Dynamics · Chapter 15
Engineering Mechanics · Dynamics

Kinetics of a Particle:
Impulse and Momentum

This chapter integrates the equation of motion over time to produce the impulse–momentum principle, a tool that links force, velocity, and elapsed time directly — without solving for acceleration first. From there it builds toward conservation of momentum, the mechanics of impact, angular momentum, steady fluid streams, and the propulsion of bodies whose mass changes as they move.

Learning objectives

  • Develop the principle of linear impulse and momentum for a particle and apply it to force–velocity–time problems.
  • Study the conservation of linear momentum for systems of particles.
  • Analyze the mechanics of impact, including the coefficient of restitution.
  • Introduce angular impulse and momentum.
  • Solve problems involving steady fluid streams and propulsion with variable mass.

15.1 Principle of Linear Impulse and Momentum

Starting from the equation of motion written as a rate of change of velocity, we integrate over a time interval. The result is a single vector equation that relates the forces acting on a particle, the time they act, and the resulting change in velocity. This avoids the two-step route of first finding acceleration and then integrating it.

Linear Momentum

The linear momentum of a particle is the product of its mass and velocity, \(\mathbf{L}=m\mathbf{v}\). Because mass is a positive scalar, momentum points in the same direction as the velocity; its magnitude carries units of mass × velocity (kg·m/s).

Linear Impulse

The linear impulse is the time integral of force, \(\mathbf{I}=\int\mathbf{F}\,dt\). It measures the cumulative effect of a force over the interval it acts and is a vector in the direction of the force, with units of force × time (N·s). Graphically, its magnitude is the area beneath the force–time curve. For a constant force this area is simply a rectangle, \(\mathbf{I}=\mathbf{F}_c(t_2-t_1)\).

Equation of Motion (rate form)
\[ \sum\mathbf{F}=m\mathbf{a}=m\frac{d\mathbf{v}}{dt} \]
Principle of Linear Impulse and Momentum
\[ m\mathbf{v}_1+\sum\int_{t_1}^{t_2}\mathbf{F}\,dt=m\mathbf{v}_2 \]

In words: the initial momentum plus the sum of all impulses acting between \(t_1\) and \(t_2\) equals the final momentum. Resolving into rectangular components gives three independent scalar equations.

Scalar Component Equations
\[ m(v_x)_1+\sum\int_{t_1}^{t_2}F_x\,dt=m(v_x)_2 \] \[ m(v_y)_1+\sum\int_{t_1}^{t_2}F_y\,dt=m(v_y)_2 \] \[ m(v_z)_1+\sum\int_{t_1}^{t_2}F_z\,dt=m(v_z)_2 \]
F t t₁ t₂ I = ∫F dt Variable force F t t₁ t₂ Fc I = Fc(t₂−t₁) Constant force
The magnitude of the impulse equals the shaded area under the force–time curve. A constant force produces a simple rectangular area. Figure 15.1 — Impulse as area under the F–t diagram (original redrawn schematic)

Impulse and Momentum Diagrams

A useful bookkeeping device is to draw three diagrams: an initial momentum diagram (\(m\mathbf{v}_1\)), an impulse diagram showing every \(\int\mathbf{F}\,dt\) acting on the particle, and a final momentum diagram (\(m\mathbf{v}_2\)). Their vector sum enforces the impulse–momentum equation, much as a free-body diagram organizes a force balance.

Watch your sign convention here — pick a positive direction for each axis before writing the equation and keep it for every term. Most early mistakes in this chapter come from flipping a sign on an impulse halfway through. Also remember: every force on the FBD creates an impulse, even forces that do no work.

15.2 Principle of Linear Impulse and Momentum for a System of Particles

Applying the equation of motion to every particle and summing eliminates the internal forces: by Newton's third law they occur in equal, opposite, collinear pairs that cancel. What remains involves only the external forces acting on the system.

Internal Forces Cancel

When the impulse–momentum equations of all particles are added, the internal-force impulses sum to zero. Only the external impulses survive, so the system's change in total momentum depends solely on external loading.

Mass-Center Momentum

The total linear momentum of a system equals that of a single fictitious particle carrying the whole mass \(m=\sum m_i\) and moving with the velocity of the mass center, \(m\mathbf{v}_G=\sum m_i\mathbf{v}_i\). This is what justifies treating a rigid body as a point mass for momentum purposes.

