB$ be along the $x$-axis.
Velocity of $B$ is $\vec{v}_B = (-100\sqrt{3} \cos 60^{\circ}) \hat{i} + (100\sqrt{3} \sin 60^{\circ}) \hat{j} = -150 \hat{i} + 150 \sqrt{3} \hat{j} \ m/s$.
Velocity of $A$ is $\vec{v}_A = (-50 \cos 30^{\circ}) \hat{i} - (50 \sin 30^{\circ}) \hat{j} = -25\sqrt{3} \hat{i} - 25 \hat{j} \ m/s$.
Relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (-150 + 25\sqrt{3}) \hat{i} + (150\sqrt{3} + 25) \hat{j} \ m/s$.
The component of $\vec{v}_{BA}$ perpendicular to $AB$ (which is along the $y$-axis) is $v_{\perp} = 150\sqrt{3} + 25 \ m/s$.
$\omega_{BA} = \frac{150\sqrt{3} + 25}{20} = \frac{150(1.732) + 25}{20} = \frac{259.8 + 25}{20} = \frac{284.8}{20} = 14.24 \ rad/s$.
Since $14.24 \ rad/s$ is not among the options,the correct choice is $D$.

(B) Let the initial velocity be $u$ at an angle $\alpha$ with the horizontal. The horizontal component is $u_x = u \cos \alpha$ and the vertical component is $u_y = u \sin \alpha$.
After time $t$,let the velocity be $v$. The horizontal component remains constant: $v_x = u \cos \alpha$.
The vertical component becomes $v_y = u \sin \alpha - gt$.
For the velocity to be at right angles to the initial direction,the dot product of the initial velocity vector $\vec{u}$ and the final velocity vector $\vec{v}$ must be zero,or more simply,the slope of the new velocity must be the negative reciprocal of the initial slope.
Alternatively,using the condition that the velocity vector $\vec{v}$ is perpendicular to $\vec{u}$,we have $\vec{u} \cdot \vec{v} = 0$.
$(u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) \cdot (u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}) = 0$
$u^2 \cos^2 \alpha + u \sin \alpha (u \sin \alpha - gt) = 0$
$u^2 \cos^2 \alpha + u^2 \sin^2 \alpha - ugt \sin \alpha = 0$
$u^2 (\cos^2 \alpha + \sin^2 \alpha) = ugt \sin \alpha$
$u^2 = ugt \sin \alpha$
$t = \frac{u}{g \sin \alpha}$

(C) The range of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Given $R = 980 \, m$ and $\theta = 45^o$,we have $980 = \frac{u^2 \sin(90^o)}{g} = \frac{u^2}{g}$.
Thus,$u^2 = 980 \times 9.8 = 9604$,so $u = 98 \, m/s$.
The time of flight is $T = \frac{2u \sin\theta}{g} = \frac{2 \times 98 \times \sin(45^o)}{9.8} = \frac{20 \times 1}{\sqrt{2}} = 10\sqrt{2} \, s$.
When fired from a car moving at $v = 18 \, km/h = 5 \, m/s$,the horizontal component of velocity increases by $\Delta v = 5 \, m/s$.
The increase in range is $\Delta R = \Delta v \times T = 5 \times 10\sqrt{2} = 50\sqrt{2} \, m$.

(B) The maximum height $H$ is given by $H = \frac{u_y^2}{2g}$.
Given $H = 20\, m$ and $g = 10\, m/s^2$,we have $20 = \frac{u_y^2}{2 \times 10}$,so $u_y^2 = 400$,which gives $u_y = 20\, m/s$.
The time of flight $T$ is given by $T = \frac{2u_y}{g} = \frac{2 \times 20}{10} = 4\, s$.
The change in momentum $\Delta \vec{P}$ is equal to the impulse provided by the gravitational force,which acts downwards throughout the motion.
$\Delta \vec{P} = \vec{F}_{ext} \times T = (mg) \times T$.
Substituting the values: $\Delta \vec{P} = 1\, kg \times 10\, m/s^2 \times 4\, s = 40\, N-s$.

