6.4 Analyze direct central impact with restitution | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Impulse, Momentum, and Impact

Using linear momentum and relative velocity to determine motion immediately after impact

When two particles collide directly, their motion during impact is along one straight line, called the line of impact. The collision is brief: forces between the particles can create large impulses, while impulses from external forces are often negligible over the same interval. For the two-particle system, this allows us to conserve linear momentum along the line. Momentum alone generally cannot determine both final velocities, so we also use the coefficient of restitution, which relates relative approach speed to relative separation speed. In this lesson, velocities are signed scalar components along the line. Positive means motion in the chosen positive direction; negative means motion the other way.

What you will learn

Describe a direct central impact and identify its line of impact.
Apply linear momentum conservation and the coefficient of restitution to solve for post-impact velocities.
Choose a consistent positive direction and interpret velocity signs correctly.
Check impact results using momentum, units, and limiting cases.

1. Define the system and choose a frame

A direct central impact occurs when the particles approach and collide along one common straight line. Model each body as a particle: only translation along that line is included, not rotation.

Use an observer fixed to the ground and a one-dimensional coordinate axis along the line of impact. Choose right as positive, for example. Let

v_A

and

v_B

be the signed velocities just before impact, and

v_A'

and

v_B'

the signed velocities just after impact. The prime marks the final state; it does not mean a derivative.

The system is the two colliding particles together. During the short impact, if the net external impulse along the line is negligible, total linear momentum of this system is conserved. The contact impulses are internal to the system: they change each particle's momentum but not the system's total momentum.

m_Av_A+m_Bv_B=m_Av_A'+m_Bv_B'

Use one coordinate axis along the line of impact.
A velocity has direction and sign; speed is its nonnegative magnitude.
Momentum conservation requires negligible net external impulse along the line during impact.

2. Relate relative speeds with restitution

The coefficient of restitution,

e

, compares the relative speed of separation after impact with the relative speed of approach before impact. Suppose A is on the left and approaches B to its right. With right positive, the approach speed is

v_A-v_B

, and the separation speed after impact is

v_B'-v_A'

.

For this arrangement, write

v_B'-v_A'=e(v_A-v_B)

. Both relative speeds in this form are positive when the particles approach before impact and separate afterward. Check that the initial velocities describe approach before using this form.

For an ordinary passive impact,

0\leq e\leq1

. If

e=0

, the particles have no relative velocity immediately after impact. If

e=1

, the relative speed of separation equals the relative speed of approach. Neither statement requires an individual particle to stop or reverse direction.

v_B'-v_A'=e(v_A-v_B)

Restitution compares relative speeds along the line of impact.
Momentum conservation and restitution together usually give two equations for two final velocities.
Check the initial directions so the approach-speed expression has the correct sign.

3. Solve and check the impact model

List the masses, signed initial velocities, and value of

e

, then write both governing equations. Solving for the signed final velocities gives direction as well as speed. A negative result means motion opposite to the positive axis; it is not a negative speed.

Momentum terms have units of kilogram metres per second. Each term in the restitution equation has units of metres per second, so both equations are dimensionally consistent. After solving, substitute the results into both equations and check that the particles separate immediately after impact.

A limiting case helps confirm the model: for equal masses, a stationary second particle, and

e=1

, the first particle stops and the second takes its initial velocity. This is consistent with momentum conservation and equal relative speeds before and after impact.

e=\frac{v_B'-v_A'}{v_A-v_B}

Solve momentum and restitution together; neither relationship alone is generally enough.
Use velocity signs to report direction, and give speed separately if requested.
Verify momentum and the relative-speed relationship after solving.

4. Choose the model conditions carefully

This impulse–momentum method applies to a short, direct impact when the relevant particle motion is along one line. It avoids finding the detailed contact force during impact. Instead, momentum conservation describes the system's overall change, and restitution describes the change in relative motion.

Confirm that all velocities use the same observer and that the particles are approaching before impact. If an external impulse along the line is not negligible, the two-particle momentum equation needs an external impulse term rather than the conservation form used here.

