---
title: "Linear momentum and impulse-momentum equation"
subject: "Fluid Machines"
chapter: 1
key: "semester_5/fluid_machines:linear-momentum-impulse-equation"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/linear-momentum-impulse-equation
site: BCT Engineering
---

# Linear momentum and impulse-momentum equation

Fluid Machines · Chapter 1: Introduction of Turbomachime and Dynamic Action of Fluid

- Asked in: year not given, 6 marks
- Video: [Fluid Mechanics: The Momentum Equation](https://www.youtube.com/watch?v=_fCv98W0L90) — Caspar Hewett
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/linear-momentum-impulse-equation
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-1

## Question

State Newton's second law in terms of linear momentum. Derive the impulse-momentum equation and the force exerted by a fluid jet on a stationary body using the control volume approach.

## Short answer

Newton's second law: force equals rate of change of momentum. For steady flow, net force on fluid equals mass flow rate times change in velocity. For a jet on a stationary body, force equals ρQ(V1x - V2x). For a flat plate normal to jet, F = ρaV²; for a curved vane turning through angle θ, F = ρaV²(1 - cosθ).

## Answer

#### Momentum principle

Newton's second law: the rate of change of linear momentum of a body is equal to the resultant force acting on it, and takes place in the direction of that force.

$$
\vec F = \frac{d(m\vec V)}{dt}
$$

If the force $F$ acts for a short time $dt$, then $F\,dt = d(mV)$. The left side is the **impulse** and the right side is the change of momentum. This is the **impulse-momentum equation**.

#### Control volume form (steady flow)

Consider a fixed control volume with one inlet (section 1) and one outlet (section 2), through which fluid of density $\rho$ flows steadily. The mass flow rate is
$m = \rho Q = \rho A_1 V_1 = \rho A_2 V_2$.

In time $dt$ the mass $dm = \rho Q\,dt$ enters with velocity $V_1$ and the same mass leaves with velocity $V_2$. Momentum change of the fluid in the given direction $x$:

$$
dm\,(V_{2x} - V_{1x}) = \rho Q\,dt\,(V_{2x}-V_{1x})
$$

Dividing by $dt$, the net external force on the fluid in the control volume (pressure forces, body force and force by the solid surface) is

$$
\sum F_x = \rho Q\,(V_{2x} - V_{1x}), \qquad \sum F_y = \rho Q\,(V_{2y} - V_{1y})
$$

#### Force of jet on a stationary body

For a free jet, the pressure is atmospheric everywhere on the control surface, so pressure forces cancel and the body force is negligible. The only force on the fluid is the force $R$ exerted by the body on the fluid. By Newton's third law the force of the jet on the body is equal and opposite:

$$
F_x = \rho Q\,(V_{1x} - V_{2x})
$$

Hence the force on a body equals the rate at which the fluid loses momentum in that direction. Direction of the force is found by the sign of $(V_{1x}-V_{2x})$.

**Special results for a jet of area $a$ and velocity $V$:**

- Flat plate normal to jet (fluid leaves along plate, so $V_{2x}=0$): $F = \rho a V^2$.
- Symmetrical curved vane turning the jet through angle $\theta$ (measured from the jet direction): $F = \rho a V^2(1-\cos\theta)$.

These expressions are the basis for the analysis of Pelton wheel buckets and other impulse machines.

## Check yourself

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

1. Newton's second law in momentum form states that force equals:
   A. Change of kinetic energy
   B. Mass times velocity
   C. Pressure times area only
   D. Rate of change of linear momentum
   Answer: D. F = d(mV)/dt: the force equals the rate of change of momentum and acts in its direction. (Key point: Newton's second law)

2. For steady flow through a control volume, the net force in x on the fluid is:
   A. rho Q (V1x - V2x)
   B. rho Q V1x
   C. rho Q (V2x + V1x)
   D. rho Q (V2x - V1x)
   Answer: D. Force on fluid = mass flow rate times (outlet velocity minus inlet velocity) in that direction. (Key point: Control volume equation)

3. Why do pressure forces drop out for a free jet on a body?
   A. Velocity is zero
   B. Gravity cancels them
   C. The jet has zero pressure
   D. Pressure is atmospheric everywhere on the control surface
   Answer: D. A free jet is at atmospheric pressure all round, so net pressure forces cancel, and body force is negligible. (Key point: Free jet force on body)

## Related in Fluid Machines

- [Force of jet on flat plate, stationary and moving](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Ajet-on-flat-plate-numerical) — Fluid Machines, chapter 1
- [Force of jet on symmetrical curved vane](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Ajet-on-curved-vane-numerical) — Fluid Machines, chapter 1

## Related in other subjects

- [Momentum equation and its applications](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Amomentum-equation-control-volume-applications) — Fluid Mechanics, chapter 4
- [Jet on moving curved vanes, force and work](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ajet-on-moving-curved-vanes) — Fluid Mechanics, chapter 7
- [Principle of impulse and momentum](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aimpulse-momentum-principle-particle) — Engineering Mechanics, chapter 4
- [General energy equation for control volume](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ageneral-energy-equation-control-volume) — Fundamental of Thermodynamics & Heat Transfer, chapter 4