6.5 Analyze introductory oblique impact | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Impulse, Momentum, and Impact

Use normal and tangential components, impulse–momentum, and restitution

An oblique impact occurs when the relative velocity of colliding bodies is not entirely along the contact normal. The normal is perpendicular to the contacting surfaces at the contact point; the tangent is along the surface. In this lesson, velocities are measured by an observer at rest relative to the ground. For smooth contact, the contact impulse acts along the normal, so it changes normal velocity but not tangential velocity. The coefficient of restitution describes the change in relative normal speed. Because impact lasts a short time, we often neglect impulses from forces such as gravity when they are small compared with the contact impulse.

What you will learn

Define the system, observer, frame, contact normal, and positive directions for an impact.
Resolve velocities into components normal and tangent to a smooth contact surface.
Apply impulse–momentum and restitution to determine velocities just after impact.
Check results using signs, units, momentum, and limiting cases.

1. Define the contact geometry and states

First identify the system. For a particle striking a fixed surface, the system is the particle. For two colliding particles, consider both together when checking total momentum, and consider each particle separately when applying the contact impulse. State the observer and frame: here, the observer is stationary relative to the ground.

Choose perpendicular unit vectors: a normal vector

\mathbf{n}

and a tangent vector

\mathbf{t}

. For a particle striking a fixed surface, take

\mathbf{n}

outward from the surface toward the particle. The positive coordinate directions are along these vectors. A negative normal velocity then means motion into the surface. For two particles, state which way the normal points and keep that convention throughout.

The initial state is immediately before contact; the final state is immediately after separation. The calculation relates velocities at those two instants, not the detailed motion during contact. Smooth contact is an idealization: it means there is no tangential contact impulse. If a tangential impulse acts, the unchanged-tangent rule does not apply.

\mathbf{v}=v_n\mathbf{n}+v_t\mathbf{t}

The normal is perpendicular to the contact; the tangent lies along it.
Use the same directions before and after impact.
Smooth contact has no tangential contact impulse.

2. Governing impact relationships

Impulse–momentum says that the impulse on a particle equals its change in momentum. Impulse is the accumulated effect of force over the short contact interval. Contact impulse is often much larger than the impulse of gravity during this interval, so gravity’s impulse may be neglected as an approximation.

For a particle striking a fixed smooth surface, resolve its velocity into normal and tangent components. The contact impulse is normal, so the tangential component is unchanged. The normal component reverses direction and is scaled by the coefficient of restitution

e

. With the outward normal positive and incoming normal velocity

v_n^-<0

, the outgoing component is

v_n^+=-e v_n^-

. Thus it points away from the surface.

For two particles, define relative velocity as the velocity of A minus the velocity of B. Choose the normal so that the incoming relative normal component

g_n=(\mathbf{v}_A^- -\mathbf{v}_B^-)\cdot\mathbf{n}

is positive. Restitution gives an outgoing relative normal component of

-e g_n

. The contact impulses on the two particles are equal and opposite. Therefore, total momentum is conserved if the external impulse on the pair is negligible.

In the introductory model,

0\le e\le1

. At

e=0

, the relative normal speed after impact is zero; at

e=1

, its magnitude is unchanged. Neither limit means that tangential velocities reverse.

e=-\frac{(\mathbf{v}_A^+-\mathbf{v}_B^+)\cdot\mathbf{n}}{(\mathbf{v}_A^--\mathbf{v}_B^-)\cdot\mathbf{n}}

Use impulse–momentum for velocity changes caused by contact.
For smooth contact, tangential velocity components are unchanged.
Restitution applies to relative normal velocity, not the full velocity vector.

3. A reliable solution method

List the masses, incoming velocities, contact normal, contact condition, and restitution value. Identify the outgoing velocity or velocities to find. Resolve velocities using the same normal and tangent directions; this separates the impact behavior from the geometry of the velocity vectors.

For a particle striking a fixed smooth surface, retain its tangent component and multiply its incoming normal component by

-e

. For two particles, let

J

be the impulse magnitude along the chosen normal. Apply impulse–momentum to both particles, then use restitution to determine

J

. With the stated convention that

g_n>0

for approach, the impulse on B is along

+\mathbf{n}

and the impulse on A is opposite.

Check that the chosen signs describe approach before impact and separation afterward. Check total momentum for an isolated pair, confirm impulse units are newton-seconds, and test the limits

e=0

and

e=1

. If a Cartesian vector is required, reconstruct it from its normal and tangent components.

