Motion In A Plane – NEET UG PhysicsPractice Questions & PYQs

Determining the relative orientation of the velocity and position vectors depends on their dot product, where a result of 0 confirms that the two vectors are perpendicular. The position vector $\overset{\rightarrow}{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}$ leads to a velocity vector $\overset{\rightarrow}{v} = (-a \omega \sin \omega t) \hat{i} + (a \omega \cos \omega t) \hat{j}$ when differentiated with respect to time. The dot product $\overset{\rightarrow}{v} \cdot \overset{\rightarrow}{r}$ is determined by summing the products of their corresponding components:

$\overset{\rightarrow}{v} \cdot \overset{\rightarrow}{r} = (-a \omega \sin \omega t)(a \cos \omega t) + (a \omega \cos \omega t)(a \sin \omega t)$

Expansion yields $-a^2 \omega \sin \omega t \cos \omega t + a^2 \omega \sin \omega t \cos \omega t$, which equals 0. This result confirms that the velocity vector is perpendicular to the position vector.

To distinguish between scalar and vector quantities, one must determine if a value possesses both a magnitude and a defined spatial direction. While scalar quantities like distance, heat, and energy are fully described by their size alone, angular momentum is inherently rotational and requires a directional component for a complete description. Because it is defined by the cross product relationship $\vec{L} = \vec{r} \times \vec{p}$, it possesses a specific orientation perpendicular to the plane of motion, confirming it is a vector quantity.

Angular speed is determined by the total angle traversed in one complete revolution, which is $2\pi$, divided by the time period $t$ taken to complete it. Since both cars cover a full circle within the same time $t$, their angular speeds are equivalent as defined by the relation $\omega = \frac{2\pi}{t}$. This demonstrates that the specific masses $m_{1}$ and $m_{2}$ and radii $r_{1}$ and $r_{2}$ do not influence the angular speed, resulting in a ratio of $1:1$.

The cross product of any vector $\overset{\rightarrow}{A}$ with a zero value yields a resultant vector that effectively lacks both magnitude and direction, formally defined as a zero vector. Because the magnitude of a cross product is proportional to the product of the magnitudes of the two inputs, multiplying by zero eliminates any potential for a non-zero magnitude. Whether the zero is considered a scalar or a vector, the operational rules of vector algebra dictate that the result of $\overset{\rightarrow}{A} \times 0$ is definitively a zero vector.

Angular velocity is defined as the angle traced by a rotating object per unit time, with one full rotation being equivalent to $2\pi$ radians. To convert a rotational frequency given in revolutions per minute into radians per second, you multiply the number of revolutions by $2\pi$ and divide by 60 to account for the number of seconds in a minute. Applying this conversion to a rate of 120 revolutions per minute results in the expression:

$\omega = \frac{2 \pi \times 120}{60}$

Performing this calculation yields an angular speed of $4 \pi$ rad/s.

When two objects move in opposite directions, the relative speed between them is the sum of their individual speeds because they are closing the gap between them more quickly. The train travels North at 10 m/s, while the parrot flies South at 5 m/s, resulting in a combined relative speed of 15 m/s. Since the parrot must traverse the entire 150 metre length of the train to cross it, the required time is determined by dividing this distance by the relative speed. Applying this to the formula for time gives $t = \frac{150}{15} = 10 \text{ s}$.

Any point on a rotating fan blade experiences centripetal acceleration directed toward the axis, defined by the square of the angular speed multiplied by the distance from the pivot point. The fan rotates at $120 \text{ rpm}$, which is equivalent to $2 \text{ rotations per second}$. Since angular velocity relates to frequency through $\omega = 2 \pi f$, the value is $\omega = 4 \pi \text{ rad s}^{-1}$.

With a blade length of $0.3 \text{ m}$, we determine the acceleration using the relationship $a = \omega^2 r$. Squaring the angular velocity and multiplying by the radius gives $a = (4 \pi)^2 × 0.3$, which evaluates to a magnitude of approximately $47.4 \text{ m s}^{-2}$.

