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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Physics » Variable Mass System

Variable Mass System

Variable mass system introduction, definition, different derivations for the variable mass system - for mass ejected and for mass accreted conclusion and FAQs.

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Newton’s second law of motion states the relationship between the mass of an object, its acceleration and the net force acting on it. But this law cannot be applied to some objects that gain or lose weight when they are in motion. Another equation needs to be derived for such objects with the help of Newton’s law of motion and the addition of the momentum factor.

Variable Mass System Meaning

A matter whose mass varies with time and is in the form of collection is called a variable mass system.

Derivation

To make the calculations simpler, all bodies are considered particles. The motion equations of the variable mass system have different derivations. The equation depends on whether the mass is leaving or entering. The derivation of the equation of the variable mass system depends on the ejection of the mass by the body or the accretion of the mass by the body.

Mass Ablation or Ejection

The derivation for the system in which the mass from the main body is being ejected is different from when it is ablated. Let a mass m travel at a velocity v at time t, leading to the initial momentum of a system to be,

p1 = mv

Let’s take the velocity of the mass dm (ablated), at a time t + dt. Then the momentum we get for the system is,

p2= (m – dm)(v+dv) – udm= mv + mdv – vdm – dmdv – udm

In the above equation, u is taken as the velocity of the ejected mass. It is taken negatively because it is in a direction that is opposite to the mass. The momentum of the system at the time dt is given as:

dp=  p2 – p1 = (mv + mdv – vdm – dmdv – udm) – (mv)= mdv – (v+dv+u)dm

Now, the ablated mass m will have the relative velocity vrel as follows,

vrel= u – (- v- dv) = v + dv + u

Thus, we can write the change in momentum as,

dp= (mv)= mdv  vreldm

Therefore, by Newton’s second law

Fext = dpdt = mdv-vrel dm/dt = m dvdt  – vrel dmdt

Thus, the final equation is,

Fext+ vrel dmdt =m dvdt

Mass Accretion or accumulation

A body is moving with the velocity v has the mass m, which varies with the time and initial t. At the same time, a particle with mass dm with velocity u also starts to move. We can write the momentum at the start point (initial momentum) as,

p1=mv+udm

Now at a time t + dt, let both the main body and the particle accrete into a body of velocity v + dv. We get the system’s new momentum as,

p2= (m+dm)(v+dv)= mv+mdv+vdm+dmdv

The values of dm and dv are very small so ignore their product. At time dt, the system momentum changes as follow,

dp= p2-p1= (mv+mdv+vdm) – (mv+udm)= mdv-(u-v)dm

Therefore, by Newton’s second law

Fext = dpdt = mdv-(u-v)dmdt = m dvdt  –(u-v) dmdt

We know that the dm is in relation with m, which has a velocity u – v . Thus, by keeping u-v to be vrel, in the equation we get,

Fext + vrel dmdt =m dvdt

Forms

Using the equation of acceleration, a = dv/dt, we can write the motion equation of variable mass system as

Fext + vrel dmdt =ma

Replace the bodies with acm if they do not come under the particles a. Thus the centre mass of the system will have the acceleration,

Fext + vrel dmdt =macm

Often the force due to thrust is said as Fthrust = vrel dmdtso.

Fext  + Fthrust  = macm

If the net force is taken as the sum of the thrust force and external force, then the equation will get back to Newton’s second law, which is

Fnet = macm

Also, the equation Fext  + Fthrust  = macm shows that the body will still have the acceleration due to the thrust at no external forces.

Conclusion

Thus, it can be concluded that the momentum is conserved in a variable mass system. The equations of motion for the variable mass system have different derivations since the equations depend on the situation of the accretion or ablation of the mass. Even if no external forces act on the body, they will accelerate because of the thrust force.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What is momentum? Give the formula and an example for momentum.

Ans. Momentum is the product of the mass and velocity of an object. An example of momentum is how quickly a car move...Read full

What is the Variable mass system, and can we apply Newton's second law of motion to it?

Ans. When a matter’s mass varies with time and in the form of collection it is called a variable-mass system. ...Read full

What is the equation of rocket science?

Ans. The rocket science equation gives us the change of velocity that the rocket experiences from burning a mass of ...Read full

Solve the equation: A 8000 kg rocket is set for firing. The exhaust speed is 800 m/s w.r.t. rocket, calculate the amount of gas ejected per second for the needed thrust to give a vertical upward acceleration of 10m/s2,?

Ans. Given, m = 8000kg, v = 800m/s, a = 10 m/l...Read full

Ans. Momentum is the product of the mass and velocity of an object. An example of momentum is how quickly a car moves down a hill. The formula for momentum is p=mv, where p is momentum, m is the mass, and v is the velocity.

Ans. When a matter’s mass varies with time and in the form of collection it is called a variable-mass system. We cannot apply Newton’s second law of motion to the variable mass system.

Ans. The rocket science equation gives us the change of velocity that the rocket experiences from burning a mass of fuel that decreases the total rocket mass from m0 down to m. Soviet physicist Konstantin Tsiolkovsky derived it. The equation is as follows: Δ v = u ln ( m0/m1 )

Ans.

Given, m = 8000kg, v = 800m/s, a = 10 m/l2, we know that

Thrust = v dm/dt

v = exhaust speed

m =amount of gas ejected

The upward force is,

f = m(a+g);

a = acceleration of rocket

g = acceleration due to gravity

= 9.8m/s2

∴ Thrust = F

v dm/dt = m(a+g) ……… (1)

Substituting the given values in equation (1), we get

800 × dm/dt = 8000 × (10+9.8)

⇒ dm/dt =198kg/s

But, if we take g=10m/s

We get  

dm/dt =198 kg/s

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