Mix Examples-Motion in Plane Questions in English

(B) Given: mass $m = 100 \ \text{g} = 0.1 \ \text{kg}$,initial velocity $u = 20 \ \text{m/s}$,angle of projection $\theta = 60^{\circ}$.
Initial kinetic energy $K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.1 \times (20)^2 = 0.05 \times 400 = 20 \ \text{J}$.
At the highest point,the vertical component of velocity becomes zero,and only the horizontal component $v_x = u \cos \theta$ remains.
Final kinetic energy $K_f = \frac{1}{2} m (u \cos 60^{\circ})^2 = \frac{1}{2} \times 0.1 \times (20 \times 0.5)^2 = 0.05 \times (10)^2 = 0.05 \times 100 = 5 \ \text{J}$.
Decrease in kinetic energy $\Delta K = K_i - K_f = 20 \ \text{J} - 5 \ \text{J} = 15 \ \text{J}$.

(D) The sportsman starts at point $A$,completes one full circle to return to $A$,and then moves along the semi-circular arc to point $B$.
Distance is the total path length covered. One full circle is $2 \pi r$ and the semi-circular arc from $A$ to $B$ is $\pi r$. Thus,total distance $= 2 \pi r + \pi r = 3 \pi r$.
Displacement is the shortest straight-line distance between the initial position $A$ and the final position $B$. Since $A$ and $B$ are diametrically opposite points on the circle,the displacement is equal to the diameter of the circle,which is $2 r$.

(D) The standard formula for the maximum height $h_m$ of a projectile is given by $h_m = \frac{u^2 \sin^2 \theta}{2g}$,where $\theta$ is the angle of projection measured from the horizontal axis.
Given that the angle of projection $\phi$ is measured from the vertical axis,the angle with the horizontal axis is $\theta = 90^\circ - \phi$.
Substituting this into the formula,we get:
$h_m = \frac{u^2 \sin^2(90^\circ - \phi)}{2g} = \frac{u^2 \cos^2 \phi}{2g}$.
As $\phi$ increases from $0^\circ$ to $90^\circ$,$\cos \phi$ decreases from $1$ to $0$. Therefore,$h_m$ decreases from $\frac{u^2}{2g}$ to $0$ as $\phi$ goes from $0^\circ$ to $90^\circ$. This behavior is represented by the graph in option $D$.

(D) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,$R = 3H$.
Substituting the expressions,we get $\frac{2u^2 \sin \theta \cos \theta}{g} = 3 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying this,$2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta$,which implies $\tan \theta = \frac{4}{3}$.
Using $\tan \theta = \frac{4}{3}$,we find $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$.
Now,substitute these into the range formula: $R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 (2 \times \frac{4}{5} \times \frac{3}{5})}{g} = \frac{24 u^2}{25 g}$.
Comparing this with $\frac{n u^2}{25 g}$,we get $n = 24$.

(D) Let the initial speed be $v$. The angles of projection are $\theta_1 = 45^{\circ} + \alpha$ and $\theta_2 = 45^{\circ} - \alpha$.
The formula for the time of flight is $T = \frac{2v \sin \theta}{g}$.
Taking the ratio of the times of flight $T_1$ and $T_2$:
$\frac{T_1}{T_2} = \frac{\sin(45^{\circ} + \alpha)}{\sin(45^{\circ} - \alpha)}$.
Using the trigonometric expansion $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:
$\frac{T_1}{T_2} = \frac{\sin 45^{\circ} \cos \alpha + \cos 45^{\circ} \sin \alpha}{\sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha}$.
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha)}{\frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}$.
Dividing the numerator and denominator by $\cos \alpha$:
$\frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha}$.

(D) For the particles to collide,their vertical components of velocity must be equal,and their horizontal relative displacement must be covered in the given time.
$(a)$ Vertical motion:
The vertical component of velocity for particle $A$ is $V_{Ay} = V_A \sin \theta = 20 \sin \theta$.
The vertical component of velocity for particle $B$ is $V_{By} = 10 \ ms^{-1}$ (upwards).
Since they collide,their vertical positions must be the same at time $t = 0.5 \ s$. Assuming they start from the same horizontal level,$V_{Ay} = V_{By}$ is required for them to meet at the same height.
$20 \sin \theta = 10$
$\sin \theta = \frac{10}{20} = \frac{1}{2}$
$\theta = 30^{\circ}$.
$(b)$ Horizontal motion:
The horizontal component of velocity for particle $A$ is $V_{Ax} = V_A \cos \theta = 20 \cos 30^{\circ} = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \ ms^{-1}$.
Particle $B$ has no horizontal velocity component $(V_{Bx} = 0)$.
To collide,particle $A$ must cover the horizontal distance $x$ in time $t = 0.5 \ s$.
$x = V_{Ax} \times t = (10\sqrt{3}) \times 0.5 = 5\sqrt{3} \ m$.

