7.4 Apply impulse–momentum to a system of particles | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Systems of Particles

Topic 7.4 · EN PH 131 Engineering Mechanics: Dynamics

When several particles interact, tracking every force during every instant can be difficult. The impulse–momentum method gives a useful system-level result: the change in the system’s total linear momentum equals the net external impulse on it. This is especially useful over a time interval that includes a collision or a short, strong push. The method does not require the system’s internal forces to be zero; instead, for the whole system, internal forces occur in equal-and-opposite pairs and do not change its total momentum. The examples below use original values and state their systems, observers, axes, and assumptions.

What you will learn

Define a system of particles and describe its motion from a stated reference frame.
Apply the impulse–momentum equation to relate a system’s initial and final total momentum.
Distinguish external impulses from internal forces and identify when external impulse can be neglected.
Calculate total momentum or center-of-mass velocity and check the result using units and direction.

1. Define the system, observer, and momentum

A system is the set of particles chosen for analysis. Its boundary matters: a force exerted by something outside the chosen set is external, while a force between particles inside it is internal. An observer describes motion relative to a reference frame. In this lesson, use a stationary frame fixed to the ground, with perpendicular horizontal

x

and vertical

y

axes. State which direction is positive on each axis before assigning signs.

For particle

i

, linear momentum is mass times velocity. Momentum is a vector, so its direction follows the particle’s velocity. The total momentum of a system is the vector sum of the momenta of all its particles. This total is not generally the momentum of any one particle.

The system’s center-of-mass velocity is the mass-weighted average of particle velocities. It gives a compact description of the system’s overall translation: total momentum equals total mass multiplied by center-of-mass velocity. This identity does not say that all particles move at the center-of-mass velocity.

\mathbf{P}=\sum_i m_i\mathbf{v}_i=M\mathbf{v}_G

State the system boundary and the reference frame before solving.
Add particle momenta as vectors; do not add speed magnitudes when directions differ.
Mass is measured in kilograms; momentum is measured in kilogram-metres per second.

2. Impulse–momentum for the whole system

Newton’s second law for a particle can be integrated over a time interval. The time integral of force is impulse, a vector that measures the force’s accumulated effect over that interval. For a system, add the particle equations. The forces between particles inside the system cancel in equal-and-opposite pairs, leaving the net external impulse as the change in total momentum.

Use the equation from an initial time

t_1

to a final time

t_2

. Include every external force acting during that interval, such as a push, weight, or support force, if its impulse matters. A constant force has impulse equal to force multiplied by elapsed time. If force varies, integrate it over time, or use the area under its force–time graph component by component.

A system can be treated as having conserved total momentum when its net external impulse over the interval is zero or small enough to neglect for the question’s accuracy. A large internal interaction, such as a collision between system particles, does not by itself violate this condition. It is the external impulse that determines whether the system’s total momentum changes.

\mathbf{P}_1+\int_{t_1}^{t_2}\sum\mathbf{F}_{\mathrm{ext}}\,dt=\mathbf{P}_2

Impulse has units of newton-seconds, equivalent to kilogram-metres per second.
Include external impulses; internal interaction forces cancel only when the interacting particles are both inside the system.
Momentum conservation is an assumption about net external impulse, not a claim that no forces act.

3. A reliable solution plan

First identify which particles belong to the system and which observer’s frame is being used. State the initial and final instants, sketch the relevant bodies and external forces when this clarifies the setup, and choose positive coordinate directions. List known masses, velocities, forces, and elapsed time, then name the unknown.

Next, decide whether the system-level impulse–momentum equation is appropriate. It is often a direct choice when the question gives a force acting over a time interval, asks for a velocity change, or describes particles interacting over a short interval. Resolve vectors into components if needed. Apply the equation separately along each axis, keeping signed components throughout.

Solve symbolically before substituting values when practical. Then check that the result has the requested units, points in a direction consistent with the signed momentum balance, and reduces sensibly if an impulse is set to zero. If external impulse is negligible, the final total momentum must match the initial total momentum.

\Delta\mathbf{P}=\mathbf{J}_{\mathrm{ext}}

Use one consistent frame and sign convention for every velocity and impulse.
A system-level equation determines total momentum; extra information may be needed to determine each particle’s velocity.
Check component signs and units before interpreting the result.

4. Reading results and common checks

An impulse changes momentum in its own direction: a positive impulse along an axis increases that component of momentum, while a negative impulse decreases it. If the total mass is known, the final center-of-mass velocity follows from the final total momentum. Individual particle velocities cannot be inferred from total momentum alone unless the problem supplies enough additional information.

