Particle Energy, Momentum and Impact

Introduction

Many mechanical systems are most efficiently analyzed by relating changes in motion to energy, momentum, or angular momentum rather than by resolving every force over time and distance. A vehicle stopping under braking, a projectile striking a target, a hammer driving a nail, and a rotating mechanism accelerating from rest are all particle-dynamics problems in which these methods are especially valuable.

This chapter develops the work–energy, power, impulse–momentum, angular-momentum, and impact methods for particles. These methods are complementary. Work–energy is usually efficient when displacement and speed are important; impulse–momentum is effective when forces act over short time intervals; angular momentum is useful when moments about a point simplify the analysis; and conservation laws are powerful when external work or impulse is absent or known.

Learning Objectives

After studying this chapter, you should be able to:

• formulate and apply the work–energy equation for particles;
• calculate instantaneous and average power;
• apply linear and angular impulse–momentum equations;
• identify when conservation of energy, momentum, or angular momentum is valid;
• analyze direct and oblique impact using the coefficient of restitution;
• solve multi-stage dynamics problems by selecting an efficient sequence of methods;
• recognize energy losses, modeling assumptions, and practical limitations.

1. Work and Kinetic Energy

1.1 Work of a force

The differential work done by a force F on a particle during a displacement dr is

where F is the force magnitude, θ is the angle between the force and the displacement, and ds is the differential path length. Work is a scalar and has units of joules (J), where 1 J = 1 N·m.

For motion from position 1 to position 2,

For a constant force,

Work is positive when the force component acts in the direction of motion, negative when it opposes motion, and zero when it is perpendicular to the displacement.

The work of a conservative force depends only on the initial and final positions. Gravity and an ideal spring are conservative. Kinetic friction, aerodynamic drag, and many applied forces are nonconservative.

1.2 Kinetic-energy theorem

Newton’s second law for a particle of constant mass is ΣF = ma. Taking the dot product with velocity v and using a = dv/dt gives the work–energy theorem:

where

Here, T is translational kinetic energy, m is mass in kilograms, and v is speed in metres per second. Both sides have units of joules.

For several forces,

The equation applies to a particle or body modeled as a particle and is valid in an inertial reference frame. It is particularly useful when the force varies with position or when the time history is not required.

1.3 Potential energy

For conservative forces, work can be represented by a change in potential energy:

For a particle near Earth’s surface,

where y is elevation measured from an arbitrary datum and g ≈ 9.81 m/s². Thus, an increase in elevation reduces the work done by gravity.

For a linear spring,

where k is spring stiffness in N/m and x is deformation from the unstretched position. The spring force is F_s = −kx.

Including nonconservative work, the mechanical-energy equation is

where U_nc is the work of nonconservative forces. If U_nc = 0, mechanical energy is conserved.

Worked Example 1: Sliding block with friction

A 20 kg block starts from rest and slides 5 m down a 30° incline. The coefficient of kinetic friction is 0.20. Determine its speed at the end. Take g = 9.81 m/s².

The gravitational work is

U_g = mg(s sin 30°) = 20(9.81)(5)(0.5) = 490.5 J.

The normal force is N = mg cos 30°, so the friction force is F_k = μ_k N. Its work is

U_f = −μ_k mg cos 30° s = −0.20(20)(9.81)(0.866)(5) = −169.9 J.

Therefore,

490.5 − 169.9 = 1/2(20)v².

Hence,

v = √(320.6/10) = 5.66 m/s.

The normal force does no work because it is perpendicular to the displacement.

2. Power and Efficiency

Power is the rate at which work is done:

For a constant force parallel to the velocity, P = Fv. The SI unit is the watt (W), with 1 W = 1 J/s. The average power over an interval is

Mechanical efficiency compares useful output power with input power:

For a sequence of components, overall efficiency is the product of component efficiencies, provided the losses are represented consistently. Efficiency cannot exceed unity in a passive mechanical system.

Power methods are valuable for motors, pumps, conveyors, vehicle traction, and lifting equipment. A force may be large while power is small if the velocity is small; conversely, a moderate force at high speed can require substantial power.

3. Linear Impulse and Momentum

3.1 Linear momentum

The linear momentum of a particle is

Momentum is a vector with units kg·m/s. Newton’s second law can be written

Integrating between times t₁ and t₂ gives the impulse–momentum equation:

The integral on the left is the resultant impulse, measured in N·s. For a constant resultant force,

Impulse–momentum is especially effective for short-duration interactions, such as impacts, explosions, and sudden braking. During a short event, slowly varying forces such as weight may be negligible compared with a large contact force, but this approximation must be justified rather than assumed automatically.