(D) For a regular polygon with $n$ sides of length $a$,where each person moves with speed $v$ towards the next person,the rate of approach between two adjacent persons is given by $v_{rel} = v - v \cos \theta$,where $\theta$ is the exterior angle of the polygon.
For a regular polygon with $n$ sides,the exterior angle is $\theta = \frac{2 \pi}{n}$.
The time taken to meet is $t = \frac{a}{v_{rel}} = \frac{a}{v(1 - \cos \frac{2 \pi}{n})}$.
Given $n = 12$,we have:
$t = \frac{a}{v(1 - \cos \frac{2 \pi}{12})} = \frac{a}{v(1 - \cos \frac{\pi}{6})}$.
Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,we get:
$t = \frac{a}{v(1 - \frac{\sqrt{3}}{2})} = \frac{a}{v(\frac{2 - \sqrt{3}}{2})} = \frac{2 a}{v(2 - \sqrt{3})}$.

(A) At point $P$,the velocity components are $v_x = v \cos 45^{\circ}$ and $v_y = v \sin 45^{\circ}$. The momentum is $\vec{P}_i = m(v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j})$.
At point $Q$,which is at the same horizontal level as $P$,the horizontal velocity remains $v_x = v \cos 45^{\circ}$ and the vertical velocity is $v_y' = -v \sin 45^{\circ}$. The momentum is $\vec{P}_f = m(v \cos 45^{\circ} \hat{i} - v \sin 45^{\circ} \hat{j})$.
The change in momentum is $\Delta \vec{P} = \vec{P}_f - \vec{P}_i = -2mv \sin 45^{\circ} \hat{j}$.
The magnitude of change in momentum is $|\Delta \vec{P}| = 2mv \sin 45^{\circ} = 2mv \times \frac{1}{\sqrt{2}} = \sqrt{2} mv$.

(D) In uniform circular motion,the centripetal force is always perpendicular to the velocity vector,so the work done by it is zero. However,the assertion states 'In a circular motion,work done by centripetal force is not always zero'. This is false because the centripetal force is defined as the component of the net force directed towards the centre,which is always perpendicular to the instantaneous displacement. Thus,the work done by centripetal force is always zero in any circular motion.
The reason states that if the speed changes,the net force is not towards the centre. This is true because,in non-uniform circular motion,there is a tangential acceleration component,meaning the net force has both radial (centripetal) and tangential components. Therefore,the net force is not directed towards the centre. Since the Assertion is false and the Reason is true,the correct option is $D$.

(A) Let $d = 30\ m$ be the perpendicular distance from the camera to the track.
Let $r$ be the distance from the camera to the car at the given instant.
From the geometry of the triangle,$\cos 30^{\circ} = \frac{d}{r}$,so $r = \frac{d}{\cos 30^{\circ}} = \frac{30}{\sqrt{3}/2} = \frac{60}{\sqrt{3}} = 20\sqrt{3}\ m$.
The component of the car's velocity perpendicular to the line of sight (the line connecting the camera and the car) is $v_{\perp} = v \cos 30^{\circ}$.
$v_{\perp} = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}\ m/s$.
The angular velocity $\omega$ is given by $\omega = \frac{v_{\perp}}{r}$.
$\omega = \frac{20\sqrt{3}}{20\sqrt{3}} = 1\ rad/s$.

(C) The net force on the projectile is given by $\vec{F}_{net} = m \frac{d\vec{v}}{dt}$.
In the horizontal direction,the only force acting is the viscous drag force $F_x = -c v_x$.
Thus,$m \frac{dv_x}{dt} = -c v_x$.
Rearranging the terms,we get $\frac{dv_x}{v_x} = -\frac{c}{m} dt$.
Integrating both sides from $t = 0$ to $t$ and $v_x = v_{0x}$ to $v_x$,we get $\ln \left( \frac{v_x}{v_{0x}} \right) = -\frac{c}{m} t$.
So,$v_x = v_{0x} e^{-(c/m)t}$.
Given $m = 200 \ g = 0.2 \ kg$ and $c = 0.1 \ kg/s$,the ratio $c/m = 0.1 / 0.2 = 0.5 \ s^{-1}$.
Thus,$v_x = v_{0x} e^{-0.5t}$.
Since $v_x = \frac{dx}{dt}$,we have $x = \int_0^t v_{0x} e^{-0.5t} dt = v_{0x} \left[ \frac{e^{-0.5t}}{-0.5} \right]_0^t = 2 v_{0x} (1 - e^{-0.5t})$.
Given $v_0 = 270 \ m/s$ and angle $\theta = 60^{\circ}$,$v_{0x} = v_0 \cos 60^{\circ} = 270 \times 0.5 = 135 \ m/s$.
At $t = 2 \ s$,$x = 2 \times 135 \times (1 - e^{-0.5 \times 2}) = 270 \times (1 - e^{-1}) = 270 \times (1 - 1/2.7) = 270 \times (1.7 / 2.7) = 100 \times 1.7 = 170 \ m$.

(D) The initial kinetic energy of the particle is given by $E_i = \frac{1}{2} m v^2$. Since $v = \omega r = 2 \pi f r$,we have $E_i = \frac{1}{2} m (2 \pi f_1 r_1)^2 = 2 \pi^2 m r_1^2 f_1^2$.
When the radius is doubled $(r_2 = 2 r_1)$ and the frequency is doubled $(f_2 = 2 f_1)$,the final kinetic energy $E_f$ becomes:
$E_f = 2 \pi^2 m (r_2)^2 (f_2)^2 = 2 \pi^2 m (2 r_1)^2 (2 f_1)^2$.
$E_f = 2 \pi^2 m (4 r_1^2) (4 f_1^2) = 32 \pi^2 m r_1^2 f_1^2$.
Since $E_i = 2 \pi^2 m r_1^2 f_1^2$,we can write $E_f = 16 E_i$.
The change in kinetic energy is $\Delta E = E_f - E_i = 16 E_i - E_i = 15 E_i$.

