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Some of the hardest-looking JEE mechanics questions involve a system whose mass changes: a rocket burning fuel, sand dropping onto a conveyor belt, a cart leaking water. Students who try F = ma directly get the wrong answer, because the mass in that equation is changing. This post shows one method that handles all of them, with worked numbers for each.
Where does this topic sit in the syllabus?
Neither the Syllabus for JEE (Main) 2026 nor the Physics syllabus in the JEE (Advanced) 2026 Information Brochure names rockets or variable mass explicitly. JEE Main lists "law of conservation of linear momentum and its applications", and JEE Advanced lists systems of particles, centre of mass, impulse and conservation of linear momentum. Variable mass problems are applications of exactly those ideas, which is how they appear. The NCERT Class 11 chapter Laws of Motion includes a simple version in Exercise 4.9, finding the initial thrust of a rocket of lift-off mass 20,000 kg. Check the syllabus for your exam year.
Why does F = ma fail when mass changes?
Newton's second law in its general form is F = dp/dt, the rate of change of momentum. For a fixed mass that becomes ma. When mass enters or leaves, the momentum carried by that mass must be counted too.
The reliable method is to pick a short time dt and compare the momentum of everything involved, the main body plus the bit of mass dm, before and after. NASA Glenn Research Center's derivation of the ideal rocket equation does exactly this: it notes that "because of the changing mass, we cannot use the standard form of Newton's second law", and writes the momentum change of the rocket plus its exhaust over dt.
How do you find rocket thrust?
Let the exhaust leave at speed u relative to the rocket, and let mass be ejected at a rate μ = |dm/dt|. Momentum balance over dt gives
M dv/dt = u μ − (external forces such as weight).
The term uμ is the thrust. NASA's rocket thrust equation writes it as F = ṁVe + (pe − p0)Ae, where the second term is a pressure correction at the nozzle exit. JEE problems ignore that term.
Worked example. A rocket of initial mass 5,000 kg ejects gas at 50 kg/s with exhaust speed 2,000 m/s relative to the rocket. Take g = 9.8 m/s². Find the thrust and the initial acceleration on vertical launch.
Thrust = uμ = 2,000 × 50 = 1.0 × 10⁵ N.
Weight = 5,000 × 9.8 = 4.9 × 10⁴ N.
Initial acceleration = (1.0 × 10⁵ − 4.9 × 10⁴) / 5,000 = 10.2 m/s².
The acceleration increases as fuel burns, because the same thrust pushes less mass.
How much speed can a rocket gain?
With no external force, M dv = −u dM. Integrating from initial mass m₀ to final mass m gives the ideal rocket equation:
Δv = u ln(m₀ / m).
Continuing the example. Suppose 3,000 kg of the 5,000 kg is fuel, so the final mass is 2,000 kg.
In free space: Δv = 2,000 × ln(5,000 / 2,000) = 2,000 × ln 2.5 ≈ 2,000 × 0.916 ≈ 1,830 m/s.
Burn time = 3,000 / 50 = 60 s.
Launching vertically in uniform gravity, with no air drag: Δv = u ln(m₀/m) − g t ≈ 1,830 − 9.8 × 60 ≈ 1,830 − 588 ≈ 1,240 m/s.
The logarithm is why rockets need so much fuel: doubling the mass ratio adds only u ln 2 to the speed.
Worked example: a conveyor belt
Sand falls vertically onto a horizontal belt moving at a constant 2.0 m/s, at a rate of 5.0 kg/s. Find the force needed to keep the belt moving, the power supplied and where the energy goes.
Each second, 5.0 kg of sand goes from zero horizontal speed to 2.0 m/s, gaining 10 kg m/s of horizontal momentum. So F = v (dm/dt) = 2.0 × 5.0 = 10 N.
Power = F v = 10 × 2.0 = 20 W.
Kinetic energy given to sand per second = ½ (dm/dt) v² = ½ × 5.0 × 4.0 = 10 W.
Only half the power becomes kinetic energy. The other 10 W is lost as heat while the sand slips on the belt before it reaches belt speed. This half-and-half split is a favourite exam question.
Which mistakes are most common?
Writing F = d(mv)/dt = m dv/dt + v dm/dt blindly. The v in the second term must be the velocity of the mass entering or leaving relative to the body. A cart leaking sand straight down through a hole in its floor keeps moving at constant speed, because the sand leaves with the cart's own velocity and carries its momentum away with no force needed.
Using exhaust speed relative to the ground in u μ. Thrust uses the speed relative to the rocket.
Forgetting gravity during the burn. The −gt term matters for vertical launches.
Assuming all conveyor power becomes kinetic energy. Half goes to heat.
For the momentum and impulse basics underneath all this, see our guides to collisions and momentum problems and centre of mass problems.
A practice routine
Solve one problem of each type: rocket thrust, rocket Δv, conveyor belt, leaking cart.
For each, write the momentum at t and t + dt before using any formula.
Do one energy check per conveyor problem: input power versus kinetic energy gained.
If you can follow the method but not set up the momentum balance yourself, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, which helps you see why each term appears.
This week, derive Δv = u ln(m₀/m) yourself from the momentum balance, without looking, then use it on one problem.
