Physics

Force and momentum Questions

Chapter: Mechanics | Total Available: 84 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

A block of mass 2.0 kg slides on a rough horizontal plane surface. Let the speed of the block at a particular instant be 10 m/s. It comes to rest after travelling a distance of 20 m. Which one of the following could be the magnitude ofthe frictional force?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. The frictional force is calculated as -5 N.

๐Ÿ“– Governing Law / Core Concept

Newton's second law of motion states: F = ma, where F is the net force acting on an object, m is its mass, and a is its acceleration.

๐Ÿงช Step-by-Step Breakdown

Given:

  • Mass (\(m\)) = 2.0 kg
  • Initial Velocity (\(u\)) = 10 m/s
  • Final Velocity (\(v\)) = 0 m/s
  • Distance (\(s\)) = 20 m

Using the third equation of motion:

\[ v^2 = u^2 + 2as \]

Substituting the known values:

\[ 0 = (10)^2 + 2a(20) \]
\[ 0 = 100 + 40a \]

Simplifying gives:

\[ 40a = -100 \]
\[ a = -2.5 \, \text{m/s}^2 \]

Now applying Newton's second law:

\[ F = ma \]
\[ F = 2 \times (-2.5) \]
\[ F = -5 \, \text{N} \]

The negative sign indicates that the frictional force acts in the direction opposite to the motion.

๐Ÿ” Option Analysis

Option (b) 20 N, (c) 40 N, and (d) 50 N are not possible as they do not satisfy the equations derived.

โšก Mnemonic / Speed-Run

To remember laws of motion: "Force equals mass times acceleration."

๐Ÿ“Š Visual Diagram Suggestion:

Visual Suggestion

A force diagram showing the block on a rough surface with arrows illustrating gravitational, normal, and frictional forces acting on it. This would aid in visualizing the situation effectively.

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy and Power).

Question 2 of 10 Authentic UPSC NDA Question

Why a fielder in the ground gradually pulls his hands backward while catching a fast moving cricket ball?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (A) is correct. This technique reduces the impact force on the fielder.

๐Ÿ“– Governing Law / Core Concept

The principle relevant here is the impulse-momentum theorem, which states:

\[ \text{Impulse} = \Delta \text{Momentum} \]

Impulse is the product of the average force and the time duration over which the force acts.

๐Ÿงช Step-by-Step Breakdown

When a fielder catches a cricket ball, he effectively receives momentum from the ball, which is equal to:

\[ p = m \cdot v \]

where \( m \) is the mass of the ball (in kg) and \( v \) is its velocity (in m/s).

By pulling his hands backward, the fielder increases the time (\( \Delta t \)) over which the ball decelerates to rest. This longer time reduces the average force (\( F \)) experienced, calculated as:

\[ F \Delta t = m v \]

Rearranging gives:

\[ F = \frac{m v}{\Delta t} \]

This indicates that increasing \( \Delta t \) (catching time) reduces force \( F \), mitigating injury and improving catching success.

๐Ÿ” Option Analysis

  • Option B: Incorrect, as increasing the acceleration is undesirable when catching a ball.
  • Option C: Incorrect, as velocity is not increased; the goal is to decelerate the ball.
  • Option D: Incorrect, as the above explanations suffice.

โšก Mnemonic / Speed-Run

Remember: More Time = Less Force when catching. This mnemonic highlights the benefit of time during collisions.

6. Visual Suggestion

6. Visual Suggestion

A diagram depicting a fielder catching a ball, showing forces acting on the ball and a timeline indicating increased catching time through a backward motion of hands.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 5 (Laws of Motion).

Question 3 of 10 Authentic UPSC NDA Question

The energy is always conserved for a system which is

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option A is correct. Energy is conserved in an isolated system.

๐Ÿ“– Governing Law / Core Concept

According to the Law of Conservation of Energy, energy cannot be created or destroyed in an isolated system.

๐Ÿงช Step-by-Step Breakdown

In an isolated system, no matter or energy enters or leaves the system. Therefore, the total energy remains constant.

