---
title: "Derive resultant force, variable mass system"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:variable-mass-resultant-force"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/variable-mass-resultant-force
site: BCT Engineering
---

# Derive resultant force, variable mass system

Applied Mechanics (Dynamics) · Chapter 3: System of particles

- Asked in: 2074 Asoj, 4 marks; 2073 Shrawan, 4 marks (asked 2 times)
- Video: [Momentum and the variable mass problem](https://www.youtube.com/watch?v=8ORz4GcT60w) — Engineering Mechanics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/variable-mass-resultant-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

Derive the expression for the resultant force on the system with variable mass.

## Short answer

The equation of motion for a variable-mass system is ∑F = m a – (dm/dt) u_rel, where u_rel = u – v is the relative velocity of incoming or ejected mass. For a rocket, dm/dt < 0 and the term –(dm/dt) u_rel becomes a forward thrust T = |dm/dt| u_rel, giving m a = ∑F + T.

## Answer

A **variable-mass system** gains or loses mass continuously. Consider a system of mass $m$ moving with velocity $\vec v$ that absorbs particles of mass $\Delta m$ moving with absolute velocity $\vec u$ during time $\Delta t$.

```text
 before:  m ---> v        dm ---> u
 after:   (m + dm) ---> v + dv
```

#### Impulse-momentum for the whole set of particles
Momentum at $t$: $m\vec v+\Delta m\,\vec u$. Momentum at $t+\Delta t$: $(m+\Delta m)(\vec v+\Delta\vec v)$. External resultant force $\sum\vec F$ acts for $\Delta t$:

$$
m\vec v+\Delta m\,\vec u+\sum\vec F\,\Delta t=(m+\Delta m)(\vec v+\Delta\vec v)
$$

Neglecting the second-order term $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$:

$$
\sum\vec F=m\frac{d\vec v}{dt}+\frac{dm}{dt}(\vec v-\vec u)
$$

Using the velocity of the absorbed particles relative to the system, $\vec u_{rel}=\vec u-\vec v$:

$$
\boxed{\sum\vec F=m\vec a-\frac{dm}{dt}\,\vec u_{rel}}
$$

or $m\vec a=\sum\vec F+\dfrac{dm}{dt}\vec u_{rel}$.

- **Mass gained, particles initially at rest** ($\vec u=0$): $\sum\vec F=\dfrac{d(m\vec v)}{dt}$.
- **Mass ejected** (rocket): $dm/dt<0$ and $\vec u_{rel}$ is the exhaust velocity relative to the body (backward), so the term $\dfrac{dm}{dt}\vec u_{rel}$ is a forward **thrust** $T=\left|\dfrac{dm}{dt}\right|u_{rel}$. Then $m\vec a=\sum\vec F+\vec T$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What defines a variable-mass system?
   A. Its velocity changes continuously
   B. Its mass changes with time as particles join or leave
   C. It has more than one external force
   D. It rotates about a fixed axis
   Answer: B. A variable-mass system gains or loses mass continuously, such as a rocket or sand-loaded belt. (Key point: Variable mass system)

2. Which term is dropped when dividing by delta t in the derivation?
   A. m dv
   B. The external impulse
   C. The second-order product delta m times delta v
   D. delta m times u
   Answer: C. Delta m delta v is of second order and vanishes in the limit. (Key point: Neglected term)

3. Which is the resulting equation of motion?
   A. Sum F = m a + (dm/dt) u_rel
   B. Sum F = m a - (dm/dt) u_rel
   C. Sum F = (dm/dt) a
   D. Sum F = m a only
   Answer: B. Sum F = m a - (dm/dt) u_rel, where u_rel = u - v is the velocity of the absorbed particles relative to the system. (Key point: Final equation)

4. For a rocket ejecting mass, what is the thrust?
   A. T = m a
   B. T = |dm/dt| u_rel, directed forward
   C. T = |dm/dt| v, directed backward
   D. T = 0 because momentum is conserved
   Answer: B. The exhaust leaves backward, so the term gives a forward thrust of magnitude |dm/dt| u_rel. (Key point: Rocket thrust)

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