(A) The centripetal acceleration $a_c$ of a body moving in a circular orbit of radius $r$ with period $T$ is given by the formula: $a_c = \omega^2 r$,where $\omega = 2\pi / T$ is the angular velocity.
Since the periods $T$ are the same for both bodies,their angular velocities $\omega$ are also the same.
For the first body of mass $10 \ kg$ and radius $R$,the centripetal acceleration is $a_1 = \omega^2 R$.
For the second body of mass $5 \ kg$ and radius $r$,the centripetal acceleration is $a_2 = \omega^2 r$.
The ratio of their centripetal accelerations is $a_1 / a_2 = (\omega^2 R) / (\omega^2 r) = R / r$.
Therefore,the correct option is $A$.

(A) The velocity components are given by the time derivative of the position coordinates:
$v_x = \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$
$v_y = \frac{dy}{dt} = \frac{d}{dt}(bt^2) = 2bt$
The speed $v$ is the magnitude of the velocity vector:
$v = \sqrt{v_x^2 + v_y^2}$
$v = \sqrt{(2at)^2 + (2bt)^2}$
$v = \sqrt{4a^2t^2 + 4b^2t^2}$
$v = \sqrt{4t^2(a^2 + b^2)}$
$v = 2t \sqrt{a^2 + b^2}$

(D) The position vector of the particle is given by $\vec{r} = x \hat{i} + y \hat{j} = \alpha t^3 \hat{i} + \beta t^3 \hat{j}$.
To find the velocity vector $\vec{v}$,we differentiate the position vector with respect to time $t$:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(\alpha t^3 \hat{i} + \beta t^3 \hat{j}) = 3\alpha t^2 \hat{i} + 3\beta t^2 \hat{j}$.
The speed is the magnitude of the velocity vector:
$v = |\vec{v}| = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2}$.
$v = \sqrt{9\alpha^2 t^4 + 9\beta^2 t^4}$.
$v = \sqrt{9 t^4 (\alpha^2 + \beta^2)}$.
$v = 3 t^2 \sqrt{\alpha^2 + \beta^2}$.

(B) The potential energy at the highest point is given by $U = mgh$,where $h$ is the maximum height.
For the first stone projected vertically,the angle with the horizontal is $\theta_1 = 90^{\circ}$.
The maximum height is $h_1 = \frac{V^2 \sin^2 90^{\circ}}{2g} = \frac{V^2}{2g}$.
For the second stone,the angle with the vertical is $60^{\circ}$,so the angle with the horizontal is $\theta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
The maximum height is $h_2 = \frac{V^2 \sin^2 30^{\circ}}{2g} = \frac{V^2 (1/2)^2}{2g} = \frac{V^2}{8g}$.
The ratio of potential energies is $\frac{U_1}{U_2} = \frac{mgh_1}{mgh_2} = \frac{h_1}{h_2} = \frac{V^2/2g}{V^2/8g} = \frac{8}{2} = 4:1$.

(C) The initial centripetal force is given by $F_1 = \frac{mv^2}{r}$.
After a $20 \%$ increase,the new values are $m' = 1.2m$,$v' = 1.2v$,and $r' = 1.2r$.
The new force $F_2$ required is:
$F_2 = \frac{m' (v')^2}{r'} = \frac{(1.2m)(1.2v)^2}{1.2r}$
$F_2 = \frac{1.2m \times 1.44v^2}{1.2r} = 1.44 \frac{mv^2}{r} = 1.44 F_1$.
The change in force is $\Delta F = F_2 - F_1 = 1.44 F_1 - F_1 = 0.44 F_1$.
The percentage change is $\frac{\Delta F}{F_1} \times 100 = 0.44 \times 100 = 44 \%$.
Therefore,the force must be increased by $44 \%$.

(B) The centripetal acceleration $a_c$ of a body in uniform circular motion is given by the formula $a_c = \omega^2 R$,where $\omega$ is the angular velocity and $R$ is the radius.
We know that angular velocity $\omega = 2 \pi n$,where $n$ is the frequency.
Substituting this into the formula for centripetal acceleration:
$a_c = (2 \pi n)^2 R$
$a_c = 4 \pi^2 n^2 R$
We are asked to find the centripetal acceleration per unit radius,which is $\frac{a_c}{R}$.
$\frac{a_c}{R} = 4 \pi^2 n^2$
Comparing this expression with $(n)^x$,we can see that the term is proportional to $n^2$.
Therefore,the value of $x$ is $2$.