A useful dimensional check is that force multiplied by time has units

(\mathrm{N})(\mathrm{s})=\mathrm{kg\,m/s}

, the units of momentum. A useful limiting check is to set external impulse to zero: initial and final system momentum should then agree. For a system with two particles, forces they exert on each other may be large, but their net contribution to the system’s total momentum change is zero.

[\mathbf{J}]=\mathrm{N\,s}=\mathrm{kg\,m/s}

Do not confuse total momentum with the momentum of one selected particle.
Do not discard an external force’s impulse just because the time interval is short; estimate its size.
Use conservation only after checking the external impulse over the stated interval.

Worked example

External impulse changes a two-particle system’s total momentum

Two particles of masses

3.0\,\mathrm{kg}

and

2.0\,\mathrm{kg}

move horizontally in a ground-fixed frame. The positive

x

direction is to the right. Initially, their velocities are

+4.0\,\mathrm{m/s}

and

-1.0\,\mathrm{m/s}

. During the interval, the system receives a net external impulse of

+6.0\,\mathrm{N\,s}

. Find the final total momentum and the final center-of-mass velocity. No information is given to determine the particles’ separate final velocities.

System and external impulse

The arrow represents the direction of the net external impulse, not a force magnitude.

Set the system and signs
Take both particles as the system and use the stated ground-fixed frame. The initial and final states are the ends of the interval during which the external impulse acts; right is positive.
Find initial total momentum
Add the signed momenta of both particles. The second particle’s leftward velocity contributes a negative term.
$P_1=(3.0)(4.0)+(2.0)(-1.0)=10.0\,\mathrm{kg\,m/s}$
Apply system impulse–momentum
The stated external impulse is the total momentum change. Add it to the initial momentum to find the final total momentum.
$P_2=P_1+J_{\mathrm{ext}}=10.0+6.0=16.0\,\mathrm{kg\,m/s}$
Find center-of-mass velocity
Divide total momentum by total mass. The positive result means the center of mass moves to the right in the chosen frame.
$v_G=\frac{P_2}{m_1+m_2}=\frac{16.0}{5.0}=3.20\,\mathrm{m/s}$

Answer: Final total momentum:

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

to the right. Final center-of-mass velocity:

3.20\,\mathrm{m/s}

to the right. The separate particle velocities are not determined by the supplied information.

Check: The impulse has momentum units and increases rightward momentum by

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

. If the external impulse were zero, the final total momentum would remain

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

.

Worked example

Use momentum conservation to find one particle’s final velocity

Two particles interact along a straight, horizontal line. In a ground-fixed frame, choose right as positive. Particle A has mass

0.40\,\mathrm{kg}

and initial velocity

+5.0\,\mathrm{m/s}

; particle B has mass

0.60\,\mathrm{kg}

and is initially at rest. After their interaction, A moves at

-1.0\,\mathrm{m/s}

. Find B’s final velocity. During the interaction, the net external impulse on the two-particle system is negligible.

Define the system and states
Choose A and B together as the system and use the ground-fixed frame. Compare the state immediately before the interaction with the state immediately after it. Right is positive.
Check the conservation condition
The problem states that net external impulse is negligible, so the system’s total momentum before and after is the same. Forces exchanged between A and B are internal to this system.
$P_1=P_2$
Write the signed momentum balance
Use the given initial velocities and A’s final velocity. Let $v_{B2}$ be B’s final signed velocity.
$(0.40)(5.0)+(0.60)(0)=(0.40)(-1.0)+(0.60)v_{B2}$
Solve and interpret
Rearrange for B’s velocity. A positive answer means B moves right in the chosen frame.
$v_{B2}=\frac{2.0+0.40}{0.60}=4.0\,\mathrm{m/s}$

Answer: Particle B’s final velocity is

4.0\,\mathrm{m/s}

to the right.

Check: Initial total momentum is

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

. Final total momentum is

(-0.40)+(2.4)=2.0\,\mathrm{kg\,m/s}

, as required. The velocity units follow from momentum divided by mass.

Worked example

A time-varying external force changes total momentum in two directions

A system of particles has total mass

4.0\,\mathrm{kg}

and initial total momentum

\mathbf{P}_1=(3.0\mathbf{i}-2.0\mathbf{j})\,\mathrm{kg\,m/s}

in a ground-fixed frame. Let

+x

point right and

+y

point up. From

t=0

to

t=2.0\,\mathrm{s}

, the net external force is

\mathbf{F}_{\mathrm{ext}}(t)=(6.0t\mathbf{i}+4.0\mathbf{j})\,\mathrm{N}

, with time in seconds. Find the final total momentum and center-of-mass velocity.