To travel directly across a stream, the swimmer must aim upstream such that their horizontal velocity component cancels the water's flow velocity. With the swimmer moving at 0.5 m/s at an angle of $120^{\circ}$ relative to the water current, we can identify the angle relative to the path perpendicular to the bank as $120^{\circ} - 90^{\circ} = 30^{\circ}$. The horizontal component of the swimmer's velocity is calculated using the sine of this angle, represented by the formula $v = 0.5 \sin(30^{\circ})$. Solving this expression confirms the river's speed is 0.25 m/s.

The motion of a rigid rod constrained to slide along perpendicular rails creates a specific geometric dependency between the velocities of its two endpoints. Because the rod maintains a constant length while moving against these axes, the relationship between the horizontal velocity of particle $A$, given as $10 \text{ m/s}$, and the vertical velocity of particle $B$, denoted as $v_y$, is governed by the inclination angle $\alpha$ through the equation $\tan(\alpha) = v_y / v_x$. By substituting the known values into this relationship, we can express the vertical velocity as $v_y = 10 \times \tan(60^{\circ})$. Evaluating this expression with the value of $\tan(60^{\circ}) \approx 1.732$ yields an upward velocity of $17.3 \text{ m/s}$ for particle $B$.

Circular motion requires a force directed toward the center to maintain the path, defined by the formula $F = \frac{mv^2}{R}$. To solve, first convert the velocity from 36 km/hr to m/s by multiplying by 5/18 to get 10 m/s. Substituting the mass of 500 kg, the velocity of 10 m/s, and the radius of 50 m into the formula gives $F = \frac{500 \times 10^2}{50}$ which results in 1000 N.

The angular displacement $x$ covered by a rotating hand in a fixed time interval is governed by the relation $x = \omega t$, where $\omega$ is the angular speed. By rearranging this expression to examine the ratio of angular speed to angular displacement, we find that $\frac{\omega}{x} = \frac{1}{t}$. Since the time duration $t$ is identical for all three hands of the clock, the ratio $\frac{\omega}{x}$ remains invariant across them. Applying this to the specific values for the second, minute, and hour hands, the relationship consistently evaluates as $\frac{\omega_1}{x_1} = \frac{\omega_2}{x_2} = \frac{\omega_3}{x_3} = \frac{1}{60} = k$.

The tension in a bob whirled in a horizontal circle provides the necessary centripetal force, defined by the formula $T = m\omega^2 r$ where $T$ is the tension, $m$ is mass, $\omega$ is angular speed, and $r$ is the radius. Because mass and radius are constant, the tension is directly proportional to the square of the angular speed, represented as $T \propto \omega^2$. When the tension is increased by a factor of 4, the square of the angular speed must also increase by 4, which means the angular speed itself must double. Since the initial speed is 10 rpm, doubling this value results in a new speed of 20 rpm.

In uniform circular motion, the speed of the particle remains steady, but velocity is a vector quantity that requires both constant magnitude and a fixed direction to be considered invariant. Because the particle follows a curved path, the direction of its velocity $\vec{v}$ shifts at every instant, meaning it is not constant. Similarly, the centripetal acceleration $\vec{a}$ is always directed toward the center of the circular path, and as the particle travels along the circumference, the orientation of this acceleration vector rotates continuously, ensuring that it is also not constant.

To determine how high a projectile climbs, we isolate the vertical component of its initial velocity, as this upward motion is what gravity gradually reduces to zero at the peak. This relationship is defined by the standard physics equation for maximum height:

$H = \frac{u^2 \sin^2 \theta}{2g}$

By substituting the initial speed of 280 m/s, the launch angle of 30 degrees, and the gravitational acceleration of 9.8 m/s² into the formula, we set up the calculation:

$H = \frac{280^2 \times (\sin 30^{\circ})^2}{2 \times 9.8}$

Because sin 30 degrees is 0.5, squaring this value gives 0.25, and squaring the speed yields 78400. Multiplying these together results in 19600, which, when divided by 19.6, equals a maximum height of 1000 m.

