Question

A rocket of mass 1000 kg is exhaust gases at a rate of 4 kg/s with a velocity 3000 m/s. The thrust developed on the rocket is
A. $12000N$
B. $120N$
C. $800N$
D. Zero

Answer 1

Hint:-The best approach is to apply momentum conservation and Energy conservation as the formula derivation of rocket propulsion problem is based on these conservation theorems.
That is,
$F = - V\dfrac{{\Delta M}}{{\Delta t}}$
where v is the exhaust velocity and $\dfrac{{\Delta M}}{{\Delta t}}$ is the rate of mass change.

Complete step-by-step solution:-
Given: Mass $ = m$=1000 kg
Rate of change of mass = $\dfrac{{\Delta M}}{{\Delta t}}$= 4 kg/s
Velocity =v =3000 m/s
According to the question, we have to calculate the Thrust or force felt by the rocket when mass is leaving at the given rate
First know the meaning of thrust, so it the force exerted on the rocket when the mass leaves in forms of gases at constant rate
Thrust on the rocket
$F = - \dfrac{{dp}}{{dt}}$
The formula above is based on newton’s second law of motion which states that rate of change of momentum will give us the force exerted on the particle
On further expanding this formula by applying some calculus we reach to the result:
$F = - m\dfrac{{dv}}{{dt}} + \left( { - v\dfrac{{dm}}{{dt}}} \right)$
Velocity at some point is constant but mass is changing at every point
So, the final result is
 $\eqalign{
  & F = - v\dfrac{{\Delta m}}{{\Delta t}} \cr
  & \cr} $
On putting the values
$\eqalign{
  & = - 3000 \times 4 \cr
  & = - 12000N \cr} $
The value we get is -12000N.

Note:- Negative sign indicates that thrust applied in a direction opposite to the direction of escaping gas. Some other formula used for Rocket propulsion problems are:
$\eqalign{
  & {v_f} = u\ln \left( {\dfrac{{{M_o}}}{{{M_f}}}} \right) \cr
  & F_{ext}^{total} = \dfrac{{d{p_{system}}}}{{dt}} \cr
  & a = \dfrac{F}{{{m_T}}} = \dfrac{F}{{M - \left( {\dfrac{{dm}}{{dt}}} \right)t}} \cr} $
 where ${v_f}$ is the final velocity, u is the initial velocity and a is the acceleration of the rocket.