How to Calculate Relativistic Momentum from Mass and Speed

An electron crosses an 8.4 m evacuated tube in just 32.3 nanoseconds. Its speed is already close to the speed of light. If you calculate its momentum using \(p=mv\), you’ll get a result that’s less than half the correct value.

Why is the ordinary momentum equation so far out?

At very high speeds, momentum increases faster than speed. To calculate relativistic momentum from an object’s rest mass and speed, use:

\[ p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} \]

Or, using the Lorentz factor:

\[ p=\gamma m_0v \]

The method is to calculate \(\gamma\) from the speed, multiply it by the rest mass \(m_0\) and velocity \(v\), and state the direction of the momentum.

The interesting part is understanding what those measurements mean. An experiment with high-speed electrons gives us a useful way to see why the calculation works and why using the wrong mass or speed produces the wrong answer.

01The relativistic momentum calculation

In ordinary mechanics, momentum is mass multiplied by velocity:

\[ p=mv \]

That works extremely well for cars, cricket balls, and other objects moving much more slowly than light.

But consider two identical electrons, one travelling at \(0.10c\) and the other at \(0.90c\).

The second electron moves nine times faster. Would you expect it to have nine times as much momentum?

Not quite. Its momentum is about 20.5 times as large.

The reason is that the relationship between momentum and velocity is no longer linear at relativistic speeds.

The correct equation is:

\[ p=\gamma m_0v \]

where:

• \(p\) is relativistic momentum, measured in \(\mathrm{kg\,m\,s^{-1}}\).
• \(m_0\) is invariant mass, also called rest mass, measured in kilograms (\(\mathrm{kg}\)).
• \(v\) is velocity, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\gamma\) is the Lorentz factor, which has no units.

The Lorentz factor is calculated using:

\[ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]

Here, \(c\) is the speed of light in a vacuum:

\[ c=3.00\times10^8\ \mathrm{m\,s^{-1}} \]

The \(\gamma\) calculation uses the magnitude of velocity, so direction does not affect its value.

Which mass should you use?

Always use the object’s rest mass, \(m_0\).

This is the mass measured in a frame where the object is stationary. Its value does not change when you observe the object from another inertial reference frame.

The quantity \(\gamma m_0\) sometimes appears in older explanations as “relativistic mass”. That terminology can be misleading because it suggests the object’s invariant mass is increasing.

It’s clearer to keep the rest mass fixed and apply the Lorentz factor separately.

An electron moving at \(0.90c\) still has the same rest mass as an electron sitting stationary in your laboratory.

What changes between those frames is its momentum and total energy.

Rest frame versus laboratory frame

Imagine an electron travelling right through a laboratory.

In the electron’s instantaneous rest frame, it is stationary, so its momentum is zero. In the laboratory frame, the electron has a non-zero velocity and momentum.

Both observers use the same electron rest mass.

For one-dimensional motion, choose a positive direction before substituting:

• Moving right: \(v_x>0\), so \(p_x>0\).
• Moving left: \(v_x<0\), so \(p_x<0\).
• Stationary in the chosen frame: \(v_x=0\), so \(p_x=0\).

The speed determines the momentum’s magnitude. The velocity direction determines the direction of the momentum vector.

A reliable four-step method

If the speed is already written as a fraction of \(c\), keep that fraction when calculating \(\gamma\).

For example, if \(v=0.75c\):

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-0.75^2}}\\ &=\frac{1}{\sqrt{0.4375}}\\ &=1.512 \end{aligned} \]

You don’t need to substitute \(3.00\times10^8\) into the Lorentz factor calculation because the two factors of \(c\) cancel.

However, when calculating momentum in SI units, you must convert the velocity into \(\mathrm{m\,s^{-1}}\).

02How an electron experiment exposed the problem

The momentum equation is useful, but why should we trust a correction that makes momentum grow faster than speed?

One route is to measure the speeds of electrons as their kinetic energies increase.

Classical mechanics predicts:

\[ K=\frac12m_0v^2 \]

where \(K\) is kinetic energy, measured in joules (\(\mathrm J\)).

Rearranging:

\[ v=\sqrt{\frac{2K}{m_0}} \]

If this equation remained exact, increasing an electron’s kinetic energy would make it travel faster and faster without a limiting speed.

