Motion In A Plane – NEET UG PhysicsPractice Questions & PYQs

The horizontal range of a projectile launched with a constant initial speed depends on the sine of twice the launch angle, defined by the formula $R = \frac{v^{2} \sin(2\theta)}{g}$. When two launch angles are complementary-meaning they sum to $90^\circ$-the sine values $\sin(2\theta)$ and $\sin(2(90^\circ-\theta))$ become identical because $\sin(180^\circ - 2\theta)$ simplifies to $\sin(2\theta)$. Since the angles $30^\circ$ and $60^\circ$ are complementary, the range for body $A$ is $\frac{v^2 \sin(60^\circ)}{g}$ and the range for body $B$ is $\frac{v^2 \sin(120^\circ)}{g}$. Because these values are mathematically equivalent, the ratio of the horizontal range of body $A$ to body $B$ is $1: 1$.

A vector is defined as a unit vector when its magnitude equals 1. The magnitude is calculated by finding the square root of the sum of the squares of its components, which implies that the sum of the squares of these components must equal 1. For the given components, this relationship is expressed as $0.5^2+(-0.8)^2+c^2=1$. Squaring these values leads to $0.25+0.64+c^2=1$, which simplifies to $0.89+c^2=1$. Subtracting 0.89 from 1 results in $c^2=0.11$, meaning the value for $c$ is $\sqrt{0.11}$.

Determining the velocity of the river stream starts by finding the effective speed at which the boat traverses the width of the river. Since the boat crosses a distance of 1 km in 15 minutes, or 1/4 hours, its resultant crossing speed is defined by $v = 1 / (1/4) = 4$ km/hr. Given that the boat's speed in still water forms the hypotenuse of the velocity vector triangle and the resultant crossing speed forms one of the legs, the river velocity $v_{\text{River}}$ is found using the Pythagorean theorem as $v_{\text{River}} = \sqrt{5^2 - 4^2}$, which results in 3 km/hr.

The relationship between linear velocity, angular velocity, and the position vector is defined by the cross product operation $\vec{v} = \vec{\omega} \times \vec{r}$. To determine the resultant vector, we construct a determinant using the unit vectors and the components of the given angular velocity and position vectors.

$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -6 & 6 \\ 3 & -4 & 1 \end{vmatrix}$

Evaluating this determinant along the first row allows us to compute the individual components of the linear velocity. For the $\hat{i}$ component, the calculation $(-6 \times 1) - (6 \times -4)$ simplifies to $18\hat{i}$. Calculating the $\hat{j}$ component requires taking the negative of the product difference $(5 \times 1) - (6 \times 3)$, which results in $13\hat{j}$. Finally, the $\hat{k}$ component is determined by $(5 \times -4) - (-6 \times 3)$, yielding $-2\hat{k}$, which combines to give the final expression $18\hat{i} + 13\hat{j} - 2\hat{k}$.

The slope of a straight line connecting the origin to a coordinate point is defined by the ratio of the y-coordinate to the x-coordinate, which corresponds to the tangent of the angle the line makes with the positive x-axis. For a particle located at the point $(\sqrt{3}, 3)$, the slope is determined by the formula $\frac{3}{\sqrt{3}}$, which simplifies to $\sqrt{3}$. Given that the tangent of 60 degrees equals $\sqrt{3}$, the path of the particle creates an angle of 60 degrees with the x-axis.

Every object moving in a circle possesses a linear velocity that relates to its angular velocity through the radius of the path, expressed by the equation $v = r\omega$. Since standard physics calculations require consistent units, convert the radius from 20 cm to 0.2 m before proceeding. Multiplying this radius by the angular velocity of 10 rad/s yields the linear velocity:

$v = 0.2 \times 10 = 2$

Consequently, the magnitude of the velocity at any point along the circle is 2 m/s.

When an object travels in a horizontal circle, the tension in the string provides the necessary centripetal force to maintain its circular path, which is calculated using the relationship $F = \frac{mv^2}{r}$. Since the string will break if the tension exceeds 25 N, we use this value as the upper limit for the centripetal force to determine the maximum speed. By substituting the mass of 0.25 kg and the radius of 1.96 m into the equation $25 = \frac{0.25 v^2}{1.96}$, we can isolate $v^2$ by calculating $v^2 = \frac{25 \times 1.96}{0.25}$. Simplifying the right side results in $v^2 = 196$, and solving for $v$ gives a maximum speed of 14 m/s.