(D) Given:
Initial velocity $\vec{u} = (\hat{i} + \hat{j}) \, m/s$
Acceleration $\vec{a} = (\hat{i} + \hat{j}) \, m/s^2$
Time $t = 10 \, s$
Using the first equation of motion for vectors:
$\vec{v} = \vec{u} + \vec{a}t$
Substituting the values:
$\vec{v} = (\hat{i} + \hat{j}) + (\hat{i} + \hat{j}) \times 10$
$\vec{v} = \hat{i} + \hat{j} + 10\hat{i} + 10\hat{j}$
$\vec{v} = 11\hat{i} + 11\hat{j} \, m/s$
The speed is the magnitude of the velocity vector:
$|\vec{v}| = \sqrt{(11)^2 + (11)^2}$
$|\vec{v}| = \sqrt{121 + 121} = \sqrt{242}$
$|\vec{v}| = 11\sqrt{2} \, m/s$

(A) The torque $\tau$ acting on the particle about point $P$ is due to the gravitational force $mg$ acting downwards at a horizontal distance $x$ from $P$.
$\tau = mgx$
Since the horizontal velocity is constant,$x = (V_0 \cos 45^o) t = \frac{V_0}{\sqrt{2}} t$.
Therefore,$\tau = mg \left( \frac{V_0}{\sqrt{2}} t \right) = \frac{mgV_0}{\sqrt{2}} t$.
We know that torque is the rate of change of angular momentum: $\tau = \frac{dL}{dt}$.
Integrating with respect to time from $t = 0$ to $t = \frac{V_0}{g}$:
$L = \int_{0}^{t} \tau dt = \int_{0}^{V_0/g} \frac{mgV_0}{\sqrt{2}} t dt$
$L = \frac{mgV_0}{\sqrt{2}} \left[ \frac{t^2}{2} \right]_{0}^{V_0/g}$
$L = \frac{mgV_0}{\sqrt{2}} \cdot \frac{1}{2} \left( \frac{V_0}{g} \right)^2$
$L = \frac{mgV_0^3}{2 \sqrt{2} g^2} = \frac{1}{2 \sqrt{2}} \frac{mV_0^3}{g}$

(B) $1$. Conservation of Mechanical Energy: The potential energy stored in the compressed spring is converted into the kinetic energy of the block as it leaves the spring.
$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$
$v = \sqrt{\frac{k}{m}} x$
Given: $k = 500\,N/m$,$x = 5.0\,cm = 0.05\,m$,$m = 500\,g = 0.5\,kg$.
$v = \sqrt{\frac{500}{0.5}} \times 0.05 = \sqrt{1000} \times 0.05 = 10\sqrt{10} \times 0.05 = 0.5\sqrt{10}\,m/s$.
$2$. Projectile Motion: The block leaves the table horizontally at height $H = 4\,m$.
The time of flight $t$ is given by $H = \frac{1}{2} gt^2 \implies t = \sqrt{\frac{2H}{g}}$.
$t = \sqrt{\frac{2 \times 4}{10}} = \sqrt{0.8} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\,s$.
$3$. Horizontal Distance $(R)$: $R = v \times t$
$R = (0.5\sqrt{10}) \times (\frac{2}{\sqrt{5}}) = 0.5 \times 2 \times \sqrt{\frac{10}{5}} = 1 \times \sqrt{2} = \sqrt{2}\,m$.
Therefore,the correct option is $B$.

(C) The position vector is $\vec{r} = (t^2 - 8t + 12)\hat{i} + t^2\hat{j}$.
Velocity $\vec{v} = \frac{d\vec{r}}{dt} = (2t - 8)\hat{i} + 2t\hat{j}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 2\hat{i} + 2\hat{j}$.
For the velocity and acceleration vectors to be perpendicular,their dot product must be zero: $\vec{v} \cdot \vec{a} = 0$.
$(2t - 8)(2) + (2t)(2) = 0$.
$4t - 16 + 4t = 0$.
$8t = 16$.
$t = 2 \text{ sec}$.

(D) The radius of the circular path is $R = 10 \, cm = 0.1 \, m$. The speed of the particle is $v = \sqrt{2} \, m/s$.
When the particle covers $\frac{3}{4}$ of the circular path,its displacement $\Delta r$ is the distance between the initial point $A$ and the final point $B$. This forms the hypotenuse of a right-angled triangle with two sides of length $R$.
Displacement $\Delta r = \sqrt{R^2 + R^2} = R\sqrt{2}$.
The distance covered is $d = \frac{3}{4} \times (2\pi R) = \frac{3\pi R}{2}$.
The time taken is $t = \frac{d}{v} = \frac{3\pi R}{2v}$.
The magnitude of average velocity is $|v_{avg}| = \frac{\text{Displacement}}{\text{Time}} = \frac{R\sqrt{2}}{\frac{3\pi R}{2v}} = \frac{2\sqrt{2}v}{3\pi}$.
Substituting $v = \sqrt{2} \, m/s$,we get $|v_{avg}| = \frac{2\sqrt{2} \times \sqrt{2}}{3\pi} = \frac{4}{3\pi} \, m/s$.