3.2 Conservation of linear momentum

For a system of particles,

where P is total system momentum and ΣJ_ext is external impulse. If the external impulse is zero or negligible,

Internal forces cancel in pairs under Newton’s third law. Momentum is therefore conserved for an isolated system, not necessarily for each individual particle.

Worked Example 2: Average stopping force

A 1200 kg car traveling at 20 m/s is brought to rest in 0.80 s. Determine the average resultant stopping force.

Using impulse–momentum,

F_avg(0.80) = 1200(0 − 20).

Thus,

F_avg = −30,000 N.

The negative sign indicates a force opposite to the initial direction. The average force magnitude is 30.0 kN. This is a resultant force; tire forces, aerodynamic drag, and road slope would be included if relevant.

4. Angular Momentum and Angular Impulse

The angular momentum of a particle about point O is

where r is the position vector from O to the particle. Its units are kg·m²/s. The scalar magnitude is H_O = rmv sin θ, where θ is the angle between r and v.

Taking moments of Newton’s law about O gives

Integration produces angular impulse–momentum:

If the net external moment impulse about O is zero, angular momentum about O is conserved. This is useful for particles moving under central forces, collisions analyzed about a pin, and mechanisms in which reaction forces pass through the selected moment center.

For planar motion about a fixed axis, a particle at perpendicular distance r from the axis has

where ω is angular velocity in rad/s. The expression is the particle form of H = Iω, with moment of inertia I = mr².

The point about which angular momentum is taken must be selected carefully. A force passing through O has zero moment about O, but forces not passing through O may produce angular impulse.

5. Conservation Laws and Method Selection

Energy, momentum, and angular momentum express different physical balances.

A multi-stage problem often requires more than one method. For example, a body may slide down a rough track, strike another body, and then compress a spring. A suitable sequence is:

1. use work–energy for the sliding stage;
2. use momentum and restitution during the short impact;
3. use work–energy for the spring-compression stage.

Do not conserve mechanical energy through an inelastic impact. Momentum may be conserved during the impact even though kinetic energy decreases.

6. Impact of Particles

Impact is a short-duration interaction during which large contact forces cause abrupt changes in velocity. The line of impact is the common normal at the contact point at the instant of contact.

6.1 Direct central impact

In a direct impact, the centers of mass move along the line of impact. Let u_A and u_B be the velocities before impact and v_A and v_B the velocities after impact, all measured along a common positive axis.

If external impulse during impact is negligible, linear momentum is conserved:

The coefficient of restitution e is defined by

Using signed velocities,

for the usual ordering in which A approaches B from behind. The range for passive materials is normally 0 ≤ e ≤ 1. A value of 1 represents a perfectly elastic impact; e = 0 represents a perfectly plastic impact in which the particles have the same velocity immediately after impact.

For two particles, the post-impact velocities are

These formulas assume a one-dimensional impact, negligible external impulse, constant masses, and a known coefficient of restitution.

6.2 Energy loss in direct impact

Kinetic energy is conserved only when e = 1. The kinetic-energy loss during a two-particle impact is

where

is the reduced mass. This expression is nonnegative for 0 ≤ e ≤ 1. Lost kinetic energy becomes deformation, heat, sound, and vibration; it is not destroyed, but it is no longer available as translational kinetic energy.

Worked Example 3: Direct impact

A 2 kg cart moving at 6 m/s strikes a stationary 3 kg cart. The coefficient of restitution is 0.40. Determine both final velocities.

Momentum gives

2(6) + 3(0) = 2v_A + 3v_B.

Restitution gives

v_B − v_A = 0.40(6 − 0) = 2.4.

Therefore v_B = v_A + 2.4. Substitution into momentum gives

12 = 2v_A + 3(v_A + 2.4),

so v_A = 0.96 m/s and v_B = 3.36 m/s.

Initial kinetic energy is 36 J. Final kinetic energy is

1/2(2)(0.96)² + 1/2(3)(3.36)² = 17.78 J.

Thus approximately 18.22 J of translational kinetic energy is lost during impact.

6.3 Oblique impact

In oblique impact, velocities are resolved into components normal and tangential to the contact surface. For smooth contacting bodies, the impulsive contact force acts only along the normal direction. Consequently:

• normal momentum changes according to the impact equations;
• tangential velocity components remain unchanged if tangential impulse is negligible;
• restitution is applied only to the normal components.

For each particle, resolve velocity as

where n and t are unit vectors normal and tangent to the contact surface. The normal impact equations are

For smooth bodies, v_At = u_At and v_Bt = u_Bt. If friction is significant, tangential impulse changes the tangential components and a friction or restitution model is required. A purely normal restitution law is then incomplete.