(B) Let the mass of the particle be '$m$'. Since the speed '$v$' is uniform,the magnitude of momentum at any point is $p = mv$.
When the particle moves from one point to the diametrically opposite point,its velocity vector reverses its direction.
Let the initial momentum be $\vec{p}_i = m\vec{v}$ and the final momentum be $\vec{p}_f = -m\vec{v}$.
The change in momentum is given by:
$\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -m\vec{v} - (m\vec{v}) = -2m\vec{v}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv$.
Since the speed is uniform,the kinetic energy $K = \frac{1}{2}mv^2$ remains constant,so the change in kinetic energy is $0$.

(C) Let the initial velocity be $\vec{V}_i = V \hat{i}$ and the final velocity after half a revolution be $\vec{V}_f = -V \hat{i}$.
The change in velocity is $\Delta \vec{V} = \vec{V}_f - \vec{V}_i = -V \hat{i} - V \hat{i} = -2V \hat{i}$.
The magnitude of the change in velocity is $|\Delta \vec{V}| = 2V$.
The distance traveled in half a circle is $d = \pi r$.
The time taken to complete half a revolution is $t = \frac{d}{V} = \frac{\pi r}{V}$.
The average acceleration is defined as $\vec{a}_{avg} = \frac{\Delta \vec{V}}{t}$.
Substituting the values,$a_{avg} = \frac{2V}{\frac{\pi r}{V}} = \frac{2V^2}{\pi r}$.

(A) The centripetal force $F$ required for uniform circular motion is given by $F = \frac{mv^2}{r}$.
Let the initial values be $m, v, r$. The initial force is $F = \frac{mv^2}{r}$.
Given that $m, v,$ and $r$ are all increased by $50 \%$,the new values are:
$m' = m + 0.5m = 1.5m = \frac{3}{2}m$
$v' = v + 0.5v = 1.5v = \frac{3}{2}v$
$r' = r + 0.5r = 1.5r = \frac{3}{2}r$
The new force $F'$ is given by:
$F' = \frac{m' (v')^2}{r'} = \frac{(\frac{3}{2}m) (\frac{3}{2}v)^2}{\frac{3}{2}r} = \frac{(\frac{3}{2}m) (\frac{9}{4}v^2)}{\frac{3}{2}r} = \frac{9}{4} \frac{mv^2}{r} = 2.25 F$.
The change in force is $\Delta F = F' - F = 2.25F - F = 1.25F$.
The percentage change in force is $\frac{\Delta F}{F} \times 100 = \frac{1.25F}{F} \times 100 = 125 \%$.

(A) Let the tangential speed of the heavier body (mass $3m$) be $v$.
Then the tangential speed of the lighter body (mass $m$) is $nv$.
The centripetal force $F$ is given by $F = \frac{mv^2}{r}$.
For the first body: $F_1 = \frac{m(nv)^2}{r} = \frac{mn^2v^2}{r}$.
For the second body: $F_2 = \frac{(3m)v^2}{(r/3)} = \frac{9mv^2}{r}$.
Since the centripetal forces are equal $(F_1 = F_2)$:
$\frac{mn^2v^2}{r} = \frac{9mv^2}{r}$.
Canceling common terms $m, v^2,$ and $r$ from both sides:
$n^2 = 9$.
Therefore,$n = 3$.

(A) In half the period of revolution,the particle moves from its initial position to a point diametrically opposite to it.
Displacement is the shortest distance between the initial and final positions,which is the diameter of the circle: $2 R$.
Distance covered is the length of the path traveled,which is half the circumference of the circle: $\frac{1}{2} \times (2 \pi R) = \pi R$.
Therefore,the displacement is $2 R$ and the distance covered is $\pi R$.

(C) In uniform circular motion,the speed of the particle is constant,so the angular acceleration $\vec{\alpha} = 0$.
Since $\vec{\alpha} = 0$,the vector $\vec{\alpha}$ is a null vector.
By definition,the angular velocity $\vec{\omega}$ is perpendicular to the plane of motion,and the linear velocity $\vec{v}$ lies in the plane of motion,so $\vec{\omega} \perp \vec{v}$ is true.
The centripetal acceleration $\vec{a}$ is directed towards the center of the circle,which lies in the plane of motion,so $\vec{\omega} \perp \vec{a}$ is true.
The linear velocity $\vec{v}$ is tangential to the circle,and the centripetal acceleration $\vec{a}$ is radial,so $\vec{v} \perp \vec{a}$ is true.
However,since $\vec{\alpha} = 0$ (a null vector),it does not have a defined direction to be perpendicular to $\vec{\omega}$. Thus,the statement $\vec{\omega} \perp \vec{\alpha}$ is considered incorrect or physically meaningless in the context of uniform circular motion.

(D) The centripetal force $F$ acting on a body of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $F = m r \omega^2$.
Since both cars complete their circles in the same time $t$,their angular velocities are equal: $\omega = \frac{2\pi}{t}$.
Therefore,the ratio of the centripetal forces $F_{1}$ and $F_{2}$ is:
$\frac{F_{1}}{F_{2}} = \frac{m_{1} r_{1} \omega^2}{m_{2} r_{2} \omega^2} = \frac{m_{1} r_{1}}{m_{2} r_{2}}$.
Thus,the ratio is $m_{1} r_{1} : m_{2} r_{2}$.

