Momentum And Force Relation
The linear momentum of a moving particle varies with time according to the relation p = 8 + 6t, where p is in kg·m/s and t is in seconds. What is the magnitude of the force acting on this particle?
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- momentum and force relation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
6 N
As stated in NCERT Class 11, Chapter 5 (Laws of Motion), the most general statement of Newton's Second Law is that force equals the time rate of change of linear momentum, F = dp/dt. Given the momentum function p = 8 + 6t, we differentiate with respect to time. The constant term 8 represents the initial momentum at t = 0 and has zero derivative, while the term 6t differentiates to 6. Therefore the force is F = dp/dt = 6 N, constant in time. The constant initial momentum does not contribute to the force because only the changing part of momentum corresponds to a net force. Option 8 N mistakes the initial momentum for the force. Option 14 N wrongly adds the two coefficients 8 and 6. Option 48 N multiplies them, which has no physical justification. This approach is more powerful than the familiar F = ma form because it remains valid even when mass changes, whereas F = ma assumes constant mass. Here, since the slope is constant, the motion is equivalent to uniform acceleration, and one could even extract the mass if the particle's velocity at any instant were additionally provided. Plausibility check: a linear momentum-time graph has a constant slope, implying a constant force, and the slope here is exactly 6 N with correct units of kg·m/s², confirming the answer.
This medium difficulty physics question is from the chapter laws of motion, covering the topic of momentum and force relation. It appeared in the 2025 exam.
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