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[parent] example of particle (Example)

Particle in Physics: Examples and Complete Worked Solutions

This companion entry develops the modeling ideas from Particle in Physics: Definition, Models, and Uses. The exercises deliberately mix calculations with model-selection questions because the central issue is not merely how to move a particle mathematically, but when a particle description is physically appropriate [1, 2, 4, 5].

All exercises are stated first. Complete worked solutions follow afterward.

PIC

Figure 1. Progression of the problem set from classical point-particle kinematics through systems of particles, relativity, quantum scales, and model selection.

Part I: Exercises

Exercise 1: state, trajectory, momentum, and force

A particle of constant mass

m  = 3 kg
(1)

moves in the plane according to

r(t) = 2tex + (t2 − 1)ey.
(2)

Find r, v, a, p, F, and the speed at t = 2 s.

PIC

Figure 2. Classical particle state for Exercise 1. Position locates the particle on its trajectory; velocity, momentum, and force describe its instantaneous dynamics.

Exercise 2: when is Earth a point particle?

Take Earth’s mean radius to be

              6
a =  6.37 × 10  m
(3)

and the characteristic orbital length scale to be one astronomical unit,

               11
L =  1.496 × 10   m.
(4)

Compute a∕L. If a rough orbital model only needs geometric finite-size effects below 10−3, is a point-particle approximation reasonable? Name two problems for which the same model would be inadequate.

Exercise 3: a baseball as a particle or rigid body

A baseball has radius approximately 3.66 cm and travels about 50 m during a fly ball. Estimate a∕L. Discuss whether a point-particle model is suitable for

  1. a rough estimate of the center-of-mass trajectory with no spin effects;
  2. a calculation in which spin, Magnus force, and rotational energy matter.

Exercise 4: a particle in phase space

A one-dimensional harmonic oscillator has

m  = 0.50 kg,
(5)

x (t) = A cos(ωt),
(6)

with

                         −1
A = 0.20 m,     ω =  4.0 s  .
(7)

At t = π∕8 s, find x, v, p, the spring constant k, and the total mechanical energy. State the corresponding one-dimensional phase-space point (x,p).

Exercise 5: particle mechanics from a Lagrangian

A particle of mass m moves vertically near Earth’s surface. Let y be positive upward and use

    1
L = --m ˙y2 − mgy.
    2
(8)

Apply the Euler–Lagrange equation to derive the equation of motion.

Exercise 6: central force and angular momentum

A particle of mass m moves under a central force

F = f (r)er.
(9)

Using

L  = r × mv,
(10)

show that the angular momentum about the force center is conserved. Explain why the trajectory must remain in a plane.

Exercise 7: center of mass of two particles

Two particles have

m1 =  2kg,     m2  = 3kg,
(11)

with positions

r1 = 1ex m,     r2 = 5 exm.
(12)

Their velocities are

v1 =  2ex + ey m ∕s,
(13)

v =  − e +  2e m ∕s.
 2      x     y
(14)

Find the center-of-mass position, total momentum, and center-of-mass velocity.

PIC

Figure 3. Two-particle geometry for Exercises 7–9. The center of mass provides a single-particle description of the translational motion of the system.

Exercise 8: external force and center-of-mass motion

For the system in Exercise 7, suppose the net external force is constant:

Fext = 6ex N.
(15)

Find the center-of-mass acceleration. Using the initial total momentum from Exercise 7, find the total momentum after 2 s.

Exercise 9: translational and internal kinetic energy

Using the masses and velocities from Exercise 7, calculate

  1. the total kinetic energy;
  2. the kinetic energy associated with center-of-mass translation;
  3. the kinetic energy remaining in motion relative to the center of mass.

Interpret the result physically.

Exercise 10: test-particle approximation

A 1000 kg spacecraft orbits Earth, whose mass is approximately

ME  = 5.972 × 1024 kg.
(16)

Compute the mass ratio m∕ME. Explain why the spacecraft is naturally treated as a test particle in a first orbital model, and state one circumstance in which that approximation would not capture all relevant physics.

Exercise 11: classify the particle model

For each description, identify the most appropriate term among point particle, test particle, tracer particle, and material point.

  1. A tiny neutrally buoyant bead used to visualize a fluid flow while its effect on the flow is neglected.
  2. An ideal positive charge used to define the direction of an Electric Field while its own field is ignored.
  3. A labeled infinitesimal parcel followed as an elastic solid deforms.
  4. A planet represented only by its mass and center-of-mass position in a two-body orbit calculation.

