6.1 Calculate particle linear momentum | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Impulse, Momentum, and Impact

EN PH 131 study topic 6.1: Calculate particle linear momentum

Linear momentum describes the motion of a particle by combining its mass with its velocity. It is a vector: its direction is the direction of the particle’s velocity, not necessarily the direction of a force or of its path length. In this lesson, the system is one particle, and an observer measures its motion in a stated reference frame. Unless a problem says otherwise, use a fixed frame with perpendicular

x

and

y

axes. Momentum depends on the measured velocity, so stating the frame and positive coordinate directions matters. The calculations below focus on finding momentum from mass and velocity, including when velocity is given by components or a direction and speed.

What you will learn

Define the system and reference frame used to describe a particle’s momentum.
Calculate linear momentum from a particle’s mass and velocity.
Resolve momentum into components and report its direction with consistent SI units.
Check a momentum result using units, signs, and limiting cases.

1. Define the system, frame, and quantities

A system is the object or collection of objects selected for study. For this topic, the system is a single particle with mass

m

. A particle model treats the object’s mass as concentrated at a point; its size and shape are not needed to calculate its linear momentum.

The observer describes the particle relative to a reference frame, which provides a position coordinate system and a way to measure time. Use a frame fixed to the ground when that is the natural choice in a problem. State which direction is positive on each axis. The particle’s velocity must be measured in that same frame.

Position is the particle’s location; displacement is the change in position; path length is the distance travelled along its path. Velocity describes the rate and direction of change of position. Momentum uses velocity, not path length, displacement, or speed alone. Speed is the magnitude of velocity and has no direction.

System: the particle whose momentum is being calculated.
Reference frame: the observer’s coordinate system for measuring velocity.
Mass is a scalar; velocity and linear momentum are vectors.

2. Governing relationship and units

For a particle of mass

m

and velocity vector

\mathbf{v}

, linear momentum is the product of mass and velocity. In a fixed Cartesian frame, calculate the velocity components first, then multiply each component by the same mass. The momentum vector therefore points in the same direction as the velocity when mass is positive.

In SI units, mass is measured in kilograms and velocity in metres per second. Momentum is measured in kilogram-metres per second. The equivalent unit

\mathrm{N\,s}

is also used, but for this calculation the mass-times-velocity form makes the units especially clear.

If only speed and direction are given, express velocity as components using the stated axes before calculating momentum. For example, a direction angle measured counterclockwise from the positive

x

-axis gives

v_x=v\cos\theta

and

v_y=v\sin\theta

. The signs of the components encode the direction.

\mathbf{p}=m\mathbf{v},\qquad \mathbf{p}=p_x\mathbf{i}+p_y\mathbf{j},\qquad p_x=mv_x,\quad p_y=mv_y

Calculate the velocity vector in the chosen frame before multiplying by mass.
Momentum components inherit the signs of the corresponding velocity components.
The dimensions are mass times length per time.

3. A reliable calculation and checks

Begin by identifying the particle, the observer’s frame, and the axes. Record the known mass and velocity information, and identify whether the requested result is the full vector, its components, its magnitude, or its direction. If direction is supplied as an angle, specify how that angle is measured.

Use the defining relationship component by component. If velocity is given by speed and angle, resolve it into signed components. If velocity is already given in component form, use those values directly. To find the magnitude of momentum, take the vector magnitude after calculating its components; do not add component magnitudes as though they were scalars.

Check that the units are

\mathrm{kg\,m/s}

, the signs match the velocity direction, and the momentum magnitude equals mass times speed. A particle at rest has zero momentum. If the mass is doubled while velocity stays fixed, momentum doubles. These checks catch common arithmetic and direction errors.

|\mathbf{p}|=m|\mathbf{v}|

Keep the same axes and signs from the stated velocity through the final momentum.
Use the vector magnitude only after finding the components.
A zero velocity gives zero momentum; increasing mass at fixed velocity increases momentum proportionally.