The geometric relationship between two vectors is defined by their scalar product, where $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos \theta$. By calculating the dot product of $\vec{A}=3 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{B}=3 \hat{i}+4 \hat{j}-5 \hat{k}$, we obtain $3(3) + 4(4) + 5(-5) = 9 + 16 - 25 = 0$. This result confirms that $\cos \theta = 0$, which implies that the vectors are perpendicular and the angle between them is $\theta = 90^{\circ}$.

When a projectile is launched over horizontal terrain, its range reaches a maximum value at an angle of 45 degrees, which reduces the general range formula to $R = \frac{u^2}{g}$. Given the maximum range of 16 km, we convert this to 16000 m to match the unit of acceleration due to gravity, which is 10 m/s². Substituting these values into the formula yields the equation $16000 = \frac{u^2}{10}$. Multiplying both sides by 10 results in $u^2 = 160000$, which gives a muzzle velocity of 400 m/s.

The magnitude of the resultant vector $\overset{\rightarrow}{C}$ formed by adding vectors $\overset{\rightarrow}{A}$ and $\overset{\rightarrow}{B}$ is determined by the law of cosines:
$C^2 = A^2 + B^2 + 2AB\cos(\theta)$
where $\theta$ is the angle between the two vectors. Substituting the given values into this relationship results in:
$3^2 + 4^2 + 2(3)(4)\cos(\theta) = 5^2$
Simplifying the squares and the product gives $9 + 16 + 24\cos(\theta) = 25$, which simplifies further to $25 + 24\cos(\theta) = 25$. Because this reduces to $24\cos(\theta) = 0$, it follows that $\cos(\theta) = 0$, confirming that the angle $\theta$ is $\frac{\pi}{2}$.

The magnitude of the cross product of two vectors is defined as $|\vec{A} \times \vec{B}| = AB \sin \theta$, while their dot product is defined as $\vec{A} \cdot \vec{B} = AB \cos \theta$. Substituting these standard definitions into the given condition results in the following equation:
$AB \sin \theta = \sqrt{3} AB \cos \theta$
By dividing both sides by $AB \cos \theta$, the expression simplifies to $\tan \theta = \sqrt{3}$. Evaluating the inverse tangent of this value confirms that the angle between the two vectors is $60^{\circ}$.

Change in velocity is a vector operation defined as the difference between the final velocity and the initial velocity, expressed as $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$. Since subtracting a vector is equivalent to adding its negative, we determine the resultant by combining the westward final velocity with a southward vector, which is the reverse of the initial northward motion. Given that these two vectors each have a magnitude of 50 km/hr and are perpendicular, the magnitude of the change is found via $\sqrt{50^2 + 50^2} = 70.7$ km/hr. Combining a westward and a southward component creates a resultant vector oriented in the South-West direction.

When a boat travels across a river, its actual path is determined by the vector sum of its velocity relative to the water and the river's flow velocity. Since these two velocities act perpendicularly to each other, the magnitude of the resultant velocity relates to the individual speeds through the Pythagorean theorem, expressed as $v_{resultant}^{2} = v_{boat}^{2} + v_{river}^{2}$. By substituting the given values of $10 \text{ km h}^{-1}$ for the resultant velocity and $8 \text{ km h}^{-1}$ for the boat's velocity into this formula, we get $10^{2} = 8^{2} + v_{river}^{2}$. This simplifies to $100 = 64 + v_{river}^{2}$, which means $v_{river}^{2} = 36$, confirming that the velocity of the river is $6 \text{ km h}^{-1}$.

When projectiles are launched with identical speeds at angles that are complementary, meaning they sum to $90^{\circ}$, the horizontal distance covered remains unchanged. This property arises from the range equation $R = \frac{v^{2} \sin(2\theta)}{g}$, which produces the same result for both $60^{\circ}$ and $30^{\circ}$ because the sine values of their doubled angles are equivalent. In contrast, time of flight and maximum height depend directly on the sine and the square of the sine of the launch angle, respectively, so these properties vary for each trajectory.