7. Applied Case: Conveyor Transfer and Collision

Background

Consider an illustrative automated packaging line. A 10 kg package slides down a 20° guideway for 4 m. The coefficient of kinetic friction is 0.15. It then strikes a 6 kg package initially at rest on a low-friction conveyor. The coefficient of restitution is 0.50. Engineers must determine whether the second package will achieve the required transfer speed of 2.0 m/s.

Analysis of the approach stage

The first stage is naturally solved by work–energy. Starting from rest,

U_g = 10(9.81)(4)sin20° = 134.2 J.

Friction does work

U_f = −0.15(10)(9.81)cos20°(4) = −55.3 J.

Thus,

78.9 = 1/2(10)u_A²,

which gives u_A = 3.97 m/s.

Analysis of the collision

During the short collision, external horizontal impulse is assumed negligible. With package B initially stationary,

10(3.97) = 10v_A + 6v_B.

Restitution gives

v_B − v_A = 0.50(3.97) = 1.985.

Solving, v_A = 1.59 m/s and v_B = 3.58 m/s.

Outcome and engineering interpretation

The second package leaves the impact at 3.58 m/s, exceeding the 2.0 m/s requirement. The first package continues forward, so downstream spacing and sensor timing must also be checked. The result depends on the assumed coefficient of restitution, friction coefficient, and negligible external impulse. In a real design, these parameters should be measured over relevant temperature, surface, and speed ranges.

Learning points

The example demonstrates method selection: work–energy avoids calculating the guideway acceleration and travel time, while momentum plus restitution resolves the impact efficiently. It also shows why energy should not be conserved through an impact with e = 0.50.

8. Challenges, Limitations, and Professional Practice

Modeling limitations

Particle models neglect body rotation, deformation, and distributed mass. They are appropriate when dimensions are small compared with the path scale or when rotation is irrelevant. During impact, however, the contact geometry may generate spin. A rigid-body impact model with angular momentum and impulse may then be necessary.

The coefficient of restitution is not a universal material constant. It can depend on impact speed, temperature, surface condition, shape, compliance, and friction. Treating it as constant is an engineering approximation.

Reference frames

Momentum and energy equations must be used consistently in an inertial frame. Kinetic energy changes under a change of reference frame, although physical predictions remain consistent when all terms are transformed correctly. For impact calculations, choose a fixed laboratory frame unless a moving frame provides a clearly justified simplification.

Measurement and simulation

High-speed video, force plates, instrumented impact rigs, and accelerometers can estimate velocities, contact duration, and impulse. Numerical simulation using finite-element contact models can represent deformation more accurately, but simulation results require validated material and contact parameters. AI-based parameter estimation may assist with experimental data processing, but engineering judgment and physical validation remain essential.

Safety and ethics

Energy and impact calculations affect vehicle crashworthiness, machine guarding, sports equipment, lifting systems, and pedestrian safety. Engineers must apply appropriate safety factors, document assumptions, consider uncertainty, and comply with relevant standards. A result based on an optimistic restitution coefficient or underestimated stopping distance can create serious hazards.

Best-Practice Workflow

1. Define the particle or system and draw a free-body or interaction diagram.
2. Select a consistent coordinate system and inertial reference frame.
3. Identify the stage: displacement-driven, time-driven, rotational, or impact.
4. List known speeds, positions, forces, masses, and material parameters.
5. Choose the shortest valid method rather than combining equations unnecessarily.
6. State neglected work or impulse explicitly.
7. Check signs, units, limiting cases, and energy behavior.
8. Interpret the result physically and assess whether the assumptions are realistic.

Chapter Summary

Work–energy relates the work of forces to the change in kinetic energy and is often the best method when positions and speeds are known. Conservative forces can be represented through potential energy, while friction and other nonconservative forces appear as work terms. Power is the rate of energy transfer and is central to machine and motor selection.

Impulse–momentum relates the impulse of a resultant force to the change in linear momentum and is especially useful for short-duration events. Angular impulse–momentum performs the same role for moments and angular momentum. Momentum or angular momentum is conserved only when the corresponding external impulse is zero or negligible.

Impact analysis combines momentum conservation with the coefficient of restitution. Direct impact uses one-dimensional equations; oblique impact requires normal and tangential components. For smooth contact, tangential velocity is unchanged, while restitution applies only in the normal direction. Mechanical energy is generally lost during impact unless the collision is perfectly elastic.

Multi-stage problems are solved most effectively by selecting a method for each stage—for example, work–energy before impact, momentum and restitution during impact, and work–energy afterward. Reliable engineering analysis depends on clear system boundaries, correct reference frames, realistic assumptions, unit checks, and validation against measurements.

Key Terms

Work: Energy transferred by a force acting through displacement.