Worked example

Momentum from two velocity components

A particle of mass

3.2\ \mathrm{kg}

moves in a ground-fixed frame with velocity

\mathbf{v}=(4.0\mathbf{i}-1.5\mathbf{j})\ \mathrm{m/s}

. Find its linear momentum, magnitude, and direction. Take

+x

to the right and

+y

upward.

Particle velocity and momentum

Both vectors point down and right; the diagram indicates direction, not scale.

Set the system and frame
The system is the 3.2 kg particle. The observer is fixed to the ground, and the stated axes define positive directions. The initial state is the given velocity; the requested final state is the particle’s momentum in this frame.
Apply the definition
The velocity components are already given, so multiply each by the particle’s mass. The negative vertical component means the resulting momentum also points partly downward.
$\mathbf{p}=3.2(4.0\mathbf{i}-1.5\mathbf{j})$
Calculate components and magnitude
The components are $12.8$ and $-4.8\ \mathrm{kg\,m/s}$ . The magnitude follows from the perpendicular components, and the direction angle is measured from $+x$ ; the negative vertical component places it below that axis.
$\mathbf{p}=(12.8\mathbf{i}-4.8\mathbf{j})\ \mathrm{kg\,m/s},\quad |\mathbf{p}|=13.7\ \mathrm{kg\,m/s},\quad \theta=-20.6^\circ$

Answer:

\mathbf{p}=(12.8\mathbf{i}-4.8\mathbf{j})\ \mathrm{kg\,m/s}

; its magnitude is

13.7\ \mathrm{kg\,m/s}

and its direction is

20.6^\circ

below

+x

.

Check: The velocity magnitude is

\sqrt{4.0^2+(-1.5)^2}=4.27\ \mathrm{m/s}

, so

m|\mathbf{v}|=3.2(4.27)=13.7\ \mathrm{kg\,m/s}

. The dimensions and downward sign agree.

Worked example

Momentum from speed and direction

A

0.80\ \mathrm{kg}

sensor cart is moving at

6.0\ \mathrm{m/s}

at

120^\circ

counterclockwise from the positive horizontal axis, as measured by a ground-fixed observer. Find its momentum vector and direction.

Oblique motion

The velocity and momentum point into the second quadrant.

Resolve the velocity
The system is the cart modelled as a particle, and the observer uses the ground-fixed axes. Since the angle is measured counterclockwise from $+x$ , the velocity is in quadrant two: its horizontal component is negative and its vertical component is positive.
$v_x=6.0\cos120^\circ,\qquad v_y=6.0\sin120^\circ$
Multiply by mass
Momentum has the same component signs as velocity. Multiplying both velocity components by $0.80\ \mathrm{kg}$ gives the required vector.
$\mathbf{p}=(-2.4\mathbf{i}+4.2\mathbf{j})\ \mathrm{kg\,m/s}$
Report direction and check
Because momentum is a positive scalar multiple of velocity, its direction remains $120^\circ$ from $+x$ . Its magnitude must equal mass times the given speed.
$|\mathbf{p}|=0.80(6.0)=4.8\ \mathrm{kg\,m/s}$

Answer:

\mathbf{p}=(-2.4\mathbf{i}+4.2\mathbf{j})\ \mathrm{kg\,m/s}

, with magnitude

4.8\ \mathrm{kg\,m/s}

and direction

120^\circ

counterclockwise from

+x

.

Check: The negative

x

and positive

y

components place the vector in quadrant two, consistent with the stated angle. The magnitude is

m v=4.8\ \mathrm{kg\,m/s}

.

Worked example

Momentum with a negative velocity

A particle of mass

1.5\ \mathrm{kg}

moves along a straight horizontal track. In a frame fixed to the track, right is positive and its velocity is

-3.0\ \mathrm{m/s}

. Calculate its linear momentum and state what happens to momentum if the particle comes to rest.

Leftward particle motion

Both vectors point left, opposite the positive horizontal direction.