Exercise 12: when the point-particle model fails

A solid disk rolls without slipping down an incline. Compare two models:

  1. the disk is replaced by a point mass at its center;
  2. the disk is treated as a rigid body with moment of inertia.

Which model is adequate if only the translational motion of a frictionless sliding object is wanted? Which is required for rolling motion, and why?

Exercise 13: relativistic particle momentum and energy

A massive particle moves at

v = 0.80c.
(17)

Calculate the Lorentz factor γ. Compare the relativistic momentum

p   = γmv
  rel
(18)

with the Newtonian momentum pN = mv. Compare the relativistic kinetic energy

K   =  (γ − 1)mc2
  rel
(19)

with the Newtonian value KN = mv2∕2, expressing both in units of mc2.

Exercise 14: de Broglie wavelength and the classical-particle limit

Use

     h-
λ =  p.
(20)

Find the de Broglie wavelength of

  1. a 0.145 kg baseball moving at 40 m/s;
  2. a nonrelativistic electron with kinetic energy 100 eV.

Use

                    − 34
h =  6.62607015  × 10    Js,
(21)

m   = 9.10938 × 10 −31kg.
  e
(22)

Explain why the wave nature of one object is ordinarily negligible while that of the other is experimentally important.

Exercise 15: photon energy and momentum

A photon has wavelength

λ = 500 nm.
(23)

Find its energy and momentum using

     hc          h
E =  --,     p = --.
     λ           λ
(24)

Give the energy in joules and electron-volts. Explain why this is a useful example of a “particle” that is not a Newtonian point mass.

Exercise 16: choose the simplest adequate model

For each spacecraft problem, identify the simplest adequate physical model and explain what extra degree of freedom or spatial information is required as the problem becomes more detailed.

  1. A first two-body estimate of orbital period around a spherical Earth.
  2. Precision orbit propagation including Earth’s J2 gravity term.
  3. Spacecraft attitude dynamics and gravity-gradient torque.
  4. Flexible solar-array vibration during attitude maneuvers.

PIC

Figure 4. Particle models are selected by physical scale and by which degrees of freedom affect the quantity being calculated. More detail is added only when the simpler model omits relevant physics.

Part II: Complete Worked Solutions

Solution 1: state, trajectory, momentum, and force

Differentiate the position:

v(t) = 2ex + 2tey.
(25)

Differentiate again:

a (t) = 2ey.
(26)

At t = 2 s,

r =  4ex + 3ey m,
(27)

v =  2ex + 4ey m ∕s,
(28)

a = 2ey m ∕s2.
(29)

The momentum is

p =  mv  = 6e  + 12e  kg m∕s,
             x      y
(30)

and the net force is

F =  ma  = 6e  N.
             y
(31)

The speed is

      -------    ---
v = √ 22 + 42 = √20 ≈  4.47 m ∕s.
(32)

This is the basic classical point-particle state: position specifies where the particle is, while momentum or velocity completes the instantaneous dynamical state.

Solution 2: when is Earth a point particle?

The size ratio is

a     6.37 × 106
--=  ----------11 ≈ 4.26 × 10−5.
L    1.496 × 10
(33)

This is much smaller than 10−3, so geometric finite-size effects are small for a rough heliocentric orbit model.

The same approximation is inadequate when finite size or internal structure matters. Examples include tidal deformation, rotational attitude, precession, and modeling Earth’s nonspherical gravity field such as the J2 term.

Solution 3: a baseball as a particle or rigid body

The scale ratio is

a    0.0366
--=  -------= 7.32 × 10−4.
L      50
(34)

For a rough center-of-mass trajectory with no spin-dependent effects, this is a natural point-particle approximation.

If spin, Magnus force, torque, or rotational kinetic energy matters, the particle model no longer contains enough degrees of freedom. At minimum, orientation and angular velocity must be added, leading toward a rigid-body model. One can still use a point-like translational state for the center of mass, but the complete model is no longer only a point particle.