System Impulse–Momentum (particles)
\[ \sum m_i(\mathbf{v}_i)_1+\sum\int_{t_1}^{t_2}\mathbf{F}_i\,dt=\sum m_i(\mathbf{v}_i)_2 \]
System Impulse–Momentum (mass center)
\[ m(\mathbf{v}_G)_1+\sum\int_{t_1}^{t_2}\mathbf{F}_i\,dt=m(\mathbf{v}_G)_2 \]

When you frame a problem as a "system," the unknown internal forces (cord tensions, contact forces between bodies) drop out — that's the whole advantage. If you actually need one of those internal forces, isolate a single body afterward and apply the principle to it alone.

15.3 Conservation of Linear Momentum for a System of Particles

When the net external impulse over the interval is zero, the system's total momentum cannot change. This is the workhorse principle for collisions and interactions, where large internal forces act over very short times.

Conservation of Linear Momentum

Condition: the sum of external impulses on the system over \(t_1\!\to\!t_2\) is zero (or negligible). Conclusion: the system's total linear momentum is unchanged, and the velocity of the mass center is constant.

\[ \sum m_i(\mathbf{v}_i)_1=\sum m_i(\mathbf{v}_i)_2 \qquad (\mathbf{v}_G)_1=(\mathbf{v}_G)_2 \]

Impulsive vs. Nonimpulsive Forces

Over a very short interaction time, forces split into two classes. Impulsive forces are large and act briefly, producing a major momentum change (the strike of a racket, an explosion). Nonimpulsive forces — weights, soft springs, anything small compared with the impulsive forces — contribute negligible impulse during that instant and may be dropped. The classification only holds for the short interval considered; over a long flight, the same weight becomes important.

A v₁ B v₂ coupling force is internal → momentum conserved
Two rail cars approach and couple. The coupling force is internal to the chosen system, so it cancels and momentum is conserved along the track. Figure 15.2 — Conservation of momentum in a coupling collision (original redrawn schematic)

15.4 Impact

Impact is the collision of two bodies over a very short time, generating large impulsive forces. During contact the bodies first deform, reach a common velocity at maximum deformation, then partly recover during restitution. The interplay of these phases is captured by a single experimental number.

Central vs. Oblique Impact

In central impact the mass centers move along the line of impact — the line through both centers, perpendicular to the plane of contact. In oblique impact one or both velocities make an angle with that line, so the problem must be resolved into line-of-impact and plane-of-contact components.

Deformation and Restitution

During contact the particles exert an equal-and-opposite deformation impulse \(\int P\,dt\) on each other, reaching a common velocity at maximum deformation. A restitution impulse \(\int R\,dt\) then pushes them apart. For real materials \(\int P\,dt>\int R\,dt\).

plane of contact line of impact A B vA vB
The line of impact passes through both mass centers and is perpendicular to the plane of contact. Impulsive forces act only along the line of impact. Figure 15.3 — Geometry of central impact (original redrawn schematic)
Coefficient of Restitution, e

\(e\) is the ratio of the relative separation velocity after impact to the relative approach velocity before impact, measured along the line of impact. It ranges from 0 to 1 and depends on the materials, geometry, and impact speed.

\[ e=\frac{(v_B)_2-(v_A)_2}{(v_A)_1-(v_B)_1} \]
Conservation of Momentum along the Line of Impact
\[ m_A(v_A)_1+m_B(v_B)_1=m_A(v_A)_2+m_B(v_B)_2 \]

Elastic and Plastic Limits

For a perfectly elastic collision \(e=1\) and no kinetic energy is lost. For a perfectly plastic (inelastic) collision \(e=0\); there is no restitution impulse, the bodies move off together, and the energy loss is a maximum. Energy lost in any impact is \(\Sigma U_{1\text{-}2}=T_2-T_1\), reappearing as heat, sound, and permanent deformation.

Oblique Impact — Four Unknowns

With the x-axis on the line of impact and the y-axis in the plane of contact, four scalar equations close the problem: momentum conserved along x for the system, the restitution relation along x, and momentum conserved along y separately for each smooth particle (no impulse acts perpendicular to the line of impact).

