Q: In the two dimensional motions:

$ y-x $ graph gives actual path of particle. The actual path of a particle in two-dimensional motion is its trajectory in the plane, which is described by the relation between x and y. A y-x graph shows this spatial curve directly, while x-t and y-t graphs only show how each coordinate varies with time.

Q: A car moves in a semicircular track of radius 700 m. If it starts from one end of the track and stops at the other end, the displacement of car is:

$ 700 m $. Displacement depends only on initial and final positions. The car starts and ends at opposite ends of a semicircle, so the straight-line separation is the diameter, which is twice the radius.

Q: The motion of a particle is described by the equation $ \boldsymbol{x}=\boldsymbol{a}+\boldsymbol{b} \boldsymbol{t}^{2} $ where $ \boldsymbol{a}= $ $ 15 \mathrm{cm} $ and $ b=3 \mathrm{cm} / \mathrm{sec}^{2} . $ Its instantaneous velocity at time 3 sec will be:

$ 36 \mathrm{cm} / \mathrm{sec} $. Instantaneous velocity is the time derivative of position. For x = a + bt^2, v = dx/dt = 2bt, so at t = 3 s with b = 3 cm/s², v = 2·3·3 = 36 cm/s.

Question 1

A block is thrown with a velocity of $ 2 m / s $ (relative to ground) on a belt, which is moving with velocity $ 4 m / s $ in opposite direction of the initial velocity of block. If the block stops slipping on the belt after $ 4 s $ of the throwing then choose the correct statement(s

Accepted Answer

Displacement with respect to ground is zero after 2.66 and displacement with respect to ground is $ 12 \mathrm{m} $ after $ 4 s $. The block starts at 2 m/s while the belt moves oppositely at 4 m/s, so the block must change velocity by 6 m/s in 4 s, giving constant acceleration. Using that motion, the block’s ground displacement becomes zero at about 2.66 s and equals 12 m after 4 s, matching the given correct statement.

Question 2

Due to air a falling body faces a resistive force proportional to square of velocity $ v, $ consequently its effective downward acceleration is reduced and is given by $ a=g-k v^{2} $ where $ k= $ $ 0.002 m^{-1} . $ The terminal velocity of the falling body is \( (\operatorname{in}

Accepted Answer

80. Terminal velocity occurs when downward acceleration is exactly balanced by air resistance, so the effective acceleration is zero. Using a = g - kv^2 and setting a = 0 gives v = sqrt(g/k), which evaluates to 80 m/s.

Question 3

In the two dimensional motions:

Accepted Answer

$ y-x $ graph gives actual path of particle. The actual path of a particle in two-dimensional motion is its trajectory in the plane, which is described by the relation between x and y. A y-x graph shows this spatial curve directly, while x-t and y-t graphs only show how each coordinate varies with time.

Question 4

A ball is thrown from rear end of the compartment of train to the front end which is moving at a constant horizontal velocity. An observer $ \boldsymbol{A} $ sitting in compartment and another observer $ B $ standing on the ground draw the trajectory. They will have

Accepted Answer

equal vertical ranges but different horizontal ranges. Both observers measure the same vertical motion because gravity is unchanged and the train’s horizontal motion does not affect vertical acceleration. Their horizontal ranges differ because observer B sees the ball’s horizontal velocity as the sum of the throw and the train’s motion, while observer A does not.

Question 5

Two forces of 3 N and 4 N act at right angles to each other on a body, as shown in the figure. The magnitude of their resultant is:

Accepted Answer

5 N. When two forces act at 90°, their resultant is found using the Pythagorean theorem. So the magnitude is $\sqrt{3^2+4^2}=5$ N.

Question 6

There are three particles shown in the figure as I, II and III. The velocity and acceleration vectors associated with the motion of three particles are shown. Which of the above could represent the velocity and acceleration vectors for a projectile following a parabolic path? I. II. III.

Accepted Answer

III only. A projectile in ideal motion has constant acceleration due to gravity, which points straight down at all times. Its velocity must be tangent to the parabolic path, so only the diagram matching both conditions can be correct.

Question 7

A passenger in a train moving at an acceleration 'a', drops a stone from the window. A person, standing on the ground, by the sides of the rails, observes the ball following:

Accepted Answer

Along a parabola with acceleration 'g". In the ground frame, the stone has an initial horizontal velocity equal to the train’s speed at the moment of release, while its vertical motion is governed only by gravity. Since horizontal acceleration is zero and vertical acceleration is g, the path is a parabola with acceleration g.

Question 8

A car moves in a semicircular track of radius 700 m. If it starts from one end of the track and stops at the other end, the displacement of car is:

Accepted Answer

$ 700 m $. Displacement depends only on initial and final positions. The car starts and ends at opposite ends of a semicircle, so the straight-line separation is the diameter, which is twice the radius.

Question 9

For the shown graph the slope of graph from point $ A $ to point $ B: $

Accepted Answer

First increases then decreases. The graph’s slope is not constant; as you move from A toward the middle, the curve becomes steeper, so the slope increases. After that, it flattens again, so the slope decreases, giving the pattern “first increases then decreases.”