For a non-isolated system, external forces can do work, resulting in energy being transformed or transferred:

  • Isolated System: \( E_{total} = E_{initial} \)
  • Non-Isolated System: \( E_{total} = E_{initial} + W_{ext} \) (where \( W_{ext} \) is work done by external forces)

๐Ÿ” Option Analysis

  • B: Non-isolated systems experience energy transfer.
  • C: Both cannot guarantee energy conservation.
  • D: None of the above contradicts conservation laws.

โšก Mnemonic / Speed-Run

Remember: "Isolated = Conserved, Non-isolated = Transferred."

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating isolated vs non-isolated systems, depicting energy flow in each type with arrows indicating energy transfer in non-isolated systems and none in isolated systems. Include labels to clarify.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter: Work, Energy and Power.

Question 4 of 10 Authentic UPSC NDA Question

Suppose, a ball of mass M is thrown upwards from a point A and it reaches up to the highest point B and returns back to point A, which one among the following is correct?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. Kinetic Energy at A equals Potential Energy at B.

๐Ÿ“– Governing Law / Core Concept

The principle governing the motion of the ball is the Conservation of Mechanical Energy, which states that the total mechanical energy (sum of kinetic energy (KE) and potential energy (PE)) in a closed system remains constant if only conservative forces, like gravity, act on the body.

๐Ÿงช Step-by-Step Breakdown

As the ball is thrown upwards from point A:

  • At point A: Kinetic Energy is maximum, Potential Energy is minimum.
  • At point B (highest point): Velocity is zero, thus Kinetic Energy is zero, and Potential Energy is maximum.

By Conservation of Energy:

\[ KE_A + PE_A = KE_B + PE_B \]

Since \( PE_A = 0 \) (taking point A as reference), and \( KE_B = 0 \) (at highest point B), we have:

\[ KE_A = PE_B \]

Thus, at point B, the entire kinetic energy at A (when the ball was thrown) has been converted into potential energy.

๐Ÿ” Option Analysis

  • Option (b): Incorrect. Kinetic Energy at A is not equal to Potential Energy at A, as KE is maximum and PE is minimum there.
  • Option (c): Incorrect. At point B, KE is zero, whereas PE is maximum.
  • Option (d): Incorrect. At point B, KE is zero while at A it is maximum.

โšก Mnemonic / Speed-Run

Remember: "Up is Potential, Down is Kinetic." Energy shifts from kinetic to potential as height increases.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating the motion of the ball with labeled points A and B, showing kinetic energy decreasing as the ball rises and potential energy increasing, highlighting the energy transformation at the highest point.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 2 (Work, Energy, and Power).

Question 5 of 10 Authentic UPSC NDA Question

Weight and mass of an object are defined with Newtons laws of motion. Which among the following is true?

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (b) is correct. Mass is a constant of proportionality.

๐Ÿ“– Governing Law / Core Concept

According to Newton's laws of motion, weight \( W \) is the gravitational force acting on an object, related to its mass \( m \) by the equation:

\[ W = mg \] \end{aligned} \]

Where:

  • \( W \) = Weight (N)
  • \( m \) = Mass (kg)
  • \( g \) = Acceleration due to gravity (m/sยฒ)

๐Ÿงช Step-by-Step Breakdown

Mass is defined as the measure of the amount of matter in an object. It remains constant regardless of location, thereby acting as a constant of proportionality in the weight equation. Weight, on the other hand, is variable - it depends on gravitational force, which changes with the distance from the center of the Earth and the local gravitational field strength.

To illustrate:

\[ m = \frac{W}{g} \] \end{aligned} \]
  • If weight changes, mass remains the same.
  • This confirms that mass is a constant of proportionality, while weight varies.

๐Ÿ” Option Analysis

Option (a) incorrectly states that weight is a constant of proportionality. Option (c) is false since mass is indeed a constant of proportionality. Option (d) mistakenly labels weight as a universal constant; however, it varies with gravitational force.

โšก Mnemonic / Speed-Run

Remember: "Mass is Matter's Measure, Weight Wavers with Gravity."