A classic trap: students apply work–energy to "shortcut" an impact. Don't — during contact you can't track how the internal deformation forces displace, so work–energy is invalid through the impact. Use momentum + restitution to get the after-velocities, then apply energy methods before and after if needed.

15.5 Angular Momentum

Angular momentum is the moment of linear momentum about a point — the rotational analogue of linear momentum. It is essential whenever a particle moves along a curved path or about a fixed axis.

Angular Momentum (Moment of Momentum)

The angular momentum \(\mathbf{H}_O\) of a particle about \(O\) is the moment of its linear momentum \(m\mathbf{v}\) about \(O\). For planar motion its magnitude is \((H_O)_z=d\,(mv)\), where \(d\) is the perpendicular distance from \(O\) to the line of \(m\mathbf{v}\). Units are kg·m²/s, and direction follows the right-hand rule.

Scalar Formulation (planar)
\[ (H_O)_z=d\,(mv) \]
Vector Formulation
\[ \mathbf{H}_O=\mathbf{r}\times m\mathbf{v}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ r_x&r_y&r_z\\ mv_x&mv_y&mv_z\end{vmatrix} \]
y x O mv d
The angular momentum about \(O\) is the moment of the linear-momentum vector \(m\mathbf{v}\); \(d\) is the perpendicular moment arm. Figure 15.4 — Scalar angular momentum about a point (original redrawn schematic)

15.6 Relation Between Moment of a Force and Angular Momentum

Taking the moment of the equation of motion about a fixed point produces a rotational counterpart of \(\mathbf{F}=\dot{\mathbf{L}}\): the resultant moment equals the time rate of change of angular momentum.

Moment–Angular-Momentum Relation

The resultant moment about a fixed point \(O\) of all forces on a particle equals the time rate of change of the particle's angular momentum about \(O\). This is Newton's second law expressed in rotational form.

\[ \sum\mathbf{M}_O=\dot{\mathbf{H}}_O \qquad \text{(compare } \sum\mathbf{F}=\dot{\mathbf{L}}\text{)} \]

Extension to a System of Particles

Summing over all particles, the internal-force moments cancel in collinear pairs, leaving \(\sum\mathbf{M}_O=\dot{\mathbf{H}}_O\) for the whole system — where \(\mathbf{H}_O\) is the total angular momentum and \(O\) is any fixed point in the inertial frame.

15.7 Principle of Angular Impulse and Momentum

Integrating the moment relation over time gives the angular analogue of the linear impulse–momentum principle. Choosing the right axis often eliminates unknown impulses whose lines of action pass through it.

Principle of Angular Impulse and Momentum
\[ (\mathbf{H}_O)_1+\sum\int_{t_1}^{t_2}\mathbf{M}_O\,dt=(\mathbf{H}_O)_2 \]

Angular Impulse

The angular impulse is the time integral of the moment, \(\int_{t_1}^{t_2}\mathbf{M}_O\,dt =\int_{t_1}^{t_2}(\mathbf{r}\times\mathbf{F})\,dt\). It captures the cumulative rotational effect of the forces about \(O\) over the interval.

Combined Scalar Equations (planar motion)
\[ m(v_x)_1+\sum\int_{t_1}^{t_2}F_x\,dt=m(v_x)_2 \] \[ m(v_y)_1+\sum\int_{t_1}^{t_2}F_y\,dt=m(v_y)_2 \] \[ (H_O)_1+\sum\int_{t_1}^{t_2}M_O\,dt=(H_O)_2 \]
Conservation of Angular Momentum

Condition: the total angular impulse about \(O\) over the interval is zero. Conclusion: angular momentum about \(O\) is conserved, \((\mathbf{H}_O)_1=(\mathbf{H}_O)_2\). This occurs notably under a central force, whose line of action always passes through \(O\) and so produces no moment.

\[ (\mathbf{H}_O)_1=(\mathbf{H}_O)_2 \]

Pick your moment axis deliberately. If you sum moments about an axis that the unknown forces (normal force, central cord tension, weight along the axis) pass through or run parallel to, those forces vanish from the equation and you isolate exactly the unknown you want. That choice is the entire skill in these problems.

15.8 Steady Flow of a Fluid Stream

Impulse–momentum ideas extend from closed particle systems to fluid that flows through a region of space called a control volume. For steady flow — equal inflow and outflow — the device must supply a force to change the fluid's momentum as it redirects the stream.

