Mechanics Archives

11. If the linear momentum of a moving object changes by two times, then i

If the linear momentum of a moving object changes by two times, then its kinetic energy will change by a factor of

This question was previously asked in

UPSC CAPF – 2022

The correct answer is B, 4.

Linear momentum (p) of an object of mass m and velocity v is given by p = mv.
Kinetic energy (k) of the same object is given by k = ½ mv².
We can express kinetic energy in terms of momentum. From p = mv, we get v = p/m. Substituting this into the kinetic energy equation:
k = ½ m(p/m)² = ½ m(p²/m²) = ½ p²/m.
So, k is proportional to p² (assuming mass m is constant).
If the initial momentum is p₁, the initial kinetic energy is k₁ = ½ p₁²/m.
If the linear momentum changes by two times, it means the new momentum p₂ = 2p₁.
The new kinetic energy k₂ = ½ p₂²/m = ½ (2p₁)²/m = ½ (4p₁²)/m = 4 (½ p₁²/m) = 4k₁.
Thus, the kinetic energy will change by a factor of 4.

This relationship (k ∝ p²) shows that kinetic energy is much more sensitive to changes in momentum than momentum is to changes in kinetic energy (or velocity).

12. Two objects, x and y, have equal mass and are moving with speeds u and

Two objects, x and y, have equal mass and are moving with speeds u and 3u respectively. Their kinetic energies k_x and k_y are related as

k_x = k_y

2k_x = k_y

9k_x = k_y

3k_x = k_y

This question was previously asked in

UPSC CAPF – 2022

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The correct answer is C, 9k_x = k_y.

The kinetic energy (k) of an object with mass (m) and speed (v) is given by the formula k = ½ mv².
For object x, mass is m and speed is u, so k_x = ½ mu².
For object y, mass is m and speed is 3u, so k_y = ½ m(3u)² = ½ m(9u²) = 9 (½ mu²).
Comparing k_y with k_x, we find k_y = 9k_x, or 9k_x = k_y.

Kinetic energy is directly proportional to the mass of the object and the square of its speed. Doubling the speed of an object quadruples its kinetic energy (if mass remains constant). Tripling the speed increases the kinetic energy by a factor of nine.

13. Which one of the following statements with regard to Newton’s third la

Which one of the following statements with regard to Newton’s third law of motion is NOT correct?

Force never occurs singly in nature

When the earth pulls a stone downwards due to gravity, the stone exerts a force on the earth

There is ‘a cause-effect relation implied in the third law

There is no cause-effect relation implied in the third law

This question was previously asked in

UPSC CAPF – 2020

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The statement “There is ‘a cause-effect relation implied in the third law” is NOT correct with regard to Newton’s third law of motion.

Newton’s third law states that for every action, there is an equal and opposite reaction. These action and reaction forces are a pair and occur simultaneously as part of a single interaction between two bodies. The law describes the nature of forces as interactions rather than implying one force causes the other in a sequential manner. There is no temporal delay or causal dependency between the action and reaction forces; they exist concurrently. Statement D, “There is no cause-effect relation implied in the third law”, is consistent with this understanding.

Statements A and B correctly describe aspects of the third law. A) Force never occurs singly in nature implies forces always come in pairs (action-reaction pairs). B) When the earth pulls a stone downwards due to gravity, the stone exerts a force on the earth is the classic example of an action-reaction pair according to the third law.

14. Which one of the following statements regarding motion is correct?

Which one of the following statements regarding motion is correct?

All the periodic motions are necessarily simple harmonic

All the simple harmonic motions are necessarily periodic motions

There is no co-relation between the simple harmonic motions and the periodicity of motion

The relation between the simple harmonic motion and periodic motion depends upon the mass of object undergoing the motion

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UPSC CAPF – 2020

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All the simple harmonic motions are necessarily periodic motions.

