Motion In A Plane – NEET UG PhysicsPractice Questions & PYQs

To find the angular acceleration of a particle moving in a circular path, we apply the rotational kinematic equation that relates angular velocities to total angular displacement. Since the particle begins from rest, its initial angular velocity is zero, while its final angular velocity $\omega$ is determined by the linear velocity $V_0$ and the radius $r$ through the relation $\omega = \frac{V_0}{r}$.

When the particle completes $n$ full rotations, the total angular displacement is $\theta = 2\pi n$ radians. By substituting these values into the kinematic formula $\omega^2 = \omega_0^2 + 2\alpha\theta$, we express the relationship as:
$\left(\frac{V_0}{r}\right)^2 = 0 + 2\alpha(2\pi n)$
Solving this equation for the angular acceleration results in:
$\alpha = \frac{V_0^2}{4 \pi n r^2}$

Angular speed represents the rate of angular displacement and is determined by the formula $\omega = \frac{2\pi}{T}$, where $T$ signifies the time period for a complete revolution. Because both particles complete their respective circular paths in the exact same amount of time, their angular speeds are equivalent regardless of the size of their orbits or their linear velocities. Dividing the angular speed of particle A by the angular speed of particle B therefore yields a ratio of $1:1$.

To achieve the shortest path when crossing a river, the swimmer must aim upstream at an angle that allows their horizontal velocity component to exactly negate the river's current. This alignment ensures the swimmer's resultant velocity vector points directly across the river, perpendicular to the banks.

With the swimmer's speed in still water at 20 m/s and the river speed at 10 m/s, the relationship between these velocities is given by $\sin \theta = \frac{10}{20}$. This calculation results in $\sin \theta = \frac{1}{2}$, confirming that the required angle is $\theta = 30^{\circ}$ west of north.

Acceleration is defined as the rate at which velocity changes, which corresponds to the second derivative of the position coordinates with respect to time. By differentiating the horizontal position $x=5t-2t^2$ and the vertical position $y=10t$ with respect to time, we obtain the velocity components $v_x=5-4t$ and $v_y=10$. Differentiating these velocity expressions again gives the acceleration components, resulting in a horizontal acceleration of $a_x = -4 \text{ m/s}^2$ and a vertical acceleration of $a_y = 0 \text{ m/s}^2$. Since both components are constant, the total acceleration of the particle at any time, including at 2 s, is $-4 \text{ m/s}^2$.

In non-uniform circular motion, a particle experiences both tangential and centripetal acceleration, which combine as vectors to form the total acceleration. The centripetal acceleration $a_c$ acts as the radial component of this total acceleration and is determined by the relation $a_c = 15 \cos 30^\circ$.

Using the circular motion formula $a_c = \frac{V^2}{R}$, we relate this radial component to the radius of 2.5 m and the particle speed V. Rearranging the formula to $V^2 = a_c R$ and substituting the given values leads to the equation $V^2 = 32.73$. Taking the square root of both sides, we find the speed is 5.7 m/s.

A particle with position vector $\vec{r} = \cos(\omega t)\hat{i} + \sin(\omega t)\hat{j}$ undergoes uniform circular motion. By differentiating the position with respect to time, we find the velocity $\vec{v} = -\omega\sin(\omega t)\hat{i} + \omega\cos(\omega t)\hat{j}$, which is perpendicular to $\vec{r}$ because their dot product is zero. Further differentiation reveals that the acceleration $\vec{a} = -\omega^2\cos(\omega t)\hat{i} - \omega^2\sin(\omega t)\hat{j}$ simplifies to $\vec{a} = -\omega^2\vec{r}$, indicating that the acceleration is always directed toward the origin.

When the magnitude of the sum of two vectors equals the magnitude of their difference, we equate the vector addition and subtraction formulas to solve for the unknown angle $\theta$. Using the standard geometric definitions, we set the squared magnitudes equal to each other:
$A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta$
By cancelling the identical $A^2$ and $B^2$ terms from both sides of the equation, we simplify the relationship to $2AB \cos \theta = -2AB \cos \theta$. Adding $2AB \cos \theta$ to both sides results in $4AB \cos \theta = 0$. Since the individual vectors possess non-zero magnitudes, the equation can only be satisfied if $\cos \theta = 0$, which corresponds to an angle of $90^{\circ}$.

Two vectors are mathematically orthogonal if their dot product is equal to zero, which requires the sum of the products of their respective components to vanish. By taking the dot product of the provided vectors $\vec{A}$ and $\vec{B}$, we obtain the expression $\cos \omega t \cos \left(\frac{\omega t}{2}\right) + \sin \omega t \sin \left(\frac{\omega t}{2}\right) = 0$.