Control Volume & Steady Flow

A control volume is a fixed region whose boundaries usually coincide with the walls and openings of a pipe, pump, or turbine. Flow is steady when the mass entering equals the mass leaving, so the momentum stored inside stays constant and only the inlet/outlet momenta change.

Mass Flow and Discharge

The mass flow \(dm/dt\) is the constant mass crossing a section per unit time. For an incompressible fluid of density \(\rho\), cross-section \(A\), and speed \(v\): \(dm/dt=\rho vA=\rho Q\), where \(Q=vA\) is the volumetric flow (discharge).

Resultant Force on the Control Volume
\[ \sum\mathbf{F}=\frac{dm}{dt}(\mathbf{v}_B-\mathbf{v}_A) \]
Resultant Moment about O
\[ \sum\mathbf{M}_O=\frac{dm}{dt}\left(\mathbf{r}_B\times\mathbf{v}_B-\mathbf{r}_A\times\mathbf{v}_A\right) \]
Mass Flow / Discharge
\[ \frac{dm}{dt}=\rho v A=\rho Q \]
vA A vB B control volume
Fluid enters at A and leaves at B with different directions; the pipe walls exert the force that changes the stream's momentum. Figure 15.5 — Steady flow through a pipe elbow control volume (original redrawn schematic)

15.9 Propulsion with Variable Mass

Some devices change mass as they move: rockets expel exhaust and lose mass; scoops and scrapers collect material and gain mass. Applying impulse–momentum to a control volume that includes both the device and the transferred mass yields the governing equations of variable-mass motion.

Thrust as an Internal Force

The force driving a rocket (its thrust) is internal to the device-plus-exhaust control volume — it acts equally and oppositely on the device and the ejected mass. The external resultant \(\sum\mathbf{F}_{cv}\) includes only drag, weight, and similar outside forces.

Control Volume Losing Mass (e.g., rocket)
\[ \sum\mathbf{F}_{cv}=m\frac{dv}{dt}-v_{D/e}\frac{dm_e}{dt} \]
Control Volume Gaining Mass (e.g., scoop)
\[ \sum\mathbf{F}_{cv}=m\frac{dv}{dt}+v_{D/i}\frac{dm_i}{dt} \]

Relative-Velocity Terms

\(v_{D/e}\) is the device's velocity relative to the ejected mass, and \(v_{D/i}\) relative to the injected mass. The last term in each equation is the thrust (loss case) or the resistive force of the collected stream (gain case). Always pair these equations with a control volume and free-body diagram.

The sign in front of the relative-velocity term is everything here. Losing mass gives a forward thrust (minus sign in the boxed form yields a positive push); gaining mass gives a retarding drag. When in doubt, build the impulse–momentum diagrams over \(dt\) and let the algebra deliver the sign — don't memorize it blindly.

v W T ve
A rocket's control volume contains the vehicle and ejected gas. Thrust is internal; only weight \(W\) and drag act externally. Figure 15.6 — Rocket as a variable-mass control volume (original redrawn schematic)

Physical meaning

Impulse is accumulated force over time and equals a vector momentum change. Momentum preserves direction, while kinetic energy does not; a collision may conserve system momentum during negligible external impulse while still dissipating kinetic energy.

Problem-solving strategy

  1. Recognize time histories, short impacts, sudden coupling, or momentum flux; do not use impulse–momentum merely because a force appears.
  2. Choose the particle, interacting system, or fixed control volume and declare axes before assigning any velocity or impulse signs.
  3. Inventory external impulses. During impact, justify which are negligible; for direct impact add restitution, and for smooth oblique impact resolve normal and tangential components.
  4. Solve vector components symbolically, then test separation, contact, and any kinematic constraints.
  5. Verify momentum, units, direction, admissibility, and nonnegative kinetic-energy loss for passive impact.

15.A Integrated Applications

EXAMPLE 1Triangular Force Pulse Reverses a Cart
A 6.00 kg calibration cart initially moves left at 1.00 m/s. A horizontal force pulse acts right, rising linearly from zero to 90.0 N during 0.200 s and then falling linearly to zero during the next 0.300 s. Find the final velocity.

Recognition strategy

A force–time history and velocities before and after identify linear impulse–momentum. The pulse area supplies impulse directly; displacement during the pulse is unnecessary.