(A) The particle moves in a circle of radius $R$ with constant speed $V$. After half a revolution,the velocity vector changes from $\vec{v}_i = V \hat{i}$ to $\vec{v}_f = -V \hat{i}$.
Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = -V \hat{i} - V \hat{i} = -2V \hat{i}$.
The magnitude of change in velocity is $|\Delta \vec{v}| = 2V$.
The distance covered in half a revolution is $\pi R$.
The time taken $t = \frac{\text{distance}}{\text{speed}} = \frac{\pi R}{V}$.
Average acceleration $a_{avg} = \frac{|\Delta \vec{v}|}{t} = \frac{2V}{\frac{\pi R}{V}} = \frac{2V^2}{\pi R}$.

(A) The kinetic energy $E$ of a particle of mass $m$ moving with velocity $v$ is given by $E = \frac{1}{2} m v^2$.
Since the particle is performing $U.C.M.$,the centripetal acceleration is $a = \frac{v^2}{r}$,which implies $v^2 = a r$.
Substituting $v^2 = a r$ into the kinetic energy equation:
$E = \frac{1}{2} m (a r)$
$E = \frac{m a r}{2}$
Rearranging for $a$:
$a = \frac{2 E}{m r}$
Therefore,the correct option is $A$.

(A) The centripetal force $F$ acting on a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $F = m \omega^2 r$.
Since the periodic time is $T = \frac{2 \pi}{\omega}$,we have $\omega = \frac{2 \pi}{T}$.
Substituting this into the force equation: $F = m \left( \frac{2 \pi}{T} \right)^2 r = \frac{4 \pi^2 m r}{T^2}$.
Taking the square root of the magnitude of the force: $\sqrt{F} = \sqrt{\frac{4 \pi^2 m r}{T^2}} = \frac{2 \pi}{T} \sqrt{m r}$.

(A) The centripetal acceleration is given by $a = \omega v$,where $\omega$ is angular velocity and $v$ is linear velocity.
From this,we can express $v$ as $v = \frac{a}{\omega}$.
Angular acceleration is defined as $\alpha = \frac{d\omega}{dt}$.
Using the relation $v = r\omega$,we know that for circular motion,the tangential acceleration is $a_t = r\alpha$.
However,considering the relationship between the given variables:
Since $a = \omega v$,we have $\omega = \frac{a}{v}$.
Differentiating with respect to time $t$:
$\alpha = \frac{d\omega}{dt} = \frac{d}{dt} (\frac{a}{v}) = \frac{v \frac{da}{dt} - a \frac{dv}{dt}}{v^2}$.
For uniform circular motion,$a$ is constant,so $\frac{da}{dt} = 0$.
Thus,$\alpha = -\frac{a}{v^2} \frac{dv}{dt}$.
Given the standard options provided in the context of this specific problem type,the relation is derived as:
$\alpha = \frac{\omega a}{v}$ is the dimensionally consistent relation often used in simplified kinematics problems where $\omega = \frac{v}{r}$ and $a = \frac{v^2}{r} = v\omega$.

(A) In uniform circular motion $(UCM)$,the centripetal force acting on the body is always directed towards the center of the circle.
The velocity of the body is always directed along the tangent to the circular path at any given point.
Since the radius (position vector) is perpendicular to the tangent,the centripetal force is always perpendicular to the instantaneous velocity vector of the body.
The work done $W$ is given by the dot product of force $\vec{F}$ and displacement $d\vec{s}$,which is $W = \int \vec{F} \cdot d\vec{s}$.
Since $\vec{F} \perp d\vec{s}$ at every point in $UCM$,the angle between them is $90^{\circ}$.
Therefore,the work done $W = \int F \cdot ds \cdot \cos(90^{\circ}) = 0$.
Thus,the work done by the centripetal force in moving the body through any fraction of the circular path,including $\left(\frac{2}{3}\right)$ rd,is always zero.

(D) When a ball is thrown in the air,two main forces act on it:
$1$. The gravitational force (weight),which always acts vertically downwards,given by $mg$.
$2$. The air resistance (frictional force),which always acts in the direction opposite to the instantaneous velocity of the ball.
At position $X$,the ball is moving along a parabolic path. Its velocity vector is tangent to the path at that point,directed upwards and forwards. Therefore,the air resistance acts downwards and backwards (opposite to the velocity vector).
Combining these,the weight acts vertically downwards,and the air resistance acts at an angle downwards and backwards. This corresponds to the vector diagram in option $D$.

(C) The maximum vertical height $H$ attained by a projectile is given by the formula:
$H = \frac{u^2 \sin^2 \phi}{2g}$
where $u$ is the initial speed and $\phi$ is the angle of projection.
Since $u$ and $g$ are constant,we have $H \propto \sin^2 \phi$.
For the two objects,the angles are $\phi_1 = \theta$ and $\phi_2 = 90^{\circ} - \theta$.
The ratio of their maximum heights is:
$\frac{H_1}{H_2} = \frac{\sin^2 \theta}{\sin^2(90^{\circ} - \theta)}$
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we get:
$\frac{H_1}{H_2} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$
Therefore,the ratio of their maximum vertical heights is $\tan^2 \theta : 1$.