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating the relationship between mass and weight, showing a scale with mass on one side and weight measurement on the other, emphasizing that while mass remains consistent, weight fluctuates with different gravitational forces.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class IX Physics, Chapter 4 (Force and Motion)

Question 6 of 10 Authentic UPSC NDA Question

Work is said to be one Joule when a force of

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (a) is correct. A force of 4 N moving an object by 25 cm does equal one Joule of work.

๐Ÿ“– Governing Law / Core Concept

Work (\( W \)) done by a constant force (\( F \)) is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
where: - \( W \) = Work (in Joules) - \( F \) = Force (in Newtons) - \( d \) = Displacement (in meters) - \( \theta \) = Angle between force and displacement direction In this case, \(\theta = 0^\circ\) (force in the same direction as displacement).

๐Ÿงช Step-by-Step Breakdown

To find the work done: - Given: \( F = 4 \, \text{N} \), \( d = 25 \, \text{cm} = 0.25 \, \text{m} \) Work calculation:
\[ W = F \cdot d \] \[ W = 4 \, \text{N} \cdot 0.25 \, \text{m} \] \[ W = 1 \, \text{J} \]

๐Ÿ” Option Analysis

- **Option (b)**: \( 2 \, \text{N} \cdot 1 \, \text{m} = 2 \, \text{J} \) (not correct). - **Option (c)**: \( 1 \, \text{N} \cdot 0.01 \, \text{m} = 0.01 \, \text{J} \) (not correct). - **Option (d)**: \( 1 \, \text{N} \cdot 0.5 \, \text{m} = 0.5 \, \text{J} \) (not correct).

โšก Mnemonic / Speed-Run

Remember: \( 1 \, \text{J} = 1 \, \text{N} \cdot 1 \, \text{m} \). Keep units consistent when calculating work; convert to meters when necessary.

6. Visual Suggestion

6. Visual Suggestion

A diagram showing a horizontal surface with an object being pushed by a force. Include vectors for force and displacement and label the angle \( \theta \) as \( 0^\circ \).

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy, and Power).

Question 7 of 10 Authentic UPSC NDA Question

A uniform motion of a car along a circular path experiences

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (b) is correct. The car experiences a change in velocity due to changing direction.

๐Ÿ“– Governing Law / Core Concept

The concept involved in this scenario is based on Newton's First Law of Motion, which states that an object in motion will remain in motion unless acted upon by an external force. Circular motion implies a constant speed but a continuously changing velocity due to the change in direction.

๐Ÿงช Step-by-Step Breakdown

A car moving in uniform circular motion maintains a constant speed; however, its direction is constantly changing. The key factors involved include:

  • Velocity (\( v \)): Change in both magnitude and direction.
  • Acceleration (\( a \)): Centripetal acceleration required to keep the car moving in a circle, given by:
    \[ a = \frac{v^2}{r} \]
    where \( r \) is the radius of the circular path.
  • Momentum (\( p \)): Defined as \( p = mv \), where \( m \) is the mass of the car. Since the direction of the velocity vector changes, the momentum changes continuously as well.

๐Ÿ” Option Analysis

Option (a): Incorrect; speed remains constant, but direction changes.
Option (c): Incorrect; a change does occur in both direction and speed.
Option (d): Incorrect; momentum cannot remain constant if velocity is changing.

โšก Mnemonic / Speed-Run

Remember: "Centripetal means center-seeking," which reminds that centripetal force is what keeps an object in a circular path, thereby influencing its velocity and momentum even when speed is uniform.

6. Visual Suggestion

6. Visual Suggestion

A circular path diagram showing a car at different points, indicating the velocity vector, acceleration vector, and centripetal force acting towards the center, helps visualize the change in direction while speed remains constant.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy and Power).

Question 8 of 10 Authentic UPSC NDA Question

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option D is correct. The upward velocity required to reach 20 m under gravitational pull is 20 m/s.