Identify the signed velocity
The system is the particle, and the observer is fixed to the track. Right is positive, so the stated negative velocity means the particle moves left. The known initial state is this velocity; the final quantity sought is momentum.
Calculate the momentum
In one dimension, retain the sign when multiplying mass by velocity. A negative result indicates leftward momentum, not negative mass.
$p_x=(1.5)(-3.0)=-4.5\ \mathrm{kg\,m/s}$
Check the limiting case
If the particle comes to rest, its velocity is zero in this same frame. The defining relationship then gives zero momentum, consistent with the particle having no motion relative to the observer.
$v_x=0\quad\Longrightarrow\quad p_x=0$

Answer: The particle’s momentum is

-4.5\ \mathrm{kg\,m/s}

along the track, meaning

4.5\ \mathrm{kg\,m/s}

to the left. At rest, its momentum is zero.

Check: The magnitude

4.5\ \mathrm{kg\,m/s}

equals

1.5\ \mathrm{kg}

times

3.0\ \mathrm{m/s}

. Its negative sign matches the leftward velocity.

Common mistakes and how to avoid them

Using speed as if it were the full momentum vector.

Correction: Speed gives only the magnitude. Use velocity components or state the direction of the momentum as well.

Dropping a negative velocity component when multiplying by mass.

Correction: Mass is positive, so each momentum component keeps the sign of its velocity component.

Adding component magnitudes to find the magnitude of momentum.

Correction: For perpendicular components, use the vector magnitude relationship, not their arithmetic sum.

Changing coordinate directions partway through a calculation.

Correction: State positive directions at the start and keep them consistent in the velocity and momentum components.

Lesson summary

A particle’s linear momentum is its mass multiplied by its velocity in a specified reference frame.
Momentum is a vector with SI units of $\mathrm{kg\,m/s}$ , directed the same way as velocity.
Calculate components using the chosen axes; retain their signs and find the vector magnitude only when needed.
Check that the direction, units, and zero-velocity limit make sense.

Check your understanding

Question 1

A

2.0\ \mathrm{kg}

particle has velocity

(-3.0\mathbf{i}+4.0\mathbf{j})\ \mathrm{m/s}

. What is its momentum?

$(-6.0\mathbf{i}+8.0\mathbf{j})\ \mathrm{kg\,m/s}$
$(6.0\mathbf{i}-8.0\mathbf{j})\ \mathrm{kg\,m/s}$
$(-1.0\mathbf{i}+2.0\mathbf{j})\ \mathrm{kg\,m/s}$
$(-6.0+8.0)\ \mathrm{kg\,m/s}$

Show answer and explanation

(-6.0\mathbf{i}+8.0\mathbf{j})\ \mathrm{kg\,m/s}

Multiply each signed velocity component by the mass. The momentum components retain the velocity signs.

Question 2

A particle’s mass is unchanged, but its velocity reverses direction while keeping the same speed. What happens to its momentum?

Its momentum keeps the same direction because mass is unchanged.
Its momentum reverses direction and keeps the same magnitude.
Its momentum becomes zero.
Its momentum doubles.

Show answer and explanation

Its momentum reverses direction and keeps the same magnitude.

Momentum points in the direction of velocity, and its magnitude is mass times speed. Reversing velocity reverses momentum without changing its magnitude.

Key terms

Particle: An object represented as a point for the motion calculation, with its mass assigned to that point.
Reference frame: The observer’s coordinate system used to measure position, time, and velocity.
Linear momentum: The vector quantity equal to a particle’s mass multiplied by its velocity.
Component: A signed part of a vector along one of the chosen coordinate axes.

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6.1 · Calculate particle linear momentum

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 6.1: Calculate particle linear momentum

What you will learn

1. Define the system, frame, and quantities

2. Governing relationship and units

3. A reliable calculation and checks

Momentum from two velocity components

Momentum from speed and direction

Momentum with a negative velocity

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review