Given information

  • Choose right as \(+x\); \(m=6.00\ \text{kg}\) and \(v_1=-1.00\ \text{m/s}\).
  • The rightward force–time graph is one triangle with base 0.500 s and height 90.0 N.
  • Rolling resistance is neglected over the short pulse, and the rail supplies no horizontal impulse.

Model and interpretation

Treat the cart as a constant-mass particle constrained to horizontal translation. The force is impulsive but finite, and its direction never reverses. Momentum, unlike kinetic energy, retains the velocity sign.

Step-by-step solution
1. Compute initial momentum. Under the right-positive convention, \(p_1=mv_1=(6.00)(-1.00)=-6.00\ \text{kg}\!\cdot\!\text{m/s}\).
2. Evaluate pulse impulse. The area under the triangular force–time graph is \(J_x=\tfrac12(0.500)(90.0)=22.5\ \text{N}\!\cdot\!\text{s}\), directed positively.
3. Apply impulse–momentum. Use \(mv_1+J_x=mv_2\): \(-6.00+22.5=6.00v_2\), with signs carried algebraically rather than using speed magnitudes.
4. Resolve the direction. Solving gives \(v_2=16.5/6.00=2.75\ \text{m/s}\); its positive sign means the pulse reverses the cart and sends it right.

Verification

Units: Both N s and kg m/s reduce to kg m/s, so impulse can be added directly to initial momentum; division by kilograms returns metres per second.

Signs and direction: Initial momentum is negative because the cart moves left, while the pulse is positive. The impulse exceeds 6.00 N s, the amount needed merely to stop the cart, so reversal is consistent.

Plausibility: After spending 6.00 N s to cancel the initial momentum, 16.5 N s remains; divided by 6.00 kg this independently gives the 2.75 m/s rightward speed.

Final result: \(v_2=+2.75\ \text{m/s}\), or 2.75 m/s rightward.

Engineering interpretation

Peak force alone does not predict the velocity change; pulse area does. Real calibration work should verify the force sensor bandwidth because an under-resolved peak can bias the integrated impulse.

EXAMPLE 2Direct Impact of Two Guided Carriers
On a straight low-friction guide, carrier A has mass 2.50 kg and moves right at 6.00 m/s while carrier B has mass 4.00 kg and moves left at 1.00 m/s. Their coefficient of restitution is 0.550. Determine both velocities immediately after direct impact.

Recognition strategy

Two post-impact velocities are unknown. During the short collision external guide impulse along the line is negligible, so momentum conservation and one restitution equation provide the required pair.

Given information

  • Take right as positive: \(u_A=+6.00\ \text{m/s}\), \(u_B=-1.00\ \text{m/s}\).
  • Masses are \(m_A=2.50\ \text{kg}\) and \(m_B=4.00\ \text{kg}\); supplied normal restitution coefficient is \(e=0.550\).
  • Impact is direct and central on a straight guide; evaluate states immediately before and after contact.

Model and interpretation

Use both carriers as the system, making their contact impulses internal. The guide is smooth in the impact direction. Restitution relates separation speed to approach speed and is applied only along this common normal.

Step-by-step solution
1. Write momentum conservation. External impulse along the guide is negligible, so \(2.5(6.00)+4(-1.00)=2.5v_A+4v_B\), or \(2.5v_A+4v_B=11.0\).
2. Write restitution. For the declared positive axis, \(e=(v_B-v_A)/(u_A-u_B)\), giving \(v_B-v_A=0.550[6-(-1)]=3.85\ \text{m/s}\).
3. Solve simultaneous equations. Substitute \(v_B=v_A+3.85\) into momentum: \(6.5v_A+15.4=11.0\), hence \(v_A=-0.677\ \text{m/s}\).
4. Recover the second velocity. Then \(v_B=v_A+3.85=3.17\ \text{m/s}\). Their relative velocity is separating, so the post-impact state is contact-admissible.

Verification

Units: Every momentum term has kg m/s, and restitution is dimensionless because it is a velocity ratio. Solved velocity components therefore retain units of m/s.

Signs and direction: Carrier A’s negative result means it rebounds left; carrier B’s positive result means it departs right. The separation difference \(v_B-v_A=3.85>0\) satisfies nonpenetration.

