Variable Mass And Newton's Second Law
Rain falls vertically and is collected by a stationary open cart, with water arriving at a steady rate of 2 kg/s landing with a downward speed of 5 m/s. What additional vertical force must the ground exert to hold the cart still?
Select the correct option:
Solution
10 N
Using the momentum form of Newton's Second Law from NCERT Class 11, Chapter 5 (Laws of Motion), force equals the rate of change of momentum, F = dp/dt, which is especially useful when mass accumulates over time. The incoming rain brings downward momentum at a rate equal to the mass arrival rate multiplied by the impact speed: dp/dt = (dm/dt) × v = 2 × 5 = 10 kg·m/s per second. To bring this downward momentum to zero each second, the ground must supply an equal upward force of 10 N in addition to supporting the static weight already present. Option 2.5 N divides instead of multiplying the rate and speed. Option 20 N doubles the result, perhaps assuming the water rebounds. Option 5 N uses only the speed and forgets the mass rate. A conceptual caution is that this analysis assumes the water sticks to the cart on impact rather than bouncing; if it rebounded elastically, the momentum change and hence the force would be larger. The same momentum-rate reasoning underlies how rockets and conveyor belts are analysed, where mass is steadily ejected or added over time. Plausibility check: the units (kg/s)·(m/s) give kg·m/s², which is newtons, and a modest rain rate producing about 10 N of extra force is physically reasonable, confirming the answer.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- variable mass and newton's second law
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
10 N
Using the momentum form of Newton's Second Law from NCERT Class 11, Chapter 5 (Laws of Motion), force equals the rate of change of momentum, F = dp/dt, which is especially useful when mass accumulates over time. The incoming rain brings downward momentum at a rate equal to the mass arrival rate multiplied by the impact speed: dp/dt = (dm/dt) × v = 2 × 5 = 10 kg·m/s per second. To bring this downward momentum to zero each second, the ground must supply an equal upward force of 10 N in addition to supporting the static weight already present. Option 2.5 N divides instead of multiplying the rate and speed. Option 20 N doubles the result, perhaps assuming the water rebounds. Option 5 N uses only the speed and forgets the mass rate. A conceptual caution is that this analysis assumes the water sticks to the cart on impact rather than bouncing; if it rebounded elastically, the momentum change and hence the force would be larger. The same momentum-rate reasoning underlies how rockets and conveyor belts are analysed, where mass is steadily ejected or added over time. Plausibility check: the units (kg/s)·(m/s) give kg·m/s², which is newtons, and a modest rain rate producing about 10 N of extra force is physically reasonable, confirming the answer.
This hard difficulty physics question is from the chapter laws of motion, covering the topic of variable mass and newton's second law. It appeared in the 2025 exam.
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