In the examples, the system is the pair of particles, the observer is ground-fixed, the positive direction is right, and the final state is immediately after contact ends. Each diagram shows the one-dimensional direction of approach without suggesting a detailed contact-force model.

This is an impulse–momentum analysis, not a force-versus-time calculation.
The simple momentum equation depends on negligible external impulse along the line.
The chosen positive direction must be used consistently for every velocity.

Worked example

A moving particle strikes a stationary particle

Particle A has mass

2.0\ \mathrm{kg}

and moves right at

4.0\ \mathrm{m/s}

. Particle B has mass

3.0\ \mathrm{kg}

and is initially at rest. They collide directly, and

e=0.50

. Find their velocities immediately after impact. The system is A and B; the observer is fixed to the ground; right is positive. Assume external impulse along the line is negligible.

Before impact

A approaches stationary B along the line of impact.

Set up momentum
The known signed velocities are $v_A=4.0\ \mathrm{m/s}$ and $v_B=0$ . The unknowns are $v_A'$ and $v_B'$ . Negligible external impulse lets us conserve momentum for the two-particle system.
$2v_A'+3v_B'=8$
Apply restitution
A approaches B at $4.0\ \mathrm{m/s}$ . The separation speed after impact is half this approach speed.
$v_B'-v_A'=0.50(4.0)=2.0$
Solve and interpret
Use $v_B'=v_A'+2.0$ in the momentum equation. Solving gives positive signed velocities, so both particles move right immediately after impact.
$v_A'=0.40\ \mathrm{m/s},\qquad v_B'=2.40\ \mathrm{m/s}$

Answer: Immediately after impact, A moves right at

0.40\ \mathrm{m/s}

and B moves right at

2.40\ \mathrm{m/s}

.

Check: Initial momentum is

2(4.0)+3(0)=8.0\ \mathrm{kg\,m/s}

. Final momentum is

2(0.40)+3(2.40)=8.0\ \mathrm{kg\,m/s}

. The separation speed is

2.40-0.40=2.0\ \mathrm{m/s}

, as required.

Worked example

Two particles approach one another

Particle A has mass

1.5\ \mathrm{kg}

and velocity

+6.0\ \mathrm{m/s}

. Particle B has mass

2.5\ \mathrm{kg}

and velocity

-2.0\ \mathrm{m/s}

. They collide directly with

e=0.80

. Find their velocities immediately after impact. The system is both particles, the observer is ground-fixed, and right is positive. Assume external impulse is negligible.

Before impact

A moves right and B moves left, so they approach.

Identify approach speed and momentum
A moves right while B moves left, so the particles approach. Their relative approach speed is $6.0-(-2.0)=8.0\ \mathrm{m/s}$ . Initial system momentum is $1.5(6.0)+2.5(-2.0)=4.0\ \mathrm{kg\,m/s}$ .
$1.5v_A'+2.5v_B'=4.0$
Use restitution
The relative separation speed is $0.80(8.0)=6.4\ \mathrm{m/s}$ .
$v_B'-v_A'=6.4$
Solve and check directions
Substitute $v_B'=v_A'+6.4$ into momentum conservation. A's negative result means it rebounds left; B's positive result means it moves right.
$v_A'=-3.0\ \mathrm{m/s},\qquad v_B'=3.4\ \mathrm{m/s}$

Answer: A rebounds left at

3.0\ \mathrm{m/s}

, and B moves right at

3.4\ \mathrm{m/s}

.

Check: Initial momentum is

1.5(6.0)+2.5(-2.0)=4.0\ \mathrm{kg\,m/s}

. Final momentum is

1.5(-3.0)+2.5(3.4)=4.0\ \mathrm{kg\,m/s}

. Separation speed is

3.4-(-3.0)=6.4\ \mathrm{m/s}

.

Worked example

Find restitution from measured velocities

Two particles approach along a straight line. Their initial velocities are

v_A=5.0\ \mathrm{m/s}

and

v_B=1.0\ \mathrm{m/s}

, with right positive. Immediately after impact, their measured velocities are

v_A'=3.0\ \mathrm{m/s}

and

v_B'=5.0\ \mathrm{m/s}

. Find

e

. The observer is ground-fixed; take both particles as the system. Assume external impulse is negligible.

Along the impact line

Before impact A is faster; after impact B is faster.

Confirm approach and separation
Before impact, A is faster than B, so approach speed is $5.0-1.0=4.0\ \mathrm{m/s}$ . After impact, B is faster, so separation speed is $5.0-3.0=2.0\ \mathrm{m/s}$ .
$e=\frac{v_B'-v_A'}{v_A-v_B}$
Calculate the ratio
Divide the measured separation speed by the approach speed.
$e=\frac{5.0-3.0}{5.0-1.0}=0.50$

Answer: The coefficient of restitution is

e=0.50

.

Check: The value lies in the ordinary passive-impact range

0\leq e\leq1

. The relative speed after impact is half the relative speed before impact, consistent with the result.

Common mistakes and how to avoid them

Using positive speeds for both particles even when they move in opposite directions.

Correction: Assign signed velocity components using one shared positive axis. For example, motion left is negative when right is positive.

Writing restitution as a ratio of one particle's final and initial speeds.

Correction: Restitution compares the particles' relative speed of separation with their relative speed of approach along the line of impact.

Assuming momentum conservation without considering external impulse.

Correction: Conserve momentum for the two-particle system only when net external impulse along the impact line is negligible during the short collision.

Treating a negative final velocity as impossible.

Correction: A negative signed velocity indicates motion opposite to the chosen positive direction. Check whether that direction fits the collision.

Lesson summary

Choose an observer and positive axis along the direct line of impact.
For negligible external impulse, conserve the two-particle system's linear momentum.
Use restitution to relate relative separation speed to relative approach speed.
Solve for signed final velocities, then verify momentum, separation, and the value of $e$ .

Check your understanding

Question 1

Two equal-mass particles approach directly. A moves right at

3.0\ \mathrm{m/s}

and B moves left at

1.0\ \mathrm{m/s}

. If

e=1

, what are their velocities immediately after impact?

A moves left at $1.0\ \mathrm{m/s}$ ; B moves right at $3.0\ \mathrm{m/s}$ .
A moves right at $1.0\ \mathrm{m/s}$ ; B moves left at $3.0\ \mathrm{m/s}$ .
Both move right at $1.0\ \mathrm{m/s}$ .
A stops; B moves right at 2.0\ $\mathrm{m/s}$ .

Show answer and explanation

A moves left at

1.0\ \mathrm{m/s}

; B moves right at

3.0\ \mathrm{m/s}

.

For equal masses in a direct impact with

e=1

, the particles exchange velocities. Momentum remains unchanged and their relative separation speed is

4.0\ \mathrm{m/s}

.

Question 2

What does

e=0

imply about relative motion immediately after a direct impact?

The particles have zero relative velocity immediately after impact.
Both particles must stop in the ground-fixed frame.
Each particle reverses its own velocity.
The system's momentum becomes zero.

Show answer and explanation

The particles have zero relative velocity immediately after impact.

With zero restitution, the relative separation speed is zero. This does not mean either particle must be stationary in the observer's frame.

Key terms

Direct central impact: A collision in which the particles' motion during impact is along one common straight line through their centres.
Line of impact: The common straight line along which particles approach and separate in a direct central impact.
Coefficient of restitution: The ratio of relative speed of separation after impact to relative speed of approach before impact.
Impulse: The effect of a force acting over a time interval, equal to the change in momentum.

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6.4 · Analyze direct central impact with restitution

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Using linear momentum and relative velocity to determine motion immediately after impact

What you will learn

1. Define the system and choose a frame

2. Relate relative speeds with restitution

3. Solve and check the impact model

4. Choose the model conditions carefully

A moving particle strikes a stationary particle

Two particles approach one another

Find restitution from measured velocities

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