Periodic motion is defined as any motion that repeats itself in a regular interval of time. Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts towards the equilibrium. Since SHM repeats itself at regular intervals (its period), it is always a periodic motion. However, not all periodic motions are SHM (e.g., a simple pendulum with large amplitude is periodic but not SHM; a uniform circular motion is periodic but not oscillatory/SHM).

SHM is characterized by a sinusoidal variation of displacement, velocity, and acceleration with time. Examples include the oscillation of a mass on a spring (Hooke’s Law) or the small-angle oscillations of a simple pendulum.

15. A snowboard pulled up by a tow rope travels at the rate of 5 m/s up a

A snowboard pulled up by a tow rope travels at the rate of 5 m/s up a mountain. If 3000 watt of power is used, what force was applied to it ?

50 N

100 N

600 N

15000 N

This question was previously asked in

UPSC CAPF – 2019

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The relationship between power (P), force (F), and velocity (v) when the force is applied in the direction of motion is given by the formula P = F * v.

– Given Power (P) = 3000 watts.
– Given velocity (v) = 5 m/s.
– We need to find the force (F).

– Rearranging the formula, F = P / v.
– Substituting the given values, F = 3000 W / 5 m/s = 600 N.
– Therefore, the force applied to the snowboard was 600 Newtons.

16. If the amplitude of oscillation of a simple pendulum is very small, th

If the amplitude of oscillation of a simple pendulum is very small, then its time period of oscillation

1. depends on the length of the pendulum, L
2. depends on the acceleration due to gravity, g
3. depends upon the mass of the bob of the pendulum, m
4. does not depend upon the amplitude of the pendulum, A

Select the correct answer using the code given below.

1, 2 and 3

1, 2 and 4

2, 3 and 4

1 and 4 only

This question was previously asked in

UPSC CAPF – 2018

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The time period of oscillation of a simple pendulum, when the amplitude is very small, depends on the length of the pendulum and the acceleration due to gravity, but not on the mass of the bob or the amplitude.

For small oscillations (amplitude A is very small), the formula for the time period (T) of a simple pendulum is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This formula explicitly shows that the time period depends on L and g, but it does not contain the mass (m) of the bob or the amplitude (A). The condition of small amplitude is crucial for this simplified formula to hold, implying independence from amplitude in this specific case.

For larger amplitudes, the time period of a simple pendulum is slightly longer and does depend on the amplitude. However, the question specifically states “amplitude… is very small”, which validates the use of the small-angle approximation formula.

17. Which one of the following statements is not true ?

Which one of the following statements is not true ?

The gravitational force of earth acting on a body of mass 1 kg is 9.8 newton

The force acting on an object of mass 5 kg moving with a uniform velocity of 10 m/s on a frictionless surface is zero

The SI unit of weight is kg

The momentum of a man having mass 100 kg walking with a uniform velocity of 2 m/s is 200 newton second

This question was previously asked in

UPSC CAPF – 2016

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Statement C is incorrect. Weight is the force exerted on a mass by gravity. The SI unit of force is the Newton (N). Kilogram (kg) is the SI unit of mass, which is a measure of the amount of matter in an object and is independent of gravity. Weight is measured in Newtons.

Distinguishing between mass and weight and knowing their respective SI units is a fundamental concept in physics.

Statement A is correct: The weight of a 1 kg mass on Earth, where the standard acceleration due to gravity (g) is approximately 9.8 m/s², is given by W = mg = 1 kg * 9.8 m/s² = 9.8 N.
Statement B is correct: If an object moves with uniform velocity, its acceleration is zero (a=0). According to Newton’s second law, the net force acting on the object is F = ma. If a=0, then F=0. On a frictionless surface, there are no opposing forces like friction.
Statement D is correct: Momentum (p) is calculated as mass (m) times velocity (v). p = 100 kg * 2 m/s = 200 kg⋅m/s. The unit kg⋅m/s is equivalent to Newton-second (N⋅s), as 1 N = 1 kg⋅m/s². So, the momentum is 200 N⋅s.