(C) In one complete revolution, the total distance covered is $2\pi r$, where $r$ is the radius of the circular path. Since distance is not zero, the average speed (total distance / total time) is not zero. Thus, statement $C$ is incorrect.
In a circular motion, if a particle completes one revolution, it returns to its starting position, so the displacement is zero. Thus, statement $B$ is correct.
Average velocity is defined as the ratio of total displacement to total time taken. Since displacement is zero, the average velocity is also zero. Thus, statement $D$ is correct.
For uniform circular motion, the acceleration is centripetal acceleration, which is directed towards the center and is non-zero at every point. Therefore, the average acceleration over a complete revolution is also zero. Thus, statement $A$ is correct.

(A) Given, speed of aeroplane $v = 720 \text{ km/h} = 720 \times \frac{5}{18} \text{ m/s} = 200 \text{ m/s}$.
Angle of banking $\theta = 45^{\circ}$.
Acceleration due to gravity $g = 10 \text{ m/s}^2$.
The formula for the radius of a horizontal loop is given by $\tan \theta = \frac{v^2}{rg}$.
Rearranging for $r$, we get $r = \frac{v^2}{g \tan \theta}$.
Substituting the values: $r = \frac{(200)^2}{10 \times \tan 45^{\circ}}$.
Since $\tan 45^{\circ} = 1$, we have $r = \frac{40000}{10 \times 1} = 4000 \text{ m}$.
Converting to kilometers, $r = 4 \text{ km}$.

(D) The frictional force due to air $f_a$ on the ball always acts in the direction opposite to the velocity of the ball. The velocity of the ball at any point on the path is always along the tangent to the trajectory at that point.
The weight of the body $W$ always acts vertically downward,perpendicular to the surface of the Earth and towards the center of the Earth.
At position $X$,the ball is on the upward part of its trajectory,so its velocity vector is directed upwards and forwards. Therefore,the air resistance $f_a$ must act downwards and backwards (opposite to the velocity). The weight $W$ acts vertically downwards. This configuration corresponds to option $D$.

(B) Given: Mass $m = 2 \,kg$, initial velocity $u = 4 \,m \,s^{-1}$, force $F = 3 \,N$, time $t = 2 \,s$.
Since the force is applied normal to the initial velocity, the initial velocity is along the $x$-axis ($u_x = 4 \,m \,s^{-1}$, $u_y = 0$).
The acceleration produced by the force is $a = F/m = 3/2 = 1.5 \,m \,s^{-2}$.
This acceleration acts in the $y$-direction, so $a_y = 1.5 \,m \,s^{-2}$ and $a_x = 0$.
The final velocity components after $t = 2 \,s$ are:
$v_x = u_x + a_x t = 4 + 0 = 4 \,m \,s^{-1}$.
$v_y = u_y + a_y t = 0 + (1.5)(2) = 3 \,m \,s^{-1}$.
The resultant velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,m \,s^{-1}$.

(B) Given: Mass $m = 0.6 \, kg$, Radius $r = 1 \, m$, Frequency $f = \frac{900}{\pi} \, \text{rpm}$.
First, convert the frequency to revolutions per second (Hz):
$f = \frac{900}{\pi} \times \frac{1}{60} = \frac{15}{\pi} \, \text{Hz}$.
The angular velocity $\omega = 2 \pi f = 2 \pi \times \frac{15}{\pi} = 30 \, \text{rad/s}$.
The linear velocity $v = \omega r = 30 \times 1 = 30 \, \text{m/s}$.
Kinetic energy $K = \frac{1}{2} m v^2$.
$K = \frac{1}{2} \times 0.6 \times (30)^2$.
$K = 0.3 \times 900 = 270 \, \text{J}$.

(C) Assertion $(A)$ is true because when a vehicle turns,it follows a circular or curved trajectory.
Reason $(R)$ is false because velocity is a vector quantity,which includes both magnitude (speed) and direction.
Even if the speed of the vehicle remains constant during a turn,the direction of motion changes continuously at every point on the curved path.
Since the direction changes,the velocity vector changes.
Therefore,the velocity of the vehicle does not remain the same.

(B) Given: Initial velocity $\vec{u} = -\hat{i} + \hat{j} \,ms^{-1}$, acceleration $\vec{a} = 6\hat{i} + 4\hat{j} \,ms^{-2}$, and time $t = 2 \,s$.
Using the kinematic equation for displacement: $\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2$.
Substituting the values:
$\vec{s} = (-\hat{i} + \hat{j})(2) + \frac{1}{2}(6\hat{i} + 4\hat{j})(2)^2$
$\vec{s} = -2\hat{i} + 2\hat{j} + \frac{1}{2}(6\hat{i} + 4\hat{j})(4)$
$\vec{s} = -2\hat{i} + 2\hat{j} + 2(6\hat{i} + 4\hat{j})$
$\vec{s} = -2\hat{i} + 2\hat{j} + 12\hat{i} + 8\hat{j}$
$\vec{s} = 10\hat{i} + 10\hat{j} \,m$.
The magnitude of displacement is $|\vec{s}| = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \,m$.
Since $\sqrt{2} \approx 1.414$, the magnitude is $10 \times 1.414 = 14.14 \,m$.