Eventually, the classical prediction would exceed \(c\).

That gives us an experimental question: does the electron actually keep gaining speed according to the classical equation, or does something different happen?

Bertozzi’s 1964 electron experiment

In 1964, physicist William Bertozzi reported an experiment using accelerated electrons to investigate the relationship between kinetic energy and speed.

The aim was to obtain direct measurements that could distinguish the classical prediction from special relativity.

The experiment used a Van de Graaff electrostatic generator and an electron linear accelerator.

Follow the path through the apparatus.

1. An electron gun produced short pulses of electrons inside the Van de Graaff apparatus.
2. The electrons passed through a short insulated metal tube near the beginning of the flight path. Some electrons were collected there, producing an electrical timing signal.
3. The remaining electrons travelled along an evacuated flight path 8.4 m long.
4. At the far end, the electrons struck an aluminium disc. The collected charge produced an arrival signal.
5. Electrical cables carried the starting and arrival signals to an oscilloscope, allowing the elapsed time to be measured.

An oscilloscope displays electrical signals against time. The horizontal separation between the start and arrival signals gave the electron bunch’s flight time after accounting for the signal transmission delays.

The basic timing principle resembles two timing gates on an athletics track. Measure the separation, measure the travel time, and calculate speed.

The difference is scale. The electrons travelled several metres in only a few tens of nanoseconds.

One nanosecond is \(10^{-9}\ \mathrm s\). Small differences in cable delay could therefore produce significant timing errors, so those delays needed careful calibration.

What should the experiment show?

Suppose the electrons receive more and more kinetic energy.

Under the classical model, the speed continues to increase according to \(v=\sqrt{2K/m_0}\).

Special relativity predicts something else: the speed approaches \(c\), while kinetic energy can continue to increase.

Here are three measurements reported in Bertozzi’s experiment, with speeds calculated from the reported flight times.

Historical measurements: 8.4 m flight path

A megaelectronvolt (\(\mathrm{MeV}\)) is a unit of energy equal to one million electronvolts.

These three runs used the linear accelerator as a field-free flight enclosure, so the distance divided by time gave the electron speed to a good approximation.

Notice what happens when kinetic energy triples from \(0.50\) to \(1.50\ \mathrm{MeV}\).

The electron speed increases from approximately \(0.867c\) to \(0.959c\). The flight time falls, but only from 32.3 ns to 29.2 ns.

The electrons are gaining substantial energy without gaining a proportional amount of speed.

What the first measurements couldn’t establish alone

Measuring speed does not automatically tell us how much energy the electrons gained.

The accelerator’s electrical potential difference provides an expected energy gain, but relying entirely on that calculation would introduce another assumption.

Bertozzi addressed this by measuring the energy transferred when electrons struck the aluminium disc.

The electrons deposited most of their kinetic energy as heat. A thermocouple attached to the disc detected its temperature rise, while the collected charge gave the number of arriving electrons.

A calibrated electrical heater provided a way to relate temperature rise to deposited energy.

The measured energies agreed with the accelerator-based estimates within the experiment’s uncertainty.

There was another limitation at higher energies. In the 4.5 MeV and 15 MeV runs, electrons received additional acceleration along part of the flight path.

In those conditions, dividing the entire 8.4 m distance by the flight time gives an average speed, not the exact final speed.

The reported high-energy timing results approached the time light would take to cross 8.4 m. They did not establish that electrons reached exactly \(c\).

The historical conclusion was that the relationship between electron speed and kinetic energy agreed with special relativity rather than the unrestricted classical prediction.

An important distinction: this experiment measured speed and energy. It did not independently measure relativistic momentum directly from the flight time.

But once the electron’s speed is known, we can use its rest mass and the relativistic momentum equation to calculate \(p\).

A momentum estimate from the historical data

Take the first measurement: an electron bunch took \(32.3\ \mathrm{ns}\) to travel 8.4 m.

Its calculated speed is:

\[ \begin{aligned} v&=\frac{8.4}{3.23\times10^{-8}}\\ &=2.60\times10^8\ \mathrm{m\,s^{-1}}\\ &=0.867c \end{aligned} \]

This gives a Lorentz factor of approximately \(2.01\).

