Physics

Motion of objects

Chapter: Mechanics

Question 1 of 5 NDA MCQ

A ball is thrown vertically upward with a speedof 40 m/s. The time taken by the ball to reach themaximum height would be approximately

Detailed Explanation

Correct Answer

✓ Option (c) is correct. The time taken by the ball to reach maximum height is approximately 4 seconds.

📖 Governing Law / Core Concept

Motion under uniform acceleration is governed by the equations of kinematics, specifically:

\[ v = u + at \]

where:

  • v = final velocity (m/s)
  • u = initial velocity (m/s)
  • a = acceleration (m/s²)
  • t = time (s)

🧪 Step-by-Step Breakdown

Given:

  • Initial velocity, \( u = 40 \, \text{m/s} \)
  • Final velocity at maximum height, \( v = 0 \, \text{m/s} \)
  • Acceleration due to gravity, \( a = -10 \, \text{m/s}^2 \)

Applying the kinematic equation:

\[ 0 = 40 + (-10)t \]

Rearranging gives:

\[ 10t = 40 \]
\[ t = \frac{40}{10} = 4 \, \text{s} \]

The ball reaches a maximum height in approximately 4 seconds.

📌 Hints / Properties Used:
  • Kinematic equation: \( v = u + at \)
  • Acceleration due to gravity is \( -10 \, \text{m/s}^2 \) for upward motion.
📊 Visual Diagram Suggestion:

A coordinate graph depicting the ball's trajectory, highlighting the initial velocity at the bottom, the apex (maximum height) at 4 seconds where velocity is zero, and the downward acceleration due to gravity.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 3 (Motion in a Straight Line).

Question 2 of 5 NDA MCQ

Which one of the following equations related to the motion of an objects is NOT correct?(Symbols carry their usual meanings)

Detailed Explanation

1. Correct Answer

Option D is correct. The equation for distance during nth second is not a fundamental equation of motion.

📖 Core Concept

This question pertains to the equations of motion which describe the relationship between displacement, initial velocity, final velocity, acceleration, and time.

💡 3. Explanation

The standard equations of motion are:
s = ut + (1/2) at²
u = v - at
v² - u² = 2as
Option D does not represent this form; the correct equation for distance during nth second involves initial velocity and various time parameters.

4. Option Analysis

  • Option A: Correct formula but does not represent distance during nth second correctly.
  • Option B: Valid equation for initial velocity in terms of final velocity and acceleration.
  • Option C: Well-established equation linking initial & final velocities with acceleration and displacement.

NDA Speed-Run / Mnemonic Shortcut

Remember the fundamental equations of motion; for nth second, use derived formulas specific to that time frame.

🎯

5. Key Takeaways

The nth second motion equation is derived from foundational equations but is not a direct fundamental equation itself.

Question 3 of 5 NDA MCQ

A uniform motion of a car along a circular path experiences

Detailed Explanation

Correct Answer

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

📖 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.

🧪 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.

🔍 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.

⚡ 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.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 4 of 5 NDA MCQ

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

Detailed Explanation

Correct Answer

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

📖 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 \(9.81 \, \text{m/s}^2\)).

🧪 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 \(20 \, \text{m/s}\).

🔍 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.

⚡ 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.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 5 of 5 NDA MCQ

 The figure shown above gives the time (t) versus position (x) graphs of three objects A, B and C. Which one of the following is the correct relation between their speeds, Vₐ, Vᵦ and V꜀ respectively at any instant (t > 0)?

Detailed Explanation

Correct Answer

✓ Option (b) is correct. The relation between the speeds is Vₐ > Vᵦ > V꜀.

📖 Governing Law / Core Concept

The speed \( V \) of an object is defined as the magnitude of its velocity, which can be calculated from the slope of the position vs. time graph:

\[ V = \frac{\Delta x}{\Delta t} \]

🧪 Step-by-Step Breakdown

From the provided graph:

C: steepest slope (highest speed)
B: moderate slope (mid speed)
A: least slope (lowest speed)

Hence, the relation is:

Vₐ > Vᵦ > V꜀

🔍 Option Analysis

  • Option (a): Incorrect, as Vₐ is not less than Vᵦ.
  • Option (c): Incorrect, all speeds are not zero as indicated by positive slopes.
  • Option (d): Incorrect, since their speeds are unequal.

⚡ Mnemonic / Speed-Run

Remember: Steep slopes indicate higher speeds. Graphically visualize it as a race — the steeper the hill, the faster the runner.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 1 (Motion in a Straight Line).

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