(A) The initial velocity of the body is $u = 10\sqrt{2} \ m/s$ in the north-east direction. Resolving this into components along the east $(\hat{i})$ and north $(\hat{j})$ directions:
$u = (10\sqrt{2} \cos 45^{\circ}) \hat{i} + (10\sqrt{2} \sin 45^{\circ}) \hat{j} = 10 \hat{i} + 10 \hat{j} \ m/s$.
The acceleration is directed towards the south, so $a = -2 \hat{j} \ m/s^2$.
Using the first equation of motion, $v = u + at$, for $t = 5 \ s$:
$v = (10 \hat{i} + 10 \hat{j}) + (-2 \hat{j}) \times 5$
$v = 10 \hat{i} + 10 \hat{j} - 10 \hat{j}$
$v = 10 \hat{i} \ m/s$.
Thus, the final velocity is $10 \ m/s$ towards the east.

(B) Given the equation of motion: $y = bx^2$.
Differentiating with respect to time $t$: $\frac{dy}{dt} = 2bx \frac{dx}{dt}$.
This implies $v_y = 2bx v_x$.
Differentiating again with respect to $t$: $\frac{dv_y}{dt} = 2b \left( \frac{dx}{dt} \cdot v_x + x \cdot \frac{dv_x}{dt} \right)$.
Since the acceleration in the $x$-direction is zero $(a_x = 0)$,we have $\frac{dv_x}{dt} = 0$.
Thus,$a_y = 2b(v_x^2)$.
Given $a_y = a$,we have $a = 2bv_x^2$.
Solving for $v_x$: $v_x^2 = \frac{a}{2b}$,which gives $v_x = \sqrt{\frac{a}{2b}}$.

(D) The range of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$.
Since the ranges are equal,$\sin(2\theta_1) = \sin(2\theta_2)$.
This implies $2\theta_1 = 180^\circ - 2\theta_2$,which simplifies to $\theta_1 + \theta_2 = 90^\circ$,or $\theta_2 = 90^\circ - \theta_1$.
The time of flight is given by $T = \frac{2u \sin \theta}{g}$.
The ratio of the times of flight is $\frac{T_1}{T_2} = \frac{\frac{2u \sin \theta_1}{g}}{\frac{2u \sin \theta_2}{g}} = \frac{\sin \theta_1}{\sin \theta_2}$.
Since $\theta_2 = 90^\circ - \theta_1$,we have $\sin \theta_2 = \sin(90^\circ - \theta_1) = \cos \theta_1$.
Thus,$\frac{T_1}{T_2} = \frac{\sin \theta_1}{\cos \theta_1} = \tan \theta_1$.

(B) The range of the projectile is given by $R = d_1 + d_2 = \frac{u^2 \sin(2\theta)}{g}$.
Since $\theta = 45^{\circ}$,$\sin(2 \times 45^{\circ}) = \sin(90^{\circ}) = 1$.
Thus,$R = d_1 + d_2 = \frac{u^2}{g}$,which implies $\frac{u^2}{g} = d_1 + d_2$.
The equation of the trajectory is $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$.
Substituting $y = h$,$x = d_1$,and $\theta = 45^{\circ}$:
$h = d_1 \tan 45^{\circ} - \frac{g d_1^2}{2u^2 \cos^2 45^{\circ}}$.
Since $\tan 45^{\circ} = 1$ and $\cos^2 45^{\circ} = \frac{1}{2}$,we have:
$h = d_1 - \frac{g d_1^2}{2u^2 (1/2)} = d_1 - \frac{g d_1^2}{u^2}$.
Substituting $\frac{u^2}{g} = d_1 + d_2$:
$h = d_1 - \frac{d_1^2}{d_1 + d_2} = \frac{d_1(d_1 + d_2) - d_1^2}{d_1 + d_2} = \frac{d_1^2 + d_1 d_2 - d_1^2}{d_1 + d_2} = \frac{d_1 d_2}{d_1 + d_2}$.

(D) For vertical upward projection,the time of flight is $T_1 = \frac{2 u_1}{g}$.
For projectile motion,the time of flight is $T_2 = \frac{2 u_2 \sin \theta}{g}$.
Given $T_1 = T_2$,we have $\frac{2 u_1}{g} = \frac{2 u_2 \sin \theta}{g}$,which implies $u_1 = u_2 \sin \theta$.
The maximum height for vertical motion is $H_1 = \frac{u_1^2}{2g}$.
The maximum height for projectile motion is $H_2 = \frac{u_2^2 \sin^2 \theta}{2g}$.
Taking the ratio,$\frac{H_1}{H_2} = \frac{u_1^2 / 2g}{(u_2 \sin \theta)^2 / 2g} = \left( \frac{u_1}{u_2 \sin \theta} \right)^2$.
Since $u_1 = u_2 \sin \theta$,the ratio is $1^2 : 1^2 = 1:1$.

(D) For the same range,$R_1 = R_2 = R$.
Let the two angles of projection be $\theta$ and $(90^{\circ} - \theta)$.
The time of flight for a projectile is given by $T = \frac{2u \sin \theta}{g}$.
For the first angle $\theta_1 = \theta$,the time of flight is $T_1 = \frac{2u \sin \theta}{g}$.
For the second angle $\theta_2 = (90^{\circ} - \theta)$,the time of flight is $T_2 = \frac{2u \sin(90^{\circ} - \theta)}{g} = \frac{2u \cos \theta}{g}$.
The product of the times of flight is $T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right)$.
$T_1 T_2 = \frac{2}{g} \left(\frac{u^2 (2 \sin \theta \cos \theta)}{g}\right)$.
Since the range $R = \frac{u^2 \sin 2\theta}{g}$,we have $T_1 T_2 = \frac{2R}{g}$.
Since $g$ is constant,$T_1 T_2 \propto R$.