Set the interval and coordinate directions
Take all particles in the stated system, observed from the ground-fixed frame. The initial state is at $t=0$ and the final state is at $t=2.0\,\mathrm{s}$ . Integrate the force components using right and up as positive.
Integrate the external force
Impulse is the time integral of force. Integrate each component over the full interval, keeping its sign.
$\mathbf{J}_{\mathrm{ext}}=\int_0^{2.0}(6.0t\mathbf{i}+4.0\mathbf{j})\,dt=(12\mathbf{i}+8.0\mathbf{j})\,\mathrm{N\,s}$
Find final total momentum
Add the impulse vector to initial total momentum component by component.
$\mathbf{P}_2=\mathbf{P}_1+\mathbf{J}_{\mathrm{ext}}=(15\mathbf{i}+6.0\mathbf{j})\,\mathrm{kg\,m/s}$
Find center-of-mass velocity
Divide each component of total momentum by the system’s total mass. Both components are positive, so the center of mass moves right and up.
$\mathbf{v}_{G2}=\frac{\mathbf{P}_2}{M}=(3.75\mathbf{i}+1.50\mathbf{j})\,\mathrm{m/s}$

Answer: Final total momentum is

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

. Final center-of-mass velocity is

(3.75\mathbf{i}+1.50\mathbf{j})\,\mathrm{m/s}

.

Check: The impulse components are positive because the force components are positive throughout the interval. Each impulse component has units of newton-seconds, and dividing momentum by kilograms gives metres per second.

Common mistakes and how to avoid them

Treating an internal force between two particles as an external impulse on their combined system.

Correction: Choose the system boundary first. Forces between particles both inside the system cancel in the total momentum balance.

Using momentum conservation whenever a collision or interaction occurs.

Correction: Check the net external impulse over the interval. Momentum is conserved only when that impulse is zero or suitably negligible.

Adding speed magnitudes instead of signed or vector momentum components.

Correction: Use the chosen axes and preserve each velocity’s direction when calculating momentum.

Assuming total momentum determines every particle’s final velocity.

Correction: The system equation determines total momentum. Individual velocities require enough additional information, such as one or more specified final velocities.

Lesson summary

Define the particles in the system, observer frame, time interval, axes, and positive directions.
The net external impulse equals the change in the system’s total momentum.
Internal forces cancel in the whole-system balance when their interacting particles are both included.
If net external impulse is negligible, total system momentum is conserved.
Use vector components, SI units, and a direction and limiting-case check.

Check your understanding

Question 1

A system has initial momentum

-5.0\,\mathrm{kg\,m/s}

along

x

. Its net external impulse is

+2.0\,\mathrm{N\,s}

. What is its final

x

-momentum?

$-7.0\,\mathrm{kg\,m/s}$
$-3.0\,\mathrm{kg\,m/s}$
$+3.0\,\mathrm{kg\,m/s}$
$+7.0\,\mathrm{kg\,m/s}$

Show answer and explanation

-3.0\,\mathrm{kg\,m/s}

Add the signed impulse to the initial momentum:

-5.0+2.0=-3.0\,\mathrm{kg\,m/s}

. The negative sign indicates momentum in the negative

x

direction.

Question 2

Two particles are both included in a system and exert forces on each other during an interval. Which statement is correct for the system’s total momentum balance?

Their internal forces contribute a net impulse equal to the larger force.
Their internal forces cancel in the system total; external impulse determines the change in total momentum.
The total momentum must be zero during the interaction.
Momentum conservation is impossible whenever particles exert forces on each other.

Show answer and explanation

Their internal forces cancel in the system total; external impulse determines the change in total momentum.

Internal interaction forces occur in equal-and-opposite pairs. For the combined system, the change in total momentum is determined by net external impulse.

Key terms

System: The selected particle or group of particles being analyzed.
Linear momentum: A particle’s mass multiplied by its velocity; it is a vector quantity.
Impulse: The time integral of force over an interval; its direction is that of the accumulated force effect.
External force: A force exerted on a system by something outside the chosen system boundary.
Center-of-mass velocity: The mass-weighted average velocity of the particles in a system.

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7.4 · Apply impulse–momentum to a system of particles

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Topic 7.4 · EN PH 131 Engineering Mechanics: Dynamics

What you will learn

1. Define the system, observer, and momentum

2. Impulse–momentum for the whole system

3. A reliable solution plan

4. Reading results and common checks

External impulse changes a two-particle system’s total momentum

Use momentum conservation to find one particle’s final velocity

A time-varying external force changes total momentum in two directions

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