Solution 4: a particle in phase space

At

t = π-s,
    8
(35)

we have

ωt = π-.
     2
(36)

Therefore

x = A cos(π∕2 ) = 0.
(37)

The velocity is

v = − Aω sin(ωt) = − (0.20)(4)(1) = − 0.80m ∕s.
(38)

Hence

p = mv  = (0.50)(− 0.80) = − 0.40kg m ∕s.
(39)

For a harmonic oscillator,

  2   k-
ω  =  m ,
(40)

so

        2
k = m ω  =  (0.50 )(16 ) = 8.0 N∕m.
(41)

At this instant the potential energy is zero, so

     1   2   1            2
E =  -mv   = --(0.50 )(0.80)  = 0.160 J.
     2       2
(42)

The phase-space point is

(x,p) = (0,− 0.40kg m ∕s).
(43)

Solution 5: particle mechanics from a Lagrangian

The Euler–Lagrange equation is

d ( ∂L )    ∂L
--  ---  −  --- = 0.
dt  ∂ ˙y     ∂y
(44)

For

    1-   2
L = 2 m ˙y −  mgy,
(45)

we obtain

∂L
--- = m y˙
∂ ˙y
(46)

and

d-(m ˙y) = m ¨y.
dt
(47)

Also,

∂L-
∂y  = − mg.
(48)

Thus

m ¨y + mg =  0,
(49)

or

¨y = − g.
(50)

The same particle dynamics that Newtonian mechanics writes as Fy = may emerges from the stationary-action formulation.

Solution 6: central force and angular momentum

Differentiate angular momentum:

dL    d
---=  --(r × mv ).
dt    dt
(51)

Using the cross-product rule,

dL
---=  v × mv  + r × ma.
dt
(52)

The first term vanishes because a vector crossed with itself is zero. Newton’s second law gives

dL-
dt =  r × F.
(53)

For a central force, F is parallel to r, so

r × F = 0.
(54)

Therefore

dL-
dt =  0,
(55)

so L is constant.

Because r is always perpendicular to the fixed vector L, the trajectory remains in the plane perpendicular to L.

Solution 7: center of mass of two particles

The total mass is

M  = m1  + m2 =  5kg.
(56)

The center of mass is

      m1r1 + m2r2
R  =  -----M------.
(57)

Therefore

     2(1)-+-3(5)
R  =      5     ex = 3.4ex m.
(58)

The total momentum is

P  = m1v1  + m2v2.
(59)

Thus

P = (4ex + 2ey ) + (− 3ex + 6ey),
(60)

so

P  = ex + 8ey kgm ∕s.
(61)

The center-of-mass velocity is

V    =  P--= 0.20e  + 1.60e  m ∕s.
  CM    M          x        y
(62)

Solution 8: external force and center-of-mass motion

The center-of-mass equation is

M  A    = F   .
     CM     ext
(63)

Hence

        6-               2
ACM   = 5 ex = 1.20exm ∕s .
(64)

The total momentum obeys

dP-
 dt =  Fext.
(65)

For constant force over 2 s,

ΔP   = F   Δt =  12e  kgm ∕s.
         ext         x
(66)

Using the initial momentum from Exercise 7,

P (2s) = 13ex + 8ey kg m∕s.
(67)

internal forces do not appear in this center-of-mass balance when they cancel pairwise.

Solution 9: translational and internal kinetic energy

The total kinetic energy is

     1-   2   1-    2
T =  2m1v 1 + 2 m2v 2.
(68)

Both particles have speed squared equal to 5, so

T =  1(2)(5) + 1(3)(5) = 12.5J.
     2         2
(69)

The center-of-mass speed squared is

V 2  = (0.20)2 + (1.60 )2 = 2.60.
  CM
(70)

Therefore

T    = 1-M V 2  = 1-(5 )(2.60) = 6.50 J.
 CM    2    CM    2
(71)

The remainder is

T       = T − T    = 12.5 − 6.5 = 6.0J.
 internal         CM
(72)

Thus the system’s kinetic energy separates into motion of the center of mass plus motion of the particles relative to that center of mass.

Solution 10: test-particle approximation

The mass ratio is

 m         1000                − 22
---- = ----------24 ≈ 1.67 × 10   .
ME     5.972 × 10
(73)

The spacecraft’s gravitational effect on Earth is negligible for a first orbital calculation, while Earth’s field strongly determines the spacecraft motion. This is exactly the logic of a test particle: it responds to the field without appreciably changing the field source.

The approximation does not describe everything. Spacecraft attitude, finite area, atmospheric drag, solar-radiation pressure, flexible structures, or mutual gravitational interactions with comparable masses require additional physics.

Solution 11: classify the particle model

The classifications are:

  1. Tracer particle: the bead is used to follow the fluid motion while its feedback is neglected.
  2. Test particle: the ideal charge responds to an electric field without altering the prescribed field appreciably.
  3. Material point: the labeled element follows a piece of a continuum through deformation.
  4. Point particle: the planet’s finite size and orientation are omitted and only its mass and center-of-mass position are retained.