๐Ÿ“– Governing Law / Core Concept

The relevant formula governing the motion of an upward-moving object is derived from the equations of motion under uniform acceleration due to gravity. Specifically, the maximum height \(h\) attained can be given by: \[ h = \frac{v^2}{2g} \] where \(v\) is the initial velocity and \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

๐Ÿงช Step-by-Step Breakdown

Given that the maximum height \(h = 20 \, \text{m}\): \[ 20 = \frac{v^2}{2 \times 9.81} \] Rearranging gives: \[ v^2 = 20 \times 2 \times 9.81 = 392.4 \] Taking the square root: \[ v = \sqrt{392.4} \approx 19.8 \, \text{m/s} \] Thus, \(v\) is approximately \(20 \, \text{m/s}\).

๐Ÿ” Option Analysis

Options A (8 m/s), B (12 m/s), and C (16 m/s) are incorrect as they would not provide enough initial velocity to reach the maximum height of 20 m given the gravitational acceleration.

โšก Mnemonic / Speed-Run

Remember \(h = \frac{v^2}{2g}\) as "Height equals velocity squared over twice gravity." This formula summarizes the energy conversion from kinetic to potential energy.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating the trajectory of the tennis ball, showing its path from the point of release to the maximum height of 20 meters, annotated with velocity and gravitational forces acting on it at various points.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter on Motion in a Straight Line.

Question 9 of 10 Authentic UPSC NDA Question

An object of mass 2000 g possesses 100 J kinetic energy. The object must be moving with a speed of

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

\[ KE = \frac{1}{2} m v^2 \] \end{align*} \] Where: - \( KE \) = Kinetic Energy (in Joules) - \( m \) = Mass (in kg) - \( v \) = Velocity (in m/s) Given: - \( KE = 100 \, J \) - \( m = 2000 \, g = 2 \, kg \) Substituting known values: \[ 100 = \frac{1}{2} \times 2 \times v^2 \] \begin{align*} 100 &= 1 v^2 \\ v^2 &= 100 \\ v &= \sqrt{100} \\ v &= 10.0 \, m/s \end{align*}
๐Ÿ“Œ Hints / Properties Used:
  • Kinetic Energy Formula: \( KE = \frac{1}{2} m v^2 \)
  • Mass Conversion: \( 1 \, g = 0.001 \, kg \)
๐Ÿ“Š Visual Diagram Suggestion:

A schematic can depict a moving object with labeled vectors showing velocity and kinetic energy, illustrating the relationship between mass and speed that results in the given kinetic energy.

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 7 (System of Particles and Rotational Motion)

Question 10 of 10 Authentic UPSC NDA Question

The work done by the force acting on an object is zero if the displacement of the object

Step-by-Step Logic & Solution

Correct Answer

โœ“ Option (c) is correct. Work done is zero when the displacement is perpendicular to the force.

๐Ÿ“– Governing Law / Core Concept

The work done \( W \) by a force \( F \) acting on an object and causing a displacement \( s \) is given by the formula:

\[ W = F \cdot s \cdot \cos(\theta) \]

Here, \( \theta \) is the angle between the force and displacement vectors.

๐Ÿงช Step-by-Step Breakdown

If the displacement is perpendicular to the force:

\[ \theta = 90^\circ \Rightarrow \cos(90^\circ) = 0 \]

Thus, substituting into the work formula:

\[ W = F \cdot s \cdot 0 = 0 \, \text{J} \]

Therefore, the work done is zero when the force acts perpendicular to the displacement.

๐Ÿ” Option Analysis

  • Option A: Incorrect. Displacement in the opposite direction of the force yields negative work.
  • Option B: Incorrect. Displacement in the same direction yields positive work.
  • Option D: Incorrect. Valid scenarios exist that contradict this.

โšก Mnemonic / Speed-Run

Remember: Perpendicular = Zero Work. This can be visualized as vectors at right angles not contributing to the work done.

6. Visual Suggestion

6. Visual Suggestion

Create a diagram with two vectors: one representing force at an angle and one showing displacement perpendicular to it. Label the angle as 90ยฐ to illustrate the concept of zero work clearly.

๐Ÿ“– Factual Verification & Reference

๐Ÿ“– Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 6 (Work, Energy, and Power).

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