Kinetic energy: Energy associated with particle motion, T = 1/2mv².

Potential energy: Stored energy associated with position or deformation.

Conservative force: Force whose work depends only on initial and final positions.

Nonconservative work: Work that changes mechanical energy, such as frictional work.

Power: Rate of doing work or transferring energy.

Efficiency: Ratio of useful output power or energy to input power or energy.

Linear momentum: Product of mass and velocity, p = mv.

Impulse: Time integral of force.

Angular momentum: Moment of linear momentum about a specified point.

Angular impulse: Time integral of a moment.

Impact: Short-duration interaction producing large contact forces.

Line of impact: Common normal direction at contact during impact.

Coefficient of restitution: Ratio of relative separation speed to relative approach speed.

Direct impact: Impact in which motion is along the line of impact.

Oblique impact: Impact involving normal and tangential velocity components.

Reduced mass: μ_r = m_Am_B/(m_A + m_B) for two particles.

Perfectly elastic impact: Impact with e = 1 and no kinetic-energy loss.

Perfectly plastic impact: Impact with e = 0 and equal post-impact velocities.

External impulse: Net impulse exerted on a selected system by its surroundings.

Review Questions

Short Answer Questions

1. State the work–energy theorem and identify its units.
2. Why does a normal force usually do no work on a particle sliding along a rigid surface?
3. Define impulse and explain its relationship to momentum change.
4. Under what condition is linear momentum conserved for a system?
5. Why is restitution applied only to normal velocity components in smooth oblique impact?

Long Answer Questions

1. Derive the work–energy equation for a particle beginning with Newton’s second law.
2. Explain the differences among work–energy, impulse–momentum, and angular impulse–momentum methods, including suitable applications for each.
3. Derive the final-velocity equations for a direct impact between two particles.
4. Explain why momentum may be conserved during an inelastic impact even though kinetic energy is not conserved.
5. Discuss the assumptions and limitations of using a constant coefficient of restitution in engineering design.

Analytical / Application Questions

1. A 5 kg block starts from rest and moves 3 m down a 25° incline with μ_k = 0.18. Use work–energy to calculate its speed.
2. A 1.5 kg ball moving at 8 m/s strikes a stationary 2.5 kg ball in a direct impact with e = 0.60. Determine both velocities after impact and the kinetic-energy loss.
3. Design an analysis sequence for a package that slides down a rough ramp, collides with a second package, and then compresses a spring. Identify the governing equation and principal assumptions for each stage.

Multiple Choice Questions with Answers

Question 1

Which equation represents the work–energy theorem for a particle?

A. ΣF = ma

B. ΣU_{1→2} = T_2 − T_1

C. ΣFΔt = mv_1 + mv_2

D. ΣM = Iα

Correct Answer: B. ΣU_{1→2} = T_2 − T_1

Question 2

The SI unit of impulse is:

A. N

B. J

C. N·s

D. W

Correct Answer: C. N·s

Question 3

A force does zero work on a particle when the force is:

A. Parallel to displacement

B. Opposite to displacement

C. Perpendicular to displacement

D. Increasing with time

Correct Answer: C. Perpendicular to displacement

Question 4

For a system with negligible external impulse, which quantity is conserved?

A. Total linear momentum

B. Each particle’s kinetic energy

C. Each particle’s velocity

D. Mechanical energy during every impact

Correct Answer: A. Total linear momentum

Question 5

A perfectly plastic direct impact has coefficient of restitution:

A. e = 1

B. e = 0.5

C. e = 0

D. e > 1 necessarily

Correct Answer: C. e = 0

Question 6

A 2 kg particle moving at 4 m/s has kinetic energy equal to:

A. 4 J

B. 8 J

C. 16 J

D. 32 J

Correct Answer: C. 16 J

Question 7

In a smooth oblique impact, which velocity component is generally unchanged by the impact impulse?

A. Normal component

B. Tangential component

C. Resultant speed

D. Relative normal component

Correct Answer: B. Tangential component

Question 8

A 10 N constant force acts parallel to a particle moving at 3 m/s. Its instantaneous power is:

A. 0.3 W

B. 3.33 W

C. 13 W

D. 30 W

Correct Answer: D. 30 W

Question 9

The coefficient of restitution is defined using relative:

A. Acceleration before and after impact

B. Separation speed and approach speed

C. Momentum and impulse

D. Kinetic energy and potential energy

Correct Answer: B. Separation speed and approach speed

Question 10

Which method is generally most efficient for finding speed after a particle moves through a known displacement under position-dependent forces?

A. Work–energy

B. Impulse–momentum only

C. Angular momentum only

D. Coefficient of restitution only

Correct Answer: A. Work–energy