Question 10

The motion of a particle is described by the equation $ \boldsymbol{x}=\boldsymbol{a}+\boldsymbol{b} \boldsymbol{t}^{2} $ where $ \boldsymbol{a}= $ $ 15 \mathrm{cm} $ and $ b=3 \mathrm{cm} / \mathrm{sec}^{2} . $ Its instantaneous velocity at time 3 sec will be:

Accepted Answer

$ 36 \mathrm{cm} / \mathrm{sec} $. Instantaneous velocity is the time derivative of position. For x = a + bt^2, v = dx/dt = 2bt, so at t = 3 s with b = 3 cm/s², v = 2·3·3 = 36 cm/s.

Question 11

If particle takes $ t $ seconds less and acquires a velocity of $ \mathrm{v} \mathrm{m} / \mathrm{s} $ more in falling through the same distance on to planets where the acceleration due to gravity are $ 2 g $ and $ 8 g $ respectively, then

Accepted Answer

$ v=4 g t $. For the same falling distance, time is inversely related to the square root of gravity, and final speed is proportional to the square root of gravity. Comparing the cases for 2g and 8g gives a speed difference tied to the time difference, leading to v = 4gt.

Question 12

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

Accepted Answer

90°. The condition that the magnitude of the sum of two vectors equals the magnitude of their difference implies that the dot product of the vectors is zero. This happens when the angle between the vectors is 90°.

Question 13

If vectors $ \mathbf{A} = \cos \omega t \mathbf{i} + \sin \omega t \mathbf{j} $ and $ \mathbf{B} = \cos \frac{\omega t}{2} \mathbf{i} + \sin \frac{\omega t}{2} \mathbf{j} $ are functions of time, then the value of $ t $ at which they are orthogonal to each other is

Accepted Answer

$ t = \frac{\pi}{\omega} $. Two vectors are orthogonal if their dot product is zero. The dot product of $ \mathbf{A} $ and $ \mathbf{B} $ is $ \cos \omega t \cos \frac{\omega t}{2} + \sin \omega t \sin \frac{\omega t}{2} = \cos \left( \omega t - \frac{\omega t}{2} \right) = \cos \frac{\omega t}{2} $. For orthogonality, $ \cos \frac{\omega t}{2} = 0 $, which gives $ \frac{\omega t}{2} = \frac{\pi}{2} $, or $ t = \frac{\pi}{\omega} $.

Question 14

Vectors $ \mathbf{A}, \mathbf{B} $ and $ \mathbf{C} $ are such that $ \mathbf{A} \cdot \mathbf{B} = 0 $ and $ \mathbf{A} \cdot \mathbf{C} = 0 $. Then the vector parallel to $ \mathbf{A} $ is

Accepted Answer

$ \mathbf{B} \times \mathbf{C} $. Since $ \mathbf{A} \cdot \mathbf{B} = 0 $ and $ \mathbf{A} \cdot \mathbf{C} = 0 $, $ \mathbf{A} $ is perpendicular to both $ \mathbf{B} $ and $ \mathbf{C} $. The cross product $ \mathbf{B} \times \mathbf{C} $ is perpendicular to both $ \mathbf{B} $ and $ \mathbf{C} $, and hence is parallel to $ \mathbf{A} $.

Question 15

A stone tied with a string, is rotated in a vertical circle. The minimum speed with which the string has to be rotated

Accepted Answer

decreases with increasing length of the string. The minimum speed required to rotate a stone in a vertical circle is given by v = √(gL), where g is the acceleration due to gravity and L is the length of the string. This shows that the minimum speed decreases as the length of the string increases.

Question 16

A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is

Accepted Answer

4. The shortest path occurs when the boat's velocity is directed perpendicular to the riverbank. The boat's speed in still water is 5 km/hr, and it takes 15 minutes (0.25 hours) to cross 1 km. Thus, the boat's effective speed perpendicular to the river is 1/0.25 = 4 km/hr. Using Pythagoras' theorem, the river's velocity is √(5² - 4²) = √9 = 3 km/hr.

Question 17

A person swims in a river aiming to reach exactly opposite point on the bank of a river. His speed of swimming is 0.5 m/s at an angle 120° with the direction of flow of water. The speed of water in stream is

Accepted Answer

0.43 m/s. To swim directly across the river, the swimmer's velocity component perpendicular to the river flow must cancel the river's velocity. Using trigonometry, the river's speed is calculated as 0.5 × sin(120°) = 0.43 m/s.

Question 18

A body is whirled in a horizontal circle of radius 20 cm. It has an angular velocity of 10 rad/s. What is its linear velocity at any point on circular path

Accepted Answer

2 m/s. The linear velocity (v) in circular motion is given by v = rω, where r is the radius and ω is the angular velocity. Substituting r = 0.2 m and ω = 10 rad/s, we get v = 0.2 × 10 = 2 m/s.

Question 19

A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be moved?

Accepted Answer

3.92 m/s. The maximum tension in the string is given as 25 N. Using the formula for centripetal force, T = mv²/r, where T is the tension, m is the mass, v is the speed, and r is the radius of the circle, we solve for v: v = √(T * r / m). Substituting T = 25 N, r = 1.96 m, and m = 0.25 kg, we get v = √(25 * 1.96 / 0.25) = 3.92 m/s.

Question 20

When a body moves with a constant speed along a circle

Accepted Answer

no work is done on it. When a body moves with constant speed along a circle, its velocity changes direction continuously, so there is acceleration (centripetal). However, no work is done because the force is perpendicular to the displacement at every point.

NEET Physics: Motion in a Plane questions with solutions

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