Plausibility: Total momentum is \(2.5(-0.677)+4(3.17)=11.0\ \text{kg}\!\cdot\!\text{m/s}\). Kinetic energy falls from 47.0 J to about 20.7 J, a nonnegative 26.3 J loss expected for \(e<1\).

Final result: Carrier velocities are \(v_A=-0.677\ \text{m/s}\) and \(v_B=+3.17\ \text{m/s}\).

Engineering interpretation

Restitution is an empirical property of the specific materials, geometry, temperature, and approach speed. Equipment design should use tested values and separately assess peak contact force and local stress.

EXAMPLE 3Anchor Load for a Turning Water Jet
Water of density 998 kg/m³ flows steadily through a horizontal nozzle at 0.0350 m³/s. The inlet velocity is 14.0 m/s in the \(+x\) direction; the outlet velocity is 10.0 m/s at \(60.0^\circ\) above \(+x\). Both sections discharge at atmospheric gauge pressure. Neglect nozzle weight and find the anchor-force vector.

Recognition strategy

A steady stream changes vector momentum as it turns. A fixed control volume and the momentum-flow equation determine the net force on the fluid, after which Newton’s third law gives the nozzle and anchor loads.

Given information

  • Use fixed axes \(+x\) right and \(+y\) upward in plan view; \(\rho=998\ \text{kg/m}^3\) and \(Q=0.0350\ \text{m}^3/\text{s}\).
  • \(\mathbf V_1=14.0\mathbf i\ \text{m/s}\) and \(\mathbf V_2=10(\cos60^\circ\mathbf i+\sin60^\circ\mathbf j)\ \text{m/s}\).
  • Steady incompressible flow, one inlet and one outlet, atmospheric gauge pressure at both sections, and negligible nozzle weight are assumed.

Model and interpretation

Choose the fluid inside the stationary nozzle as the control volume and use absolute velocities. With stated zero gauge pressures, the wall resultant is the external force producing momentum-flux change. The fluid’s reaction on the nozzle is opposite, and the anchor balances it.

Step-by-step solution
1. Compute mass flow. The steady mass rate is \(\dot m=\rho Q=(998)(0.0350)=34.93\ \text{kg/s}\).
2. Resolve outlet velocity. \(\mathbf V_2=(5.000\mathbf i+8.660\mathbf j)\ \text{m/s}\), so \(\mathbf V_2-\mathbf V_1=(-9.000\mathbf i+8.660\mathbf j)\ \text{m/s}\).
3. Find wall force on fluid. The momentum equation gives \(\mathbf F_{\text{wall}}=\dot m(\mathbf V_2-\mathbf V_1)=(-314.4\mathbf i+302.5\mathbf j)\ \text{N}\).
4. Determine required anchor force. Fluid acts oppositely on the nozzle, \((314.4\mathbf i-302.5\mathbf j)\ \text{N}\), so equilibrium requires \(\mathbf F_{\text{anchor}}=(-314.4\mathbf i+302.5\mathbf j)\ \text{N}\).

Verification

Units: Mass flow in kg/s times velocity change in m/s gives kg m/s², or newtons. Both reported Cartesian components and the resultant therefore have force units.

Signs and direction: The stream loses x-momentum and gains y-momentum, so wall force on fluid is left and upward. The anchor direction matches that wall force because it must oppose the fluid reaction on the nozzle.

Plausibility: The component magnitude scale \(\dot m\Delta V\) is several hundred newtons. The resultant \(\sqrt{314.4^2+302.5^2}=436\ \text{N}\) is consistent with that scale and the resolved components.

Final result: \(\mathbf F_{anchor}=(-314\mathbf i+303\mathbf j)\ \text{N}\), with magnitude 436 N.

Engineering interpretation

This momentum load is only one anchor design input. Actual qualification must include pressure thrust if section pressures differ, nozzle and pipe weight, startup transients, vibration, joint loads, and code-required factors.

Chapter recap

  • Use impulse–momentum for time-based force effects and keep every quantity vectorial.
  • Conserve system momentum only along directions with negligible external impulse.
  • Pair momentum with restitution for impact and with absolute momentum flux for steady streams.
  • Check post-impact separation, energy loss, constraints, and practical load limitations.

Before moving on

You should be able to integrate a force–time curve, select a valid system, solve direct impact, apply angular impulse–momentum, and audit a steady-flow momentum balance.