Using the electron rest mass:

\[ \begin{aligned} p&=\gamma m_0v\\ &\approx(2.01)(9.11\times10^{-31})\\ &\quad\times(2.60\times10^8)\\ &\approx4.75\times10^{-22}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \]

The classical calculation gives only about \(2.37\times10^{-22}\ \mathrm{kg\,m\,s^{-1}}\).

The relativistic result is approximately twice as large.

This momentum is calculated from the historical speed measurement, rather than being a momentum value directly reported by the timing apparatus.

04Rest mass, momentum and energy are different quantities

One of the easiest mistakes in relativistic momentum questions is confusing rest mass with energy.

The distinction is especially important when a question gives an electron’s kinetic energy rather than its speed.

For an object of rest mass \(m_0\), its rest energy is:

\[ E_0=m_0c^2 \]

Here, \(E_0\) is the energy measured in the object’s rest frame.

In a laboratory where the object is moving, its total energy is:

\[ E=\gamma m_0c^2 \]

Its kinetic energy is the difference between total energy and rest energy:

\[ K=(\gamma-1)m_0c^2 \]

All three energies are measured in joules, although particle physicists commonly use electronvolts.

The rest mass stays invariant, but the object’s total energy depends on the observer’s inertial reference frame.

A useful comparison at \(0.80c\)

At \(v=0.80c\):

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac53 \]

For a particle of rest mass \(m_0\):

These expressions use \(m_0c\) as a momentum unit and \(m_0c^2\) as an energy unit.

Notice that \(1.667m_0\) is not a new invariant mass.

It is the factor \(\gamma m_0\) appearing in the momentum and total-energy equations.

When is \(p=m_0v\) acceptable?

When \(v\) is much smaller than \(c\), the Lorentz factor approaches 1.

That makes the classical equation a useful approximation.

For example:

These values are calculated from the relativistic equation, not experimental measurements.

There is no sharp speed at which classical momentum suddenly becomes invalid. The appropriate approximation depends on the precision required.

If an HSC question specifically asks for relativistic momentum, use the relativistic equation even when the correction is small.

For a fuller explanation of the increasing momentum near light speed, see Relativistic Momentum: How HSC Physics Explains the Speed Limit.

05How a magnetic field can reveal momentum

A timing experiment gives speed. A magnetic field provides another way to investigate momentum.

Imagine an electron entering a region containing a uniform magnetic field perpendicular to its motion.

The magnetic force acts perpendicular to the electron’s instantaneous velocity. Its path curves rather than remaining straight.

Under ideal conditions, the magnetic field changes the direction of momentum without changing the electron’s speed.

The radius of the circular path is related to momentum by:

\[ p=|q|Br \]

where:

• \(p\) is the momentum magnitude in \(\mathrm{kg\,m\,s^{-1}}\).
• \(|q|\) is the magnitude of the particle’s electric charge in coulombs (\(\mathrm C\)).
• \(B\) is magnetic flux density in tesla (\(\mathrm T\)).
• \(r\) is the path radius in metres (\(\mathrm m\)).

This relationship assumes a uniform magnetic field perpendicular to the motion, negligible energy loss, and an approximately circular path.

For the same charge and magnetic field, a particle with greater momentum follows a larger-radius path.

This gives an experimental cross-check: calculate momentum using the measured speed and rest mass, then calculate the expected curvature.

If a student incorrectly substitutes classical momentum, the predicted radius can be substantially too small.

The next questions will use this idea alongside energy conservation and collisions.

06Questions and solutions

For all questions, use \(c=3.00\times10^8\ \mathrm{m\,s^{-1}}\). Assume inertial laboratory frames, and neglect effects such as air resistance and radiation losses unless the question states otherwise.

Question 1

A particle has a rest mass of \(2.00\times10^{-27}\ \mathrm{kg}\) and travels east at \(0.80c\).

Calculate its relativistic momentum, including direction.

Solution 1

The particle’s relativistic momentum is \(8.00\times10^{-19}\ \mathrm{kg\,m\,s^{-1}}\) east.

Take east as positive.

First convert the speed:

\[ \begin{aligned} v&=0.80(3.00\times10^8)\\ &=2.40\times10^8\ \mathrm{m\,s^{-1}} \end{aligned} \]

Calculate the Lorentz factor:

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =\frac53 \]

Then:

\[ \begin{aligned} p&=\gamma m_0v\\ &=\frac53(2.00\times10^{-27})\\ &\quad\times(2.40\times10^8)\\ &=8.00\times10^{-19}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \]

Momentum is in the same direction as velocity, so the answer is east.

