6.3 Apply conservation of linear momentum to a particle system | 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 6.3

Linear momentum depends on both mass and velocity. For a system of particles, total linear momentum is the vector sum of each particle’s momentum. During an interaction, particles may exert large forces on one another, but those forces are internal to the complete system. Conservation is justified when the net external impulse on the chosen system is zero, or when it is small enough to neglect over the interval. The key steps are to define the system and its initial and final states, then assess the external impulse.

What you will learn

Define a particle system, observer, reference frame, coordinate axes, and positive directions.
Use the impulse–momentum relationship to decide whether a system’s linear momentum is conserved.
Apply conservation of linear momentum to one-dimensional and two-dimensional particle interactions.
Check momentum results using units, vector directions, and limiting cases.

1. Define the system, frame, and momentum

A system is the collection of particles whose motion you are analyzing. For two interacting carts, the system might include both carts; for an object separating into pieces, it might include all the pieces. The observer is the person measuring motion. The reference frame is the coordinate system used for those measurements. In this lesson, use an observer fixed to the ground, with positive

x

to the right and positive

y

upward.

A particle’s linear momentum is its mass multiplied by its velocity. Velocity is a vector, so momentum has the same direction as velocity. The system’s total momentum is the vector sum of the individual momenta. Add components separately: all

x

components together and all

y

components together.

Choose the system and the initial and final instants before writing an equation. The initial state is just before the interaction; the final state is just after it. A short interval is common in collisions, but short duration alone does not prove that momentum is conserved.

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

State which particles belong to the system.
Momentum is a vector; use signed components or vector addition.
Use the same observer and coordinate directions for initial and final states.

2. Decide whether momentum is conserved

The impulse–momentum principle relates a system’s change in total momentum to the net external force accumulated over time. An impulse points in the force’s direction and has units of newton-seconds, equivalent to kilogram-metres per second.

For a system of particles, forces that members of the system exert on one another are internal. They occur in equal-and-opposite pairs, so their contributions cancel in the total-system momentum balance. External forces, such as a push from outside the system, can change the system’s total momentum.

If the net external impulse is zero, total momentum before and after is equal. If the external impulse is not zero but is known, use the impulse–momentum relationship rather than assuming conservation. In a short interaction, an external force may have a negligible impulse compared with the particles’ momenta; state that approximation explicitly.

Momentum conservation does not mean each particle keeps its own momentum. Particles can exchange momentum while the system total stays unchanged. Also, momentum conservation alone may not determine every final velocity. You may need additional information, such as a shared final velocity or a specified direction.

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

Assess external impulse over the chosen interval, not just the size of a force at one instant.
Internal forces cancel in the total-system balance.
Conserve momentum only when external impulse is zero or explicitly neglected.

3. Set up components and solve

For one-dimensional motion, choose one positive direction and give velocities signs consistent with it. If right is positive, a leftward velocity has a negative

x

component. A speed is a nonnegative magnitude; it does not include direction, so do not use it in place of a signed velocity without specifying that direction.

For two-dimensional motion, apply the momentum balance in both coordinate directions. If a velocity is given by a magnitude and angle, resolve it into components before adding momenta. A particle can have a positive component along one axis and a negative component along the other.

A reliable sequence is: identify the system and observer; mark the initial and final states; list masses and velocities; choose axes and signs; assess external impulse; write the vector balance; solve components; and check units and direction. Interpret a negative answer relative to the chosen axis rather than changing its sign by guesswork.

\sum_i m_i\mathbf{v}_{i,1}+\mathbf{J}_{\mathrm{ext}}=\sum_i m_i\mathbf{v}_{i,2}

Use mass in kilograms and velocity in metres per second.
A negative component means motion opposite the positive coordinate direction.
Momentum and impulse have units of kilogram-metres per second.

Worked example

One-dimensional interaction

A

2.0\,\mathrm{kg}

cart moves right at

3.0\,\mathrm{m/s}

and catches a

1.0\,\mathrm{kg}

cart moving right at

1.0\,\mathrm{m/s}

. They latch together. Treat both carts as the system and neglect external impulse during the short interaction. Find their common velocity just after latching.

Two carts, positive direction right

The system includes both carts; their final velocity is common.

Define system and signs
The system is both carts, observed from the ground. Let right be positive $x$ . The initial state is before latching and the final state is after; both carts then share one velocity.
Apply momentum conservation
The external impulse is neglected, so the system’s total momentum is the same in both states. After latching, the final momentum is the combined mass multiplied by the common velocity.
$m_1v_{1,1}+m_2v_{2,1}=(m_1+m_2)v_2$
Substitute and solve
The given rightward velocities are positive. The positive result means the joined carts move right.
$v_2=\frac{(2.0)(3.0)+(1.0)(1.0)}{2.0+1.0}=2.33\,\mathrm{m/s}$

Answer: The carts move together at

2.33\,\mathrm{m/s}

to the right.

Check: Initial momentum is

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

to the right. Final momentum is

(3.0\,\mathrm{kg})(2.33\,\mathrm{m/s})=6.99\,\mathrm{kg\,m/s}

to the right, matching within rounding. The units reduce to

\mathrm{kg\,m/s}

. If the carts had equal initial velocities, the common velocity would equal that shared velocity, as expected.

Worked example

A two-dimensional separation

A

4.0\,\mathrm{kg}

object is initially at rest and separates into two pieces. A

1.5\,\mathrm{kg}

piece moves at

8.0\,\mathrm{m/s}

at

30^\circ

above positive

x

. Neglect external impulse during separation. Find the velocity of the

2.5\,\mathrm{kg}

piece.