This expression directly matches the trigonometric identity for the cosine of a difference, specifically $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Applying this identity simplifies the equation to $\cos \left(\omega t - \frac{\omega t}{2}\right) = \cos \left(\frac{\omega t}{2}\right) = 0$. Given that the cosine function is zero at $\frac{\pi}{2}$, we set the internal term $\frac{\omega t}{2} = \frac{\pi}{2}$, which leads to the time value $t = \frac{\pi}{\omega}$.

The minimum distance between two objects in relative motion occurs when the relative position vector is perpendicular to the relative velocity vector. Since ship A travels West at 10 km/h and ship B travels North at 10 km/h, the magnitude of the relative velocity of B with respect to A is given by $\sqrt{10^2 + 10^2} = 10\sqrt{2}$ km/h.

Viewed from the frame of ship A, ship B moves along a straight line path starting from a position 100 km South of A. At the point of closest approach, the distance from A to this path is the perpendicular projection of the initial displacement, which is $100 \cos 45^{\circ} = 50\sqrt{2}$ km. Dividing this effective distance by the relative velocity magnitude provides the duration required to reach the point of closest approach, resulting in $t = \frac{50\sqrt{2}}{10\sqrt{2}} = 5$ hours.

The velocity of the particle is found by differentiating the position vector $\overset{\rightarrow}{R} = 4 \sin(2 \pi t) \hat{i} + 4 \cos(2 \pi t) \hat{j}$ with respect to time $t$. This differentiation results in $\overset{\rightarrow}{v} = 8 \pi \cos(2 \pi t) \hat{i} - 8 \pi \sin(2 \pi t) \hat{j}$. Calculating the magnitude of this velocity vector yields $\sqrt{(8 \pi \cos(2 \pi t))^2 + (-8 \pi \sin(2 \pi t))^2} = 8 \pi$ meters per second. Since the statement claiming the velocity magnitude is 8 meters per second contradicts this derived value of $8 \pi$ meters per second, that statement is incorrect.

The path of the particle is a circle of radius 4 meters because the distance from the origin $\sqrt{(4 \sin(2 \pi t))^2 + (4 \cos(2 \pi t))^2}$ simplifies to 4 at any time $t$. Furthermore, the acceleration vector is obtained by differentiating the velocity vector, giving $\overset{\rightarrow}{a} = -16 \pi^2 \sin(2 \pi t) \hat{i} - 16 \pi^2 \cos(2 \pi t) \hat{j}$. This expression is equivalent to $-4 \pi^2 \overset{\rightarrow}{R}$, indicating the acceleration is directed towards the center, opposite to the position vector. Finally, the magnitude of the centripetal acceleration is $16 \pi^2$ meters per square second, which is consistent with the relation $\frac{v^2}{R} = \frac{(8 \pi)^2}{4} = 16 \pi^2$.

Average velocity describes the net displacement per unit of time, making it necessary to focus solely on the start and finish positions of the particle rather than any intermediate path. The position vector is $\vec{r}_i = 2\hat{i} + 3\hat{j}$ at $t = 0$ s and becomes $\vec{r}_f = 13\hat{i} + 14\hat{j}$ at $t = 5$ s, so the displacement $\Delta \vec{r}$ is calculated as the difference between these vectors:
$\Delta \vec{r} = (13 - 2)\hat{i} + (14 - 3)\hat{j} = 11\hat{i} + 11\hat{j}$
meters. With a total time interval of $\Delta t = 5$ s, the average velocity is calculated by dividing this displacement by the duration:
$\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j})$
meters per second.

Two projectiles launched at the same angle follow the exact same path if the curvature of their trajectories is identical. The shape of a parabolic trajectory depends on how quickly gravity pulls the projectile downward relative to its forward motion, which is mathematically dictated by the ratio $\frac{g}{u^2}$. Because the launch angle $\theta$ is fixed, the trajectory equation $y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}$ remains consistent only if the ratio of gravitational acceleration to the square of the launch velocity is equal for both projectiles. Equating these ratios for Earth and the planet provides the relationship $\frac{9.8}{5^2} = \frac{g_{planet}}{3^2}$. Multiplying both sides by $9$ gives $g_{planet} = \frac{9.8 \times 9}{25}$, resulting in a value of $3.528 \text{ m/s}^2$, which rounds to $3.5 \text{ m/s}^2$.

Projectile motion involves independent horizontal and vertical components, where gravity impacts only the vertical direction. Because there are no horizontal forces, the horizontal velocity stays constant throughout the path. At the point where the projectile returns to its starting height, the vertical velocity magnitude is identical to the launch speed, but the direction is inverted.

With an initial velocity of $2\hat{i}+3\hat{j}$ m/s, the horizontal part remains $2\hat{i}$ m/s for the entire flight. At point B, located at the same height as the start, the vertical velocity $3\hat{j}$ reverses to become $-3\hat{j}$ m/s. Consequently, the velocity at point B is $2\hat{i}-3\hat{j}$ m/s.