The tempting classical calculation would give \(4.80\times10^{-19}\ \mathrm{kg\,m\,s^{-1}}\), which is too small because it omits the Lorentz factor.

Question 2

Two particles, A and B, move directly towards one another in an inertial laboratory frame.

• Particle A has rest mass \(2m\) and travels right at \(0.60c\).
• Particle B has rest mass \(m\) and travels left at an unknown speed.
• The total momentum of the two-particle system is zero.

Here, \(m\) is a positive reference rest mass.

Determine:

a) the speed of particle B as a fraction of \(c\)

b) which particle has greater kinetic energy, and why.

Solution 2

Particle B must travel left at approximately \(0.832c\), and it has greater kinetic energy than A despite their momenta having equal magnitudes.

Part a: Find B’s speed.

Take right as positive.

Particle A has:

\[ \gamma_A=\frac{1}{\sqrt{1-0.60^2}}=1.25 \]

Its momentum is:

\[ \begin{aligned} p_A&=\gamma_A(2m)(0.60c)\\ &=1.50mc \end{aligned} \]

For the total momentum to be zero:

\[ p_B=-1.50mc \]

The momentum magnitudes must match, but particle B has only half A’s rest mass.

Let \(\beta_B=v_B/c\) denote B’s positive speed as a fraction of \(c\).

Using \(p=\gamma mv\):

\[ \frac{\beta_B}{\sqrt{1-\beta_B^2}} =1.50 \]

Square both sides:

\[ \frac{\beta_B^2}{1-\beta_B^2}=2.25 \]

Rearrange:

\[ \begin{aligned} \beta_B^2&=2.25(1-\beta_B^2)\\ 3.25\beta_B^2&=2.25\\ \beta_B^2&=\frac9{13} \end{aligned} \]

Therefore:

\[ \beta_B=\frac3{\sqrt{13}}\approx0.832 \]

Particle B travels at \(0.832c\) left.

Part b: Compare the kinetic energies.

Use:

\[ K=(\gamma-1)m_0c^2 \]

For particle A:

\[ \begin{aligned} K_A&=(1.25-1)(2m)c^2\\ &=0.500mc^2 \end{aligned} \]

For B, the Lorentz factor is:

\[ \begin{aligned} \gamma_B &=\frac{1}{\sqrt{1-\frac9{13}}}\\ &=\frac{\sqrt{13}}2\\ &\approx1.803 \end{aligned} \]

Therefore:

\[ \begin{aligned} K_B&=(1.803-1)mc^2\\ &=0.803mc^2 \end{aligned} \]

Particle B has approximately 1.61 times as much kinetic energy as particle A.

The tempting assumption is that cancelling momenta require equal kinetic energies.

They don’t. Opposite momenta cancel because momentum is a vector, while kinetic energy is a scalar quantity. Different rest masses and velocities can produce equal momentum magnitudes without equal kinetic energies.

Question 3

A beam of electrons enters a uniform magnetic field at \(0.80c\), travelling horizontally to the right.

The field has a magnetic flux density of \(0.250\ \mathrm T\) directed into the page.

The electron rest mass is \(9.11\times10^{-31}\ \mathrm{kg}\), and the electron charge is \(-1.60\times10^{-19}\ \mathrm C\).

An experiment measures a circular path radius of approximately \(9.1\ \mathrm{mm}\).

a) Calculate the radius predicted using relativistic momentum.

b) Calculate the radius predicted using classical momentum.

c) Determine which model better matches the observation, and state whether the electron curves upwards or downwards.

Use \(p=|q|Br\).

Solution 3

The relativistic model predicts a radius of \(9.11\ \mathrm{mm}\), the classical model predicts \(5.47\ \mathrm{mm}\), and the electron curves downwards. The observed radius agrees with the relativistic prediction.

Part a: Relativistic radius.

First:

\[ v=0.80c=2.40\times10^8\ \mathrm{m\,s^{-1}} \]

The Lorentz factor is:

\[ \gamma=\frac{1}{\sqrt{1-0.80^2}} =1.667 \]

Calculate momentum:

\[ \begin{aligned} p&=\gamma m_0v\\ &=(1.667)(9.11\times10^{-31})\\ &\quad\times(2.40\times10^8)\\ &=3.64\times10^{-22}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \]

Rearrange the magnetic-radius relationship:

\[ r=\frac{p}{|q|B} \]

Substitute:

\[ \begin{aligned} r&=\frac{3.64\times10^{-22}} {(1.60\times10^{-19})(0.250)}\\ &=9.11\times10^{-3}\ \mathrm m\\ &=9.11\ \mathrm{mm} \end{aligned} \]

Part b: Classical radius.

The classical momentum would be:

\[ \begin{aligned} p_N&=m_0v\\ &=(9.11\times10^{-31})\\ &\quad\times(2.40\times10^8)\\ &=2.19\times10^{-22}\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} r_N&=\frac{2.19\times10^{-22}} {(1.60\times10^{-19})(0.250)}\\ &=5.47\times10^{-3}\ \mathrm m\\ &=5.47\ \mathrm{mm} \end{aligned} \]

Part c: Interpret the observation and direction.

The measured radius of \(9.1\ \mathrm{mm}\) is consistent with the relativistic prediction.

The classical model predicts a much tighter curve because it underestimates the electron’s momentum.

For the direction, the magnetic force is:

\[ \mathbf F=q(\mathbf v\times\mathbf B) \]

A positive charge moving right in a field directed into the page would experience an upward force.

An electron has negative charge, so the force is reversed. It initially curves downwards.

This is why an experimental question must be answered with both the physical model and the correct direction. A negative charge reverses the force direction, not the sign of the radius.

Question 4

An electron initially travels at \(0.60c\).

It passes through an accelerator stage that increases its kinetic energy by exactly the amount of kinetic energy it already had. Its final kinetic energy is therefore twice its initial kinetic energy.

The electron’s rest mass remains unchanged.

Before and after the acceleration stage, the electron passes through separate identical field-free timing sections, each 8.4 m long. Its speed is constant within each timing section.

Determine:

a) the final electron speed as a fraction of \(c\)

b) the ratio of final momentum to initial momentum

c) the percentage decrease in flight time through the timing section.

Explain why doubling kinetic energy does not double speed or momentum.

Solution 4

The final speed is approximately \(0.745c\), momentum increases by a factor of 1.49, and the flight time decreases by approximately 19.5%.

Part a: Final speed.

Initially:

\[ \gamma_i=\frac{1}{\sqrt{1-0.60^2}}=1.25 \]

The kinetic energy is:

\[ \begin{aligned} K_i&=(\gamma_i-1)m_0c^2\\ &=0.25m_0c^2 \end{aligned} \]

Doubling it gives:

\[ K_f=0.50m_0c^2 \]

Using \(K=(\gamma-1)m_0c^2\):

\[ \gamma_f-1=0.50 \]

Therefore:

\[ \gamma_f=1.50 \]

Now rearrange the Lorentz factor:

\[ \frac vc=\sqrt{1-\frac1{\gamma^2}} \]

Substitute:

\[ \begin{aligned} \frac{v_f}{c} &=\sqrt{1-\frac1{1.50^2}}\\ &=0.745 \end{aligned} \]

The final speed is \(0.745c\), or approximately \(2.24\times10^8\ \mathrm{m\,s^{-1}}\).

Part b: Momentum ratio.

Initially:

\[ \begin{aligned} p_i&=\gamma_i m_0v_i\\ &=(1.25)m_0(0.60c)\\ &=0.750m_0c \end{aligned} \]

Finally:

\[ \begin{aligned} p_f&=\gamma_fm_0v_f\\ &=(1.50)m_0(0.745c)\\ &=1.118m_0c \end{aligned} \]

The ratio is:

\[ \begin{aligned} \frac{p_f}{p_i} &=\frac{1.118m_0c}{0.750m_0c}\\ &=1.49 \end{aligned} \]

Momentum increases by approximately 49%, not 100%.

Part c: Compare flight times.

For a fixed distance \(L\), flight time is:

\[ t=\frac Lv \]

The initial flight time is:

\[ \begin{aligned} t_i&=\frac{8.4}{0.60(3.00\times10^8)}\\ &=4.67\times10^{-8}\ \mathrm s \end{aligned} \]

The final flight time is:

\[ \begin{aligned} t_f&=\frac{8.4}{0.745(3.00\times10^8)}\\ &=3.76\times10^{-8}\ \mathrm s \end{aligned} \]

The percentage decrease is:

\[ \begin{aligned} \%\text{ decrease} &=\frac{t_i-t_f}{t_i}\times100\\ &\approx19.5\% \end{aligned} \]

The central mistake would be assuming \(K\), \(p\), and \(v\) are proportional to one another.

They aren’t. At relativistic speeds, kinetic energy depends on \(\gamma-1\), while momentum depends on \(\gamma v\).

A substantial energy gain can therefore produce a relatively small speed increase but a much larger momentum increase.

Question 5

Extension challenge: Two collisions with the same energy

Two isolated collision experiments are conducted in the same inertial laboratory frame.

All particles move along a single horizontal axis. Each incoming particle has invariant rest mass \(m\).

In both experiments, the particles collide and combine into one object. No radiation, particles, or other energy escape, and no external impulse acts on the system.

Experiment A

Two particles approach each other at equal speeds:

• Particle 1 travels at \(+0.60c\).
• Particle 2 travels at \(-0.60c\).

Experiment B

One moving particle collides with a stationary particle:

• Particle 1 travels at \(+\frac{\sqrt5}{3}c\), approximately \(+0.745c\).
• Particle 2 is stationary.

The combined object may contain internal energy produced by the collision.

For this question, use conservation of total relativistic energy and momentum.

For a single object:

\[ E=\gamma m_0c^2,\qquad p=\gamma m_0v \]

The total energy and momentum of a single combined object of invariant mass \(M\) obey:

\[ E^2=(pc)^2+(Mc^2)^2 \]

Here, \(E\) and \(p\) represent the final object’s total energy and signed momentum.

Determine:

a) the total initial energy of each experiment, measured in the laboratory frame

b) the final velocity of the combined object in each experiment

c) the invariant rest mass of each combined object.

A student argues that the two final objects must have the same rest mass because both experiments begin with the same total energy.

Explain exactly why that argument fails.

Solution 5

Both experiments begin with total energy \(2.50mc^2\), but they produce different final invariant masses.

Experiment A produces a stationary object with rest mass \(2.50m\). Experiment B produces an object moving at approximately \(0.447c\), with rest mass \(\sqrt5m\), or approximately \(2.236m\).

The difference comes from momentum conservation: Experiment B must retain some of its total energy as motion of the combined object.

Part a: Calculate initial total energy.

For Experiment A, each particle moves at \(0.60c\).

Therefore:

\[ \gamma_A=\frac{1}{\sqrt{1-0.60^2}} =1.25 \]

Each particle has total energy:

\[ E_1=1.25mc^2 \]

So the two-particle system has:

\[ E_A=2(1.25mc^2)=2.50mc^2 \]

For Experiment B, the moving particle has:

\[ \beta=\frac{\sqrt5}{3} \]

Its Lorentz factor is:

\[ \begin{aligned} \gamma_B &=\frac{1}{\sqrt{1-\frac59}}\\ &=\frac32 \end{aligned} \]

The moving particle’s energy is \(1.50mc^2\).

The stationary particle has total energy equal to its rest energy, \(mc^2\).

Therefore:

\[ \begin{aligned} E_B&=1.50mc^2+mc^2\\ &=2.50mc^2 \end{aligned} \]

The initial total energies are equal.

This confirms the student’s starting observation. The error lies in what the student concludes from it.

Part b: Calculate the final velocities.

For Experiment A, momentum is:

\[ \begin{aligned} p_A &=(1.25)m(+0.60c)\\ &\quad+(1.25)m(-0.60c)\\ &=0 \end{aligned} \]

The total initial momentum is zero.

Because momentum is conserved, the combined object must have zero momentum.

It is therefore stationary in the laboratory frame:

\[ V_A=0 \]

Experiment B is different.

The stationary particle contributes no momentum, but the moving particle contributes:

\[ \begin{aligned} p_B &=\frac32m\left(\frac{\sqrt5}{3}c\right)\\ &=\frac{\sqrt5}{2}mc \end{aligned} \]

This