18. A piece of stone tied to a string is made to revolve in a circular orb

A piece of stone tied to a string is made to revolve in a circular orbit of radius r with other end of the string as the centre. If the string breaks, the stone will :

move away from the centre.

move towards the centre.

move along a tangent.

stop.

This question was previously asked in

UPSC CAPF – 2016

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When an object is revolving in a circular orbit, its velocity at any point is directed tangentially to the circle at that point. The string provides the centripetal force required to keep the object moving in a circular path, constantly changing its direction. If the string breaks, this centripetal force is removed. According to Newton’s first law of motion (the law of inertia), an object in motion will continue in a straight line with constant speed unless acted upon by an external force. Therefore, the stone will move in a straight line along the direction of its instantaneous velocity at the moment the string breaks, which is tangential to the circular orbit.

Understanding inertia and tangential velocity in circular motion is key to predicting the motion of an object when the centripetal force is removed.

This phenomenon demonstrates Newton’s first law and the nature of velocity in circular motion. The stone does not move away from the center (that would imply a radially outward force, which doesn’t exist upon breaking), nor does it move towards the center (as the inward centripetal force is gone), nor does it stop (unless there is friction or air resistance acting).

19. Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁

Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively. Their speeds are such that each car makes a complete circle in the same time ‘t’. The ratio of angular speed of the first to that of the second car is :

m₁ : m₂

1 : 1

r₁ : r₂

1 : 2

This question was previously asked in

UPSC CAPF – 2015

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The correct answer is 1 : 1.

Angular speed (ω) is defined as the rate of change of angular displacement. For an object moving in a circle, the angular speed is given by ω = 2π / T, where T is the time period (time taken to complete one revolution). The problem states that both cars make a complete circle in the same time ‘t’. Therefore, the time period T is the same for both cars.
For the first car, ω₁ = 2π / t.
For the second car, ω₂ = 2π / t.
The ratio of their angular speeds is ω₁ : ω₂ = (2π / t) : (2π / t) = 1 : 1.
The masses (m₁ and m₂) and radii (r₁ and r₂) of the cars are irrelevant for determining the ratio of angular speeds when the time period is given to be the same.

Linear speed (v) is related to angular speed and radius by v = rω. If the angular speeds are the same but the radii are different, the linear speeds will be different. Centripetal acceleration (a_c = rω²) and centripetal force (F_c = m a_c = m rω²) would also depend on the mass and radius even if the angular speed is constant. This question specifically asks for the ratio of angular speed, which is determined solely by the time period in this case.

20. The weight of any object is felt due to gravity of Earth. When any obj

The weight of any object is felt due to gravity of Earth. When any object goes inside the Earth or above the Earth, weight decreases. It will weigh minimum when an object is placed at :

The Equator and Moon

North Pole and Saturn

South Pole and 1 Km beneath the Earth

Centre of Earth and in an orbiting satellite

This question was previously asked in

UPSC CAPF – 2014

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The weight of an object is given by W = mg, where m is its mass and g is the acceleration due to gravity. The acceleration due to gravity varies with location.

At the center of the Earth, the acceleration due to gravity (g) is zero because the gravitational forces from all parts of the Earth cancel out. In an orbiting satellite, an object is in a state of continuous freefall around the Earth, resulting in apparent weightlessness, which means the object feels minimum (essentially zero apparent) weight.

– Gravity is maximum on the Earth’s surface. It decreases as you move away from or towards the center (until the center).
– On the surface, gravity is slightly higher at the poles than at the equator due to Earth’s shape (oblate spheroid) and centrifugal force from rotation.
– The Moon has about 1/6th the gravity of Earth’s surface, but gravity is not zero.
– Saturn has a different and much stronger gravitational field compared to Earth.
– 1 km beneath the Earth’s surface, gravity has slightly decreased but is still significant.