(B) The power delivered by a force $\vec{F}$ is given by $P = \vec{F} \cdot \vec{v}$,where $\vec{v}$ is the velocity vector.
Here,the force is gravity,$\vec{F} = m\vec{g} = -mg\hat{j}$.
At point $O$ (start),the velocity $\vec{v}$ has an upward vertical component. Thus,$\vec{F} \cdot \vec{v} = (-mg\hat{j}) \cdot (v_x\hat{i} + v_y\hat{j}) = -mgv_y$. Since $v_y > 0$,the power is negative. Statement $A$ is correct.
At point $A$ (highest point),the velocity is purely horizontal,so $\vec{v} = v_x\hat{i}$. Thus,$P = (-mg\hat{j}) \cdot (v_x\hat{i}) = 0$. Statement $D$ is correct.
At point $O'$ (end),the velocity $\vec{v}$ has a downward vertical component. Thus,$v_y < 0$. Therefore,$P = -mgv_y > 0$. The power is positive. Statement $C$ is correct.
Since statement $C$ is correct,statement $B$ (which says power at $O'$ is negative) is the wrong statement.

(C) Let the mass of each ball be $m$ and the projection speed be $u$.
For the first ball projected vertically upwards, the maximum height reached is $H_1 = \frac{u^2}{2g}$. The potential energy at the highest point is $PE_1 = mgH_1 = mg \left(\frac{u^2}{2g}\right) = \frac{1}{2}mu^2$.
For the second ball projected at an angle $\theta = 60^{\circ}$ with the vertical, the angle with the horizontal is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$. The maximum height reached is $H_2 = \frac{u^2 \sin^2 \alpha}{2g} = \frac{u^2 \sin^2 30^{\circ}}{2g} = \frac{u^2 (1/2)^2}{2g} = \frac{u^2}{8g}$.
The potential energy at the highest point is $PE_2 = mgH_2 = mg \left(\frac{u^2}{8g}\right) = \frac{1}{8}mu^2$.
The ratio of their potential energies is $\frac{PE_1}{PE_2} = \frac{\frac{1}{2}mu^2}{\frac{1}{8}mu^2} = \frac{8}{2} = \frac{4}{1}$.

(A) Initial velocity $\vec{u} = 30 \hat{i} \; m/s$.
Final velocity $\vec{v} = 40 \hat{j} \; m/s$.
Change in velocity $\Delta \vec{v} = \vec{v} - \vec{u} = 40 \hat{j} - 30 \hat{i}$.
Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-30)^2 + (40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \; m/s$.
Average acceleration $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{50 \; m/s}{10 \; s} = 5 \; m/s^2$.

(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2\theta)}{g}$,where $u$ is the initial velocity,$\theta$ is the angle of projection,and $g$ is the acceleration due to gravity.
Since all bodies are projected with the same velocity $u$,the range depends on $\sin(2\theta)$.
For $P$: $\theta = 15^{\circ}$,so $2\theta = 30^{\circ}$,$\sin(30^{\circ}) = 0.5$.
For $Q$: $\theta = 30^{\circ}$,so $2\theta = 60^{\circ}$,$\sin(60^{\circ}) \approx 0.866$.
For $R$: $\theta = 45^{\circ}$,so $2\theta = 90^{\circ}$,$\sin(90^{\circ}) = 1$.
For $S$: $\theta = 60^{\circ}$,so $2\theta = 120^{\circ}$,$\sin(120^{\circ}) = \sin(60^{\circ}) \approx 0.866$.
Comparing the values,$\sin(30^{\circ})$ is the smallest. Therefore,body $P$ has the shortest range.

(A) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Given $R = 5\sqrt{3}\,m$,$u = 10\,m/s$,and taking $g = 10\,m/s^2$:
$5\sqrt{3} = \frac{10^2 \sin 2\theta}{10} \implies 5\sqrt{3} = 10 \sin 2\theta \implies \sin 2\theta = \frac{\sqrt{3}}{2}$.
Thus,$2\theta = 60^{\circ}$ or $2\theta = 120^{\circ}$,which gives $\theta_1 = 30^{\circ}$ and $\theta_2 = 60^{\circ}$.
The time of flight is given by $T = \frac{2u \sin \theta}{g}$.
For $\theta_1 = 30^{\circ}$,$T_1 = \frac{2 \times 10 \times \sin 30^{\circ}}{10} = 2 \times 0.5 = 1\,s$.
For $\theta_2 = 60^{\circ}$,$T_2 = \frac{2 \times 10 \times \sin 60^{\circ}}{10} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}\,s$.
The time interval between the balls striking the ground is $\Delta t = T_2 - T_1 = (\sqrt{3} - 1)\,s$.