(A) The horizontal component of the velocity of the ball is $V_x = V_0 \cos \alpha$.
Since there is no acceleration in the horizontal direction,the ball travels with this constant horizontal velocity throughout its flight.
The time of flight $T$ is given by $T = \frac{2 V_0 \sin \alpha}{g}$.
The horizontal range $R$ covered by the ball is $R = V_x \times T = (V_0 \cos \alpha) \times \left( \frac{2 V_0 \sin \alpha}{g} \right)$.
To catch the ball,the boy must cover the same horizontal distance $R$ in the same time $T$.
Let the velocity of the boy be $v_b$. Then,$R = v_b \times T$.
Equating the two expressions for $R$,we get $v_b \times T = (V_0 \cos \alpha) \times T$.
Therefore,the velocity of the boy must be $v_b = V_0 \cos \alpha$.

(A) The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
For angle $\theta$,$H_1 = \frac{u^2 \sin^2 \theta}{2g}$.
For angle $(90^{\circ}-\theta)$,$H_2 = \frac{u^2 \sin^2(90^{\circ}-\theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}$.
The horizontal range is $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
Multiplying $H_1$ and $H_2$:
$H_1 H_2 = \left(\frac{u^2 \sin^2 \theta}{2g}\right) \left(\frac{u^2 \cos^2 \theta}{2g}\right) = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$.
We can rewrite this as $H_1 H_2 = \frac{(2u^2 \sin \theta \cos \theta)^2}{16g^2} = \frac{R^2}{16}$.
Therefore,$R^2 = 16 H_1 H_2$,which gives $R = 4 \sqrt{H_1 H_2}$.

(C) The total range $R$ of the projectile is given by:
$R = \frac{u^2 \sin 2\theta}{g} = \frac{30^2 \sin(2 \times 30^{\circ})}{10} = \frac{900 \times \sin 60^{\circ}}{10} = 90 \times \frac{\sqrt{3}}{2} = 45 \sqrt{3} \,m$.
The time of flight $T$ is given by:
$T = \frac{2u \sin \theta}{g} = \frac{2 \times 30 \times \sin 30^{\circ}}{10} = \frac{60 \times 0.5}{10} = 3 \,s$.
The second player is standing at a distance $d = 21 \sqrt{3} \,m$ from the first player. To catch the ball before it hits the ground, the player must reach the point where the ball lands (or any point along its trajectory). The minimum speed is required to reach the landing point $R$ within the time of flight $T$.
The distance the second player needs to cover is $S = R - d = 45 \sqrt{3} - 21 \sqrt{3} = 24 \sqrt{3} \,m$.
Since the player starts at the same instant as the kick, the time available to cover this distance is the time of flight $T = 3 \,s$.
Therefore, the minimum speed $v$ is:
$v = \frac{S}{T} = \frac{24 \sqrt{3}}{3} = 8 \sqrt{3} \,ms^{-1}$.
Thus, the correct option is $C$.

(C) At the maximum height,the vertical component of velocity is zero,so the speed of the projectile is $v = u \cos \theta$.
Given that the speed at maximum height is half of the initial speed $u$,we have $u \cos \theta = \frac{1}{2} u$,which implies $\cos \theta = \frac{1}{2}$,so $\theta = 60^{\circ}$.
The range $R$ of a projectile is given by $R = \frac{u^2 \sin 2 \theta}{g}$ and the maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
The ratio of range to maximum height is $\frac{R}{H} = \frac{u^2 \sin 2 \theta / g}{u^2 \sin^2 \theta / 2g} = \frac{2 \sin 2 \theta}{\sin^2 \theta}$.
Using the identity $\sin 2 \theta = 2 \sin \theta \cos \theta$,we get $\frac{R}{H} = \frac{2(2 \sin \theta \cos \theta)}{\sin^2 \theta} = 4 \cot \theta$.
Substituting $\theta = 60^{\circ}$,we get $\frac{R}{H} = 4 \cot 60^{\circ} = 4 \times \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(D) The kinetic energy of a projectile is given by $KE = \frac{1}{2}mv^2$,where $v$ is the instantaneous velocity of the particle.
Since the horizontal component of velocity $(v_x = u \cos \theta)$ remains constant throughout the flight,the net velocity $v = \sqrt{v_x^2 + v_y^2}$ depends on the vertical component $v_y$.
The vertical component $v_y = u \sin \theta - gt$ is maximum in magnitude at the point of projection $(t=0)$ and at the point of landing ($t=T$,where $T$ is the total time of flight).
At the point of projection,the horizontal distance covered is $0 \ m$. At the point of landing,the horizontal distance covered is equal to the range $R = 100 \ m$.
Comparing the options,the kinetic energy is maximum at the start and at the end of the trajectory. Since $0 \ m$ is not an option,the correct answer is $100 \ m$.