When a ball is launched as a projectile, the only force acting on it is gravity, which influences only the vertical component of its velocity. Because gravity has no effect on horizontal motion, the horizontal component of velocity remains constant throughout the flight. At the highest point, the vertical component of velocity becomes zero, so the resultant velocity is equal to the constant horizontal component. Using the formula $v \cos \theta$, the calculation is $20 \times \cos(60^{\circ}) = 20 \times 0.5 = 10$ m/s.

Projectile motion is governed by the principle that the horizontal velocity remains constant throughout the entire path because no acceleration acts in that direction, whereas the vertical velocity component decreases and eventually becomes zero at the maximum height. Since the ball is projected at an angle of $60^{\circ}$ with the vertical, the angle of projection relative to the horizontal is $30^{\circ}$. Applying the formula for the horizontal velocity component, the speed at the peak is calculated as $10 \cos 30^{\circ}$, which results in $5 \sqrt{3} \text{ ms}^{-1}$.

The scalar product of two vectors is determined by the sum of the products of their corresponding components, while the magnitude of the vector product is derived from the length of the resultant vector obtained via the cross product determinant. The scalar product of $\vec{F}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{r}=3\hat{i}+2\hat{j}-2\hat{k}$ results from the operation $(2)(3) + (1)(2) + (-1)(-2)$, which totals 10. Computing the vector product using the determinant:

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 3 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 2) - \hat{j}(-4 + 3) + \hat{k}(4 - 3) = \hat{j} + \hat{k}$

The magnitude of the resulting vector $\hat{j} + \hat{k}$ is calculated as $\sqrt{0^2 + 1^2 + 1^2}$, which simplifies to $\sqrt{2}$.

When an object is launched at an angle, its vertical motion behaves independently of its horizontal progress. The peak elevation reached depends entirely on the initial vertical component of the launch velocity, which gravity reduces until it momentarily reaches zero at the highest point of the trajectory. We determine this vertical velocity by multiplying the total speed by the sine of the launch angle.

Given an initial speed of 20 m/s at an angle of 30 degrees, the initial vertical velocity is calculated as $20 \sin(30^{\circ}) = 10$ m/s. Utilizing the standard kinematic relationship $H = \frac{u^2 \sin^2 \theta}{2 g}$, we substitute the square of the vertical component to get 100 divided by 20, which is twice the acceleration due to gravity of 10 m/s squared. Carrying out this arithmetic reveals the maximum height to be 5 m.

The speed $v$ of a particle completing a circular path of radius $R$ in time $T$ is determined by the circumference divided by the time period, expressed as $v = \frac{2\pi R}{T}$. For a projectile launched with speed $v$ at an angle $\theta$ to the horizontal, the maximum height $H$ is defined by:
$H = \frac{v^2 \sin^2 \theta}{2g}$
Given that $H = 4R$, we equate $4R = \frac{v^2 \sin^2 \theta}{2g}$, which leads to the expression $\sin^2 \theta = \frac{8gR}{v^2}$. Substituting the circular speed $v = \frac{2\pi R}{T}$ into this equation results in $\sin^2 \theta = \frac{8gR}{(2\pi R/T)^2}$, which simplifies to $\sin^2 \theta = \frac{8gR T^2}{4\pi^2 R^2}$. Upon further simplification, this becomes $\sin^2 \theta = \frac{2g T^2}{\pi^2 R}$, and the projection angle is consequently $\theta = \sin^{-1} \left( \frac{2g T^2}{\pi^2 R} \right)^{1 / 2}$.

When an object is projected up a frictionless incline, gravity acts as the decelerating force with a magnitude of $g \sin \theta$. By setting the final velocity to zero in the kinematic formula $v^2 = u^2 - 2ax$, we find that the stopping distance is given by $x = \frac{u^2}{2 g \sin \theta}$. Comparing the two scenarios reveals that the ratio of the distances is inversely proportional to the sines of the angles, resulting in $\frac{x_1}{x_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.