Define system and axes
The system is both pieces, viewed from the ground. Let positive $x$ point right and positive $y$ point up. The object is initially at rest, so its initial total momentum is zero.
$\mathbf{P}_1=\mathbf{0}$
Resolve the known velocity
With negligible external impulse, the two final momenta must add to zero. Resolve the first piece’s velocity into the chosen axes; both components are positive.
$\mathbf{v}_1=(8.0\cos30^\circ,\,8.0\sin30^\circ)=(6.93,\,4.00)\,\mathrm{m/s}$
Solve the vector balance
The second piece must have momentum equal and opposite to the first piece’s momentum. Divide those components by its mass to obtain its velocity components.
$\mathbf{v}_2=-\frac{1.5}{2.5}(6.93,\,4.00)=(-4.16,\,-2.40)\,\mathrm{m/s}$
Interpret the direction
Both components are negative, so the second piece moves left and downward. Its speed is about $4.80\,\mathrm{m/s}$ , directed $30^\circ$ below the negative $x$ direction.
$|\mathbf{v}_2|=\sqrt{(-4.16)^2+(-2.40)^2}=4.80\,\mathrm{m/s}$

Answer: The

2.5\,\mathrm{kg}

piece has velocity

(-4.16\,\mathbf{i}-2.40\,\mathbf{j})\,\mathrm{m/s}

, or

4.80\,\mathrm{m/s}

directed

30^\circ

below the negative

x

axis.

Check: The first piece’s momentum components are

(10.39,6.00)\,\mathrm{kg\,m/s}

, while the second piece’s are approximately

(-10.40,-6.00)\,\mathrm{kg\,m/s}

. They sum to the initial zero momentum within rounding. Both velocity components have units of metres per second.

Worked example

Accounting for an external impulse

A

0.50\,\mathrm{kg}

puck moves right at

2.0\,\mathrm{m/s}

. During a brief interval, an external force supplies an impulse of

1.0\,\mathrm{N\,s}

to the left. Find the puck’s final velocity along the same line.

Puck with leftward impulse

The leftward arrow indicates the direction of the external impulse.

Set the system and sign
The system is the puck, observed from the ground. Take right as positive. The initial velocity is positive, while the external impulse is negative because it points left.
$p_1=(0.50)(2.0)=1.0\,\mathrm{kg\,m/s}$
Use impulse–momentum
The external impulse is not negligible, so momentum is not conserved. Add the signed impulse to the initial momentum to find the final momentum.
$p_2=p_1+J_{\mathrm{ext}}=1.0-1.0=0\,\mathrm{kg\,m/s}$
Find the final velocity
Divide the final momentum by the puck’s mass. The puck is momentarily at rest at the final instant; this result does not say what happens after that instant.
$v_2=\frac{p_2}{m}=0\,\mathrm{m/s}$

Answer: The puck’s final velocity is

0\,\mathrm{m/s}

.

Check: The leftward impulse cancels the initial rightward momentum exactly. Since

1\,\mathrm{N\,s}=1\,\mathrm{kg\,m/s}

, units are consistent. An impulse smaller than the initial momentum would leave the puck moving right; one larger in magnitude would reverse its direction.

Common mistakes and how to avoid them

Conserving momentum even though a significant external impulse acts.

Correction: Use the impulse–momentum balance, or justify why the external impulse is negligible for the interval.

Adding speeds as if they were signed momentum components.

Correction: Assign each velocity a direction and use signed components, or add the vectors directly.

Including only one member of an interacting group while claiming internal forces cancel.

Correction: Include all interacting particles in the system if you want their mutual forces to be internal.

Assuming momentum conservation means every particle keeps its initial momentum.

Correction: Conservation applies to the system total; particles may exchange momentum.

Lesson summary

Choose and describe the particle system, observer, reference frame, axes, and initial and final states.
Total system momentum is the vector sum of the momenta of its particles.
Internal forces cancel in the whole-system balance; external impulse changes total momentum.
Conserve linear momentum only when net external impulse is zero or can reasonably be neglected.
Use signed components, consistent SI units, and direction checks to interpret the result.

Check your understanding

Question 1

A two-particle system has zero initial momentum and negligible external impulse. One particle’s final momentum is

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

to the right. What is the other particle’s final momentum?

$6\,\mathrm{kg\,m/s}$ to the right
$6\,\mathrm{kg\,m/s}$ to the left
Zero
It cannot be determined from the information given

Show answer and explanation

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

to the left

The total final momentum must equal the initial zero momentum, so the second particle’s momentum must be equal in magnitude and opposite in direction.

Question 2

A system’s initial momentum is

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

right, and it receives an external impulse of

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

left. What is its final momentum?

$5\,\mathrm{kg\,m/s}$ right
$3\,\mathrm{kg\,m/s}$ right
$3\,\mathrm{kg\,m/s}$ left
$4\,\mathrm{kg\,m/s}$ right

Show answer and explanation

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

right

Taking right as positive gives final momentum

4-1=3\,\mathrm{kg\,m/s}

, directed right.

Key terms

Linear momentum: A vector equal to a particle’s mass multiplied by its velocity.
System: The chosen collection of particles whose total motion is being analyzed.
External impulse: The time accumulation of the net force exerted on a system from outside that system.
Conservation of linear momentum: The system’s total linear momentum remains unchanged when its net external impulse is zero.

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6.3 · Apply conservation of linear momentum to a particle system

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 6.3

What you will learn

1. Define the system, frame, and momentum

2. Decide whether momentum is conserved

3. Set up components and solve

One-dimensional interaction

A two-dimensional separation

Accounting for an external impulse

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