Law of conservation of linear momentum for a system of particles is the underlying principle
that governs the operation of a rocket. In this regard, in rocket propulsion, the system is the rocket and its ejected fuel (gases). As a rocket moves, because the gases are ejected with certain linear momentum, the rocket receives an equal magnitude of momentum in the opposite direction. Thus the rocket accelerates as a result of 'thrust' from the exhaust gases. If the gases are ejected with a speed V e relative to the rocket, it can be proved that M dV = V e dm where dV is a small increase in the speed of the rocket as a result of burning of fuel of mass dm. M is the mass of rocket and its remaining fuel after an amount of fuel having a mass dm has been ejected. Integration of this equation with suitable limits will result in a basic expression for rocket propulsion.
Consider a rocket having an initial mass 1000 kg which also includes the mass of fuel. The rocket is fired and it burns fuel at a constant rate 25 kg/s. the speed of exhaust gases relative to the rocket is a constant 2000 m/s.

Answer these questions:
(i) Thrust provided by the rocket engine is –
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) : Thrust on rocket v = 2000 m/s
f = v
= 2000 × 25 = 5 × 10 4 N.
(ii) : Initial acceleration = 
a =
= 50 m/s 2 (iii) : at t = 20 sec,
m f (final mass of rocket) = 1000 – 20 × 25 = 500 kg.
and v f = u log e
= 2000 log e 
= 2000 × log e 2
= 2000 × 0.693 = 1386 m/s
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