The terms overlap conceptually, but each emphasizes a different modeling role.

Solution 12: when the point-particle model fails

For frictionless sliding in which only translation matters, a point mass at the center of mass can be sufficient. Its translational kinetic energy is

         1
Ttrans = -M v2.
         2
(74)

For rolling, the body also has rotational kinetic energy,

Trot = 1-Iω2,
       2
(75)

and the no-slip constraint couples translation and rotation:

v = R ω.
(76)

A point particle has no orientation and no moment of inertia, so it cannot represent this rotational degree of freedom. A rigid-body model is required.

Solution 13: relativistic particle momentum and energy

The Lorentz factor is

     -----1------
γ =  ∘ -----2--2.
       1 − v ∕c
(77)

For v = 0.80c,

γ =  √---1-----=  -1-- = 1.667.
       1 − 0.64    0.60
(78)

Therefore

prel=  γ = 1.667.
pN
(79)

The relativistic kinetic energy is

Krel = (γ − 1)mc2 = 0.667mc2.
(80)

The Newtonian value is

K   = 1-m (0.80c)2 = 0.320mc2.
 N    2
(81)

Thus the Newtonian kinetic energy is less than half the relativistic value in this case. The particle concept survives, but Newtonian particle dynamics no longer does.

Solution 14: de Broglie wavelength and the classical-particle limit

For the baseball,

p = mv  = (0.145)(40) = 5.80kg m ∕s.
(82)

Therefore

    6.62607015--×-10−34            − 34
λ =         5.80        ≈ 1.14 × 10    m.
(83)

This wavelength is fantastically smaller than any ordinary baseball-scale geometry, so wave interference is unobservable in ordinary motion.

For a nonrelativistic electron,

     -p2-
K =  2m  ,
        e
(84)

so

    ∘  ------
p =    2meK.
(85)

With

                          − 17
K =  100eV  = 1.60218 × 10    J,
(86)

we obtain

     √--h----            −10
λ =    2meK   ≈ 1.23 × 10   m.
(87)

Thus

λ ≈ 0.123 nm.
(88)

That scale is comparable with atomic spacings, so diffraction and interference become physically important.

Solution 15: photon energy and momentum

For

λ = 500 × 10− 9m,
(89)

we have

     hc-            −19
E  =  λ ≈  3.97 × 10   J.
(90)

Using

1 eV = 1.602176634  × 10−19J,
(91)

this is

E  ≈ 2.48 eV.
(92)

The momentum is

p =  h-≈ 1.33 × 10− 27 kgm ∕s.
     λ
(93)

A photon carries definite energy and momentum and can produce localized detection events, so particle language is useful. It is nevertheless not a Newtonian point mass: it is a quantum excitation of the electromagnetic field and has zero rest mass.

Solution 16: choose the simplest adequate model

For a first two-body orbital-period calculation, Earth and spacecraft can both be treated as point masses. The essential degrees of freedom are their center-of-mass positions and velocities.

For precision orbit propagation including J2, the spacecraft may still be a point particle, but Earth’s gravitational field must include information about Earth’s extended, nonspherical mass distribution. Thus the field model becomes more detailed even though the spacecraft model need not.

For attitude dynamics and gravity-gradient torque, the spacecraft must have orientation, angular velocity, and inertia. A rigid-body model is therefore required.

For flexible solar-array vibration, rigid-body motion is not enough. Structural deformation introduces distributed or modal degrees of freedom, so a flexible multibody or continuum model is required.

This sequence illustrates the central rule of the particle article: use the simplest model that retains the degrees of freedom that affect the phenomenon being calculated.

Summary

The exercises show that “particle” is not one fixed physical category. It is a modeling level. A classical point particle may be characterized by position and momentum; a many-particle system can often be reduced to center-of-mass motion; a test particle probes an externally prescribed field; a relativistic particle requires new momentum and energy relations; and quantum particles demand a state description whose characteristic wavelength may make the classical trajectory picture inadequate.

The practical question is always the same:

What is the simplest particle or extended-body model that retains the degrees of freedom needed for the required accuracy?

References

[1]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[2]   H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

[3]   L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Butterworth-Heinemann, 1976.

[4]   D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

[5]   M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Westview Press, 1995.


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Physics Classification: 01.55.+b (General physics)
 45.20.-d (Formalisms in classical mechanics)
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