---
title: "Force from changing mass; acceleration increases"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:variable-mass-force-acceleration-increases"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/variable-mass-force-acceleration-increases
site: BCT Engineering
---

# Force from changing mass; acceleration increases

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

- Asked in: 2075 Chaitra, 6 marks
- Video: [Rocket Propulsion Physics & Mass Flow Rate - Newton's 3rd Law of Motion](https://www.youtube.com/watch?v=kvAf-Xouj8s) — The Organic Chemistry Tutor
- 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-force-acceleration-increases
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

Derive an expression for the force exerted on the system due to change in mass over time. Show that the final acceleration increases when system loses mass.

## Short answer

For mass ejection, m a = ∑F – |dm/dt| u_rel, where –|dm/dt| u_rel is a forward thrust T. With constant thrust and external force, acceleration a = (∑F + T)/m(t) increases as mass m(t) decreases, so the final acceleration is larger.

## Answer

#### Force due to change in mass
Let a system of mass $m$ moving with velocity $\vec v$ lose (eject) a small mass $\Delta m$ in time $\Delta t$, the ejected particles leaving with absolute velocity $\vec u$. An external force $\sum\vec F$ acts.

Momentum at $t$: $m\vec v$. At $t+\Delta t$: $(m-\Delta m)(\vec v+\Delta\vec v)+\Delta m\,\vec u$. Impulse-momentum:

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

Neglecting $\Delta m\,\Delta\vec v$ and dividing by $\Delta t$ (with $\dot m=-\Delta m/\Delta t$ the rate of change of the system mass, so $\dot m<0$):

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

With $\vec u_{rel}=\vec u-\vec v$ (exhaust velocity relative to the body) and $\left|\dot m\right|=\Delta m/\Delta t$:

$$
m\vec a=\sum\vec F-\left|\dot m\right|\vec u_{rel}
$$

Here $\vec u_{rel}$ points backward, so $\vec T=-\left|\dot m\right|\vec u_{rel}$ is a forward **thrust** force. Hence $m\vec a=\sum\vec F+\vec T$ with $T=\left|\dot m\right|u_{rel}$.

#### Why the acceleration increases when mass is lost
Along the line of motion:

$$
a=\frac{\sum F+T}{m(t)},\qquad m(t)=m_0-\left|\dot m\right|t
$$

Take a constant burn rate and constant exhaust speed, so $T$ is constant, and assume the external force (weight, drag) is roughly constant. The numerator then stays the same while the denominator $m(t)$ decreases, so $a$ increases with time:

$$
\frac{da}{dt}=\frac{(\sum F+T)\left|\dot m\right|}{m^{2}}>0
$$

(when $\sum F+T>0$). The final acceleration is therefore larger than the initial one. For a rocket in free space ($\sum F=0$): $dv=u_{rel}\dfrac{-dm}{m}$, giving $v=v_0+u_{rel}\ln\dfrac{m_0}{m}$, so the velocity grows faster as $m$ becomes small.

## Check yourself

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

1. What is the system's momentum at t + delta t when it ejects delta m?
   A. (m + delta m)(v + delta v)
   B. m v + delta m u
   C. (m - delta m)(v + delta v) + delta m u
   D. (m - delta m) v
   Answer: C. The remaining mass moves at v + delta v while the ejected mass delta m moves at absolute velocity u. (Key point: Momentum after ejection)

2. What is the resulting equation for mass loss?
   A. m a = Sum F + |m dot| u_rel
   B. m a = Sum F - |m dot| u_rel, where -|m dot| u_rel is a forward thrust
   C. m a = Sum F
   D. m a = |m dot| u_rel - m g only
   Answer: B. u_rel points backward, so -|m dot| u_rel is forward thrust T, giving m a = Sum F + T. (Key point: Equation of motion)

3. Why does acceleration increase as the system loses mass?
   A. The thrust grows with time
   B. The external force grows
   C. The denominator m(t) shrinks while the numerator (Sum F + T) stays roughly constant
   D. Velocity increases so a must increase
   Answer: C. a = (Sum F + T)/m(t); with constant burn and exhaust speed, a smaller m gives a larger a. (Key point: Why a increases)

4. For a rocket in free space, what is v?
   A. v = v0 + u_rel ln(m0/m)
   B. v = v0 + u_rel (m0 - m)
   C. v = v0 - u_rel ln(m0/m)
   D. v = v0 + u_rel m0/m
   Answer: A. Integrating dv = u_rel(-dm/m) gives v = v0 + u_rel ln(m0/m). (Key point: Rocket velocity)

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