Two vectors are parallel when their cross product equals zero. The relationship between these vectors is defined by the vector triple product identity:
$\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C}$
Substituting the given conditions $\vec{A} \cdot \vec{B} = 0$ and $\vec{A} \cdot \vec{C} = 0$ into this identity simplifies the expression to $\vec{A} \times (\vec{B} \times \vec{C}) = 0$. This outcome confirms that the vector $\vec{A}$ is parallel to the vector $\vec{B} \times \vec{C}$.

The horizontal range of a projectile is expressed as $R = \frac{2u^2 \sin\theta \cos\theta}{g}$, whereas the vertical maximum height is represented by $H = \frac{u^2 \sin^2\theta}{2g}$. Setting these two expressions equal as required by the problem yields $\frac{2u^2 \sin\theta \cos\theta}{g} = \frac{u^2 \sin^2\theta}{2g}$. By cancelling the common variables $u^2$ and $g$ from both sides and multiplying by two, the relationship simplifies to $4 \sin\theta \cos\theta = \sin^2\theta$. Dividing both sides by $\sin\theta \cos\theta$ results in $4 = \tan\theta$, confirming that the angle of projection is $\theta = \tan^{-1}(4)$.

The motion of a particle under constant acceleration is governed by the vector equation $\vec{v} = \vec{u} + \vec{a}t$, where the final velocity is the sum of the initial velocity and the product of acceleration and time. With an initial velocity of $\vec{u} = 2\hat{i} + 3\hat{j}$ and an acceleration of $\vec{a} = 0.3\hat{i} + 0.2\hat{j}$ acting for 10 seconds, the final velocity vector is calculated by adding the velocity gained over this interval to the starting velocity. Evaluating this expression results in $\vec{v} = (2\hat{i} + 3\hat{j}) + (0.3\hat{i} + 0.2\hat{j})(10)$, which simplifies to $\vec{v} = 5\hat{i} + 5\hat{j}$ after applying the time constant to each component. The magnitude of the resultant velocity is then determined by the square root of the sum of the squares of these components, $\sqrt{5^2 + 5^2}$, yielding a final value of $5\sqrt{2}$ units.

In uniform circular motion, a particle traveling at a constant speed experiences no tangential acceleration, meaning its total acceleration consists entirely of the centripetal acceleration directed toward the center of the circle. This magnitude is defined by the angular velocity $\omega$ and the radius $R$ as $a = \omega^2 R$. The angular velocity derived from the time period $T = 0.2\pi \text{ s}$ is $\omega = \frac{2\pi}{0.2\pi} = 10 \text{ rad/s}$. Using the radius converted to SI units of $R = 0.05 \text{ m}$, the calculation becomes $a = (10)^2 \times 0.05$, which results in a total acceleration of $5 \text{ m/s}^2$.

The elevation angle at the highest point of a projectile path, when viewed from the launch site, creates a right triangle where the vertical side is the maximum height $H$ and the base is half the total horizontal range $R/2$. The tangent of this angle $\phi$ is defined as $\tan \phi = \frac{H}{R/2}$, which simplifies to $\frac{2H}{R}$. Substituting the standard formulas for maximum height, $H = \frac{u^2 \sin^2 \theta}{2g}$, and range, $R = \frac{2 u^2 \sin \theta \cos \theta}{g}$, allows the initial velocity and acceleration due to gravity to cancel out completely.

This yields the expression $\tan \phi = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta}$, which simplifies further to $\frac{1}{2} \tan \theta$. Given the projectile is fired at an angle of $45^{\circ}$, substituting this into the derived formula provides $\tan \phi = \frac{1}{2} \tan 45^{\circ} = 1/2$. Thus, the elevation angle is determined to be $\phi = \tan^{-1}(1/2)$.

The horizontal distance covered by a projectile reaches its theoretical limit when the launch angle is adjusted to 45 degrees. At this specific orientation, the trigonometric component of the standard range formula effectively becomes one, allowing the maximum distance to be determined using the simplified relationship $R_{\max} = \frac{u^2}{g}$.

Substituting the provided values into this equation, we square the initial velocity of 20 m/s to obtain 400 and divide by the gravitational acceleration of 10 m/s². This division confirms the maximum range of the missile is 40 m.

Average acceleration is the rate of change of velocity, which is a vector quantity defined as the change in velocity divided by the time interval. Because the body changes direction from East to North, we determine the magnitude of this change using the Pythagorean theorem for perpendicular vectors: $\Delta v = \sqrt{30^2 + 40^2} = 50$ m/s. By dividing this total change in velocity by the time interval of 10 s, we find the average acceleration: $a_{av} = \frac{50}{10} = 5$ m/s$^2$.