J=\frac{(1+e)g_n}{1/m_A+1/m_B}

Resolve velocity before applying the impact rules.
Confirm the approach and separation signs using the chosen normal.
Keep units and vector directions consistent when reconstructing velocity.

Worked example

Particle striking a smooth vertical wall

A 0.20 kg particle moves toward a fixed smooth vertical wall with horizontal velocity 4.0 m/s toward the wall and vertical velocity 3.0 m/s upward. The coefficient of restitution is 0.50. Find its velocity just after impact.

Particle approaching wall

The particle approaches the wall horizontally; the contact impulse points away from it.

Define system and directions
Use the ground-fixed frame. The system is the particle; the initial state is just before contact and the final state just after. Positive normal $n$ points away from the wall, and positive tangent $t$ points upward. The particle moves toward the wall, so its incoming normal component is negative.
$v_n^-=-4.0\ \mathrm{m/s},\quad v_t^-=3.0\ \mathrm{m/s}$
Apply smooth-contact rules
The smooth wall gives no tangential impulse, so the tangent component is unchanged. Restitution reverses and scales the normal component. The outgoing normal component is positive, directed away from the wall.
$v_n^+=-e v_n^-=2.0\ \mathrm{m/s},\quad v_t^+=v_t^-=3.0\ \mathrm{m/s}$
Reconstruct velocity
Combine the perpendicular components along the defined normal and tangent. The speed is the magnitude of this velocity vector.
$\mathbf{v}^+=(2.0\mathbf{n}+3.0\mathbf{t})\ \mathrm{m/s},\quad |\mathbf{v}^+|=\sqrt{13}\ \mathrm{m/s}$

Answer: The particle leaves at 2.0 m/s away from the wall and 3.0 m/s upward. Its speed is approximately 3.61 m/s.

Check: The normal velocity points away from the wall, and the tangential velocity is unchanged. The outgoing normal speed is half the incoming normal speed, as required by

e=0.50

.

Worked example

Two particles with an oblique line of impact

Particle A has mass 2.0 kg and velocity

(5.0\mathbf{i}+2.0\mathbf{j})

m/s. Particle B has mass 1.0 kg and velocity

(1.0\mathbf{i}-1.0\mathbf{j})

m/s. Their smooth line of impact has unit normal

\mathbf{n}=0.6\mathbf{i}+0.8\mathbf{j}

, directed from A toward B. The coefficient of restitution is 0.50. Find both outgoing velocities.

Define system and approach speed
Use the ground-fixed frame and take both particles as the system for the momentum check. The initial and final states are immediately before and after contact. Define relative velocity as A minus B and use the stated normal. Its positive normal component indicates approach under this convention.
$g_n=(\mathbf{v}_A^- -\mathbf{v}_B^-)\cdot\mathbf{n}=4.8\ \mathrm{m/s}$
Find the impulse
Let $J$ be the impulse magnitude on B along $\mathbf{n}$ ; A receives the opposite impulse. Combining impulse–momentum with restitution determines the impulse magnitude.
$J=\frac{(1+0.50)(4.8)}{1/2.0+1/1.0}=4.8\ \mathrm{N\cdot s}$
Apply impulse–momentum
Subtract the impulse per unit mass along the normal from A’s velocity and add it to B’s velocity. The normal impulse leaves both particles’ tangential velocity components unchanged.
$\mathbf{v}_A^+=\mathbf{v}_A^- -\frac{J}{m_A}\mathbf{n},\quad \mathbf{v}_B^+=\mathbf{v}_B^- +\frac{J}{m_B}\mathbf{n}$
Substitute and check separation
Substitution gives the outgoing Cartesian components. The final relative normal component is negative, so the particles are separating.
$\mathbf{v}_A^+=(3.56\mathbf{i}+0.08\mathbf{j})\ \mathrm{m/s},\quad \mathbf{v}_B^+=(3.88\mathbf{i}+2.84\mathbf{j})\ \mathrm{m/s}$

Answer: Particle A leaves at

(3.56\mathbf{i}+0.08\mathbf{j})

m/s, and particle B leaves at

(3.88\mathbf{i}+2.84\mathbf{j})

m/s.

Check: Initial total momentum is

(11\mathbf{i}+3\mathbf{j})

kg·m/s. The final total is also

(11\mathbf{i}+3\mathbf{j})

kg·m/s. The final relative normal component is

-2.4

m/s, equal to

-e g_n

.

Worked example

Particle striking an inclined smooth plane

A particle approaches a fixed smooth plane with velocity

(-6.0\mathbf{i})

m/s. The plane’s outward unit normal is

\mathbf{n}=\cos30^\circ\mathbf{i}+\sin30^\circ\mathbf{j}

. The coefficient of restitution is 0.25. Find the velocity after impact.

Approach to inclined plane

The incoming velocity has a negative component along the outward normal; the contact impulse points outward.

Set frame and contact axes
Observe from the ground. The particle is the system; compare the instants just before and after contact. Use the given outward normal and choose the perpendicular tangent $\mathbf{t}=-\sin30^\circ\mathbf{i}+\cos30^\circ\mathbf{j}$ . The negative dot product shows the particle moves into the plane.
$\mathbf{v}^-\cdot\mathbf{n}=-6\cos30^\circ<0$
Resolve velocity and apply impact rules
The incoming normal component is $-6\cos30^\circ$ m/s and the tangent component is $6\sin30^\circ$ m/s. Retain the tangent component and reverse and scale the normal component by $e$ .
$v_n^+=0.25(6\cos30^\circ)=1.299\ \mathrm{m/s},\quad v_t^+=6\sin30^\circ=3.0\ \mathrm{m/s}$
Reconstruct Cartesian components
Combine the outgoing components along the stated unit vectors. The resulting normal component is positive, so the particle moves away from the plane.
$\mathbf{v}^+=v_n^+\mathbf{n}+v_t^+\mathbf{t}=(-0.375\mathbf{i}+3.25\mathbf{j})\ \mathrm{m/s}$

Answer: The outgoing velocity is approximately

(-0.375\mathbf{i}+3.25\mathbf{j})

m/s.

Check: Taking the dot product of the reported vector with

\mathbf{n}

gives approximately 1.299 m/s, the positive outgoing normal component. Its tangent component is 3.0 m/s, unchanged from before impact.

Common mistakes and how to avoid them

Applying restitution to the whole velocity vector.

Correction: Apply restitution only to the relative normal component. For smooth contact, tangential velocity components are unchanged.

Using the restitution sign without checking the normal direction.

Correction: Define the positive normal and verify that the incoming relative normal component represents approach before using the restitution equation.

Treating impulse as a force with units of newtons.

Correction: Impulse has units N·s, equivalently kg·m/s. It equals the change in momentum.

Assuming total kinetic energy is conserved for every impact.

Correction: For an isolated pair with negligible external impulse, conserve momentum. Restitution describes normal impact behavior; it does not say kinetic energy is conserved.

Lesson summary

Specify the system, ground-fixed observer, initial and final instants, and contact normal.
Resolve velocity into normal and tangent components.
For smooth contact, the impulse is normal and tangential velocity is unchanged.
Use restitution for relative normal velocity and impulse–momentum for velocity changes.
Check approach and separation signs, momentum, SI units, and limiting values of restitution.

Check your understanding

Question 1

A particle approaches a fixed smooth wall with normal velocity component

-5

m/s. If

e=0.4

, what is its outgoing normal component, taking positive direction away from the wall?

$-2$ m/s
$2$ m/s
$5$ m/s
$12.5$ m/s

Show answer and explanation

2

m/s

The outgoing component is

-e v_n^-=-0.4(-5)=2

m/s, directed away from the wall.

Question 2

For a smooth impact against a fixed surface, which component of a particle’s velocity remains unchanged?

The normal component
The tangential component
Both components reverse
Neither component can be determined

Show answer and explanation

The tangential component

The smooth-contact impulse has no tangential component, so impulse–momentum leaves the tangential velocity unchanged.

Question 3

Two particles approach with relative normal speed 3.0 m/s, and their coefficient of restitution is 0.6. What is their relative normal velocity after impact, using the same normal convention?

1.8 m/s approaching
-1.8 m/s separating
3.6 m/s separating
-3.0 m/s approaching

Show answer and explanation

-1.8 m/s separating

Restitution gives outgoing relative normal velocity

-e g_n=-0.6(3.0)=-1.8

m/s, so the bodies separate.

Key terms

Oblique impact: An impact in which relative velocity is not entirely along the contact normal.
Contact normal: A direction perpendicular to the contacting surfaces at the point of contact.
Coefficient of restitution: A dimensionless measure of separating relative normal speed compared with approaching relative normal speed.
Impulse: The accumulated effect of a force over a time interval; it equals the change in momentum.

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6.5 · Analyze introductory oblique impact

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Use normal and tangential components, impulse–momentum, and restitution

What you will learn

1. Define the contact geometry and states

2. Governing impact relationships

3. A reliable solution method

Particle striking a smooth vertical wall

Two particles with an oblique line of impact

Particle striking an inclined smooth plane

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review