Motion In A Plane – NEET UG PhysicsPractice Questions & PYQs

Determining the geometric path of a particle described by parametric equations requires eliminating the time-dependent variable by relating $x$ and $y$ through a trigonometric identity. Given the equations $x = a \sin \omega t$ and $y = a \cos \omega t$, we isolate the trigonometric functions by writing $\frac{x}{a} = \sin \omega t$ and $\frac{y}{a} = \cos \omega t$. Squaring both sides and adding them yields $\frac{x^2}{a^2} + \frac{y^2}{a^2} = \sin^2 \omega t + \cos^2 \omega t$, which simplifies to:
$x^2 + y^2 = a^2$
Since this result defines the relationship between coordinates for a circle centered at the origin with radius $a$, the particle follows a circular path.

The first equation of motion in vector form, $\vec{v} = \vec{u} + \vec{a}t$, allows us to find the final velocity by combining the initial velocity with the change over time caused by constant acceleration. Starting with an initial velocity of $\vec{u} = 3\hat{\imath} + 4\hat{\jmath}$ and an acceleration of $\vec{a} = 0.4\hat{\imath} + 0.3\hat{\jmath}$, applying the acceleration over 10 seconds provides an additional velocity component of $4\hat{\imath} + 3\hat{\jmath}$. The total velocity after this interval becomes $\vec{v} = 7\hat{\imath} + 7\hat{\jmath}$. Because speed corresponds to the magnitude of the velocity vector, we calculate it as $\sqrt{7^2 + 7^2}$, simplifying to $7\sqrt{2}$ units.

The triangle law of vector addition dictates that when two vectors are connected head-to-tail, the resultant vector is drawn from the starting tail of the first to the ending head of the second. By following the sequence shown in the diagram, vector $\overset{\rightarrow}{d}$ connects to the tail of vector $\overset{\rightarrow}{e}$, creating a path that begins at the start of $\overset{\rightarrow}{d}$ and ends at the tip of $\overset{\rightarrow}{e}$. Since $\overset{\rightarrow}{f}$ completes this path by bridging the initial position of $\overset{\rightarrow}{d}$ directly to the final position of $\overset{\rightarrow}{e}$, the resulting relationship is expressed as $\overset{\rightarrow}{d}+\overset{\rightarrow}{e}=\overset{\rightarrow}{f}$.

In projectile motion, the velocity vector is decomposed into horizontal and vertical components, with the horizontal component remaining constant throughout the flight. Because the vertical velocity reaches zero at the maximum height, the speed of the projectile at that point is determined entirely by the initial horizontal component, $u \cos \theta$. Equating this to half the initial speed gives $u \cos \theta = u/2$, which simplifies to $\cos \theta = 1/2$ and results in a projection angle of $60^{\circ}$.

The instantaneous velocity of a particle is represented by the slope of the tangent to its distance-time graph, which is calculated as $v = \frac{ds}{dt}$. This relationship means the velocity is greatest at the location where the curve is the steepest. Because the slope reaches its peak steepness at point C, the particle achieves its maximum instantaneous velocity at that point.

The change in momentum is determined by calculating the vector difference between the final and initial states, where momentum is defined as the product of mass and velocity. Since the horizontal component of velocity remains constant throughout projectile motion, the horizontal momentum does not change, meaning the net variation depends entirely on the vertical components.

When the particle returns to level ground, its vertical velocity has reversed from $v\sin\theta$ directed upwards to $v\sin\theta$ directed downwards. This results in a vertical momentum change of $mv\sin\theta - (-mv\sin\theta) = 2mv\sin\theta$. Using the given projection angle of $45^\circ$, the magnitude of the change is $2mv \cdot \frac{1}{\sqrt{2}}$, which simplifies to $mv\sqrt{2}$.

When the magnitude of the sum of two vectors equals the magnitude of their difference, squaring both sides of the equation $\left|\overset{\rightarrow}{A}+\overset{\rightarrow}{B}\right|=\left|\overset{\rightarrow}{A}-\overset{\rightarrow}{B}\right|$ is the most direct way to isolate the term containing the angle between them. Expanding this results in $\left|\overset{\rightarrow}{A}\right|^{2} + \left|\overset{\rightarrow}{B}\right|^{2} + 2\overset{\rightarrow}{A}\cdot\overset{\rightarrow}{B} = \left|\overset{\rightarrow}{A}\right|^{2} + \left|\overset{\rightarrow}{B}\right|^{2} - 2\overset{\rightarrow}{A}\cdot\overset{\rightarrow}{B}$. Canceling identical terms on both sides leaves $4\overset{\rightarrow}{A}\cdot\overset{\rightarrow}{B} = 0$, which simplifies to $AB\cos\theta = 0$ and confirms that $\theta = 90^{\circ}$.

The horizontal range of a projectile, expressed as $R = \frac{u^{2} \sin 2\alpha}{g}$ where $\alpha$ is the angle of projection, is governed by the sine of double the projection angle. Substituting the first angle, $45^{\circ}-\theta$, into this formula yields $R_1 = \frac{u^{2} \sin(90^{\circ}-2\theta)}{g}$, which simplifies to $\frac{u^{2} \cos 2\theta}{g}$ because $\sin(90^{\circ}-x) = \cos x$. Similarly, substituting the second angle, $45^{\circ}+\theta$, yields $R_2 = \frac{u^{2} \sin(90^{\circ}+2\theta)}{g}$, which also simplifies to $\frac{u^{2} \cos 2\theta}{g}$ since $\sin(90^{\circ}+x) = \cos x$. Because the resulting horizontal ranges are identical for both angles of projection, their ratio is $1:1$.

Uniform circular motion involves an object traveling in a circular path at a constant speed, where the continuous change in velocity direction requires a radial acceleration. The stone completes 22 revolutions in 44 seconds, which establishes a time period $T = 44/22 = 2$ seconds per revolution. From this period, the angular velocity is calculated using $\omega = 2\pi/T$, resulting in $\omega = 2\pi/2 = \pi$ radians per second.

The magnitude of this acceleration is determined by the formula $a = r\omega^2$. Given the radius $r = 1$ meter and the angular velocity $\omega = \pi$ radians per second, the calculation yields $a = 1 \times \pi^2 = \pi^2$ meters per second squared. In this scenario, the acceleration is $\pi^2$ meters per second squared and is directed along the radius away from the center.

Two vectors are perpendicular only when their dot product equals zero, as the cosine of the 90-degree angle between them is zero. Setting the dot product of the vectors $(2 \hat{i} + 3 \hat{j} + 8 \hat{k})$ and $(-4 \hat{i} + 4 \hat{j} + \alpha \hat{k})$ to zero gives the equation $(2)(-4) + (3)(4) + (8)(\alpha) = 0$.

Computing these products results in $-8 + 12 + 8 \alpha = 0$, which simplifies to $4 + 8 \alpha = 0$. Isolating $\alpha$ leads to $8 \alpha = -4$, resulting in the final value of $\alpha = -1 / 2$.

The vector product of any two vectors produces a resultant vector that is geometrically perpendicular to the plane containing those original vectors. Because the vector $\vec{B} \times \vec{A}$ is inherently perpendicular to both $\vec{B}$ and $\vec{A}$, the dot product of this result with $\vec{A}$ involves two vectors that are orthogonal to each other. By the definition of the dot product, the operation between any two perpendicular vectors always yields a value of zero, making $(\vec{B} \times \vec{A}) \cdot \vec{A}$ equal to zero.

Consider the geometry formed by the paths of the two boys as they move toward the point of interception. Since the boy starting at $B$ runs perpendicular to the line segment $AB$ of length $a$, the path taken by the boy from $A$ forms the hypotenuse of a right-angled triangle. If both meet at a specific point after time $t$, the distance covered by the boy from $A$ is $vt$ and the distance covered by the boy from $B$ is $v_1t$.

According to the Pythagorean theorem, the square of the distance covered by the boy from $A$ equals the sum of the squares of the initial separation and the distance traveled by the boy from $B$, which is represented by:
$(vt)^{2} = a^{2} + (v_{1}t)^{2}$
Rearranging the terms to isolate the time variable gives $(v^{2} - v_{1}^{2})t^{2} = a^{2}$, and solving for $t$ results in $t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$.

The relationship between the cross product and the dot product of two vectors is defined by the angle $\theta$ between them, specifically through the ratio of their magnitudes. Given that the cross product magnitude $\left| \vec{A} \right| \left| \vec{B} \right| \sin \theta$ equals $\sqrt{3}$ times the dot product $\left| \vec{A} \right| \left| \vec{B} \right| \cos \theta$, one can simplify the equation to $\tan \theta = \sqrt{3}$, which establishes that the angle between the vectors is $60^{\circ}$. The magnitude of the resultant vector is then determined by the law of cosines for vector addition:
$\left| \vec{A} + \vec{B} \right| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
Substituting $\cos 60^{\circ} = 0.5$ into this expression simplifies the term $2AB \cos \theta$ to $AB$, leading to the final value of $\left( A^2 + B^2 + AB \right)^{1/2}$.

Since the particle experiences constant tangential acceleration along a circular path, we can apply the linear kinematic equation $v^2 = u^2 + 2as$ by treating the arc length of the circle as the distance traveled. Given that the particle completes two full revolutions, the total angular displacement is $\theta = 4\pi$ radians. With the radius $r = 20/\pi$ meters, the total distance $s$ covered is the product of the radius and the angular displacement, yielding $s = r\theta = 80$ meters.

With the initial velocity $u = 0$ and the final velocity $v = 80$ m/s, we substitute these values into the kinematic equation to find the acceleration:
$80^2 = 0^2 + 2 \times a \times 80$
Solving this expression for the tangential acceleration $a$ gives $6400 = 160a$, which simplifies to $a = 40$ m/s$^2$.

When two vectors are perpendicular, their dot product equals zero. Given that the sum $\vec{F}_1 + \vec{F}_2$ is perpendicular to the difference $\vec{F}_1 - \vec{F}_2$, we set their dot product to zero as $(\vec{F}_1 + \vec{F}_2) \cdot (\vec{F}_1 - \vec{F}_2) = 0$. Expanding this product leads to $F_1^2 - \vec{F}_1 \cdot \vec{F}_2 + \vec{F}_2 \cdot \vec{F}_1 - F_2^2 = 0$, where the middle terms cancel out because the dot product is commutative. This leaves $F_1^2 - F_2^2 = 0$, which mathematically confirms that the two forces must be equal in magnitude.

Newton's second law connects force, mass, and acceleration, meaning we obtain the acceleration vector by dividing the force vector by the mass. With a mass of 3 kg and a force $\vec{F} = 6t^2 \hat{i} + 4t \hat{j}$, the acceleration vector is $\vec{a} = 2t^2 \hat{i} + \frac{4}{3} t \hat{j}$. Because velocity is the integral of acceleration with respect to time, we integrate this expression from $t = 0$ to $t = 3$. Integrating the components, the velocity is determined by the expression $\left[ \frac{2}{3} t^3 \hat{i} + \frac{2}{3} t^2 \hat{j} \right]_0^3$. Substituting the upper limit $t = 3$ yields the final velocity $18 \hat{i} + 6 \hat{j}$.

The motion of an object in the vertical direction is governed entirely by the acceleration due to gravity and remains independent of any horizontal velocity components. Since both particles start from the same height with an initial vertical velocity $u_y = 0$, they both experience identical free-fall motion. Although particle B is projected with a horizontal speed of 5 m/sec, this velocity has no vertical component to influence its downward descent, which ensures that both particles strike the ground simultaneously.

Angular velocity is fundamentally defined by the relationship $\omega = \frac{2\pi}{t}$, where $t$ represents the time taken to complete one full revolution. Since the problem states that both particles move with the same time period, they share an identical value for $t$. Because the angular velocity depends exclusively on this time interval, equal time periods necessitate equal angular velocities for both particles. Consequently, the ratio of their angular velocities simplifies to $\frac{\omega_1}{\omega_2} = 1$, which makes the mass and the radius of the circular paths irrelevant to the final result.

The magnitude of the resultant vector $\left| \overset{\rightarrow}{A} + \overset{\rightarrow}{B} \right|$ is equal to the sum of the individual magnitudes $\left| \overset{\rightarrow}{A} \right| + \left| \overset{\rightarrow}{B} \right|$ only when the two vectors are parallel and point in the exact same direction. Because the vectors share an identical orientation, there is no angular separation between them, which means the angle is $\theta = 0^{\circ}$.

This scenario is governed by the law of conservation of mechanical energy, which dictates that the initial gravitational potential energy at a given height is converted entirely into kinetic energy. Regardless of whether an object falls freely or slides down a frictionless inclined plane, the work done by gravity depends only on the vertical displacement, leading to the same final energy state expressed by $\frac{1}{2}mv^2 = mgh$. Since both the man and the bag start from rest at the same elevation, they must reach the ground with identical final speeds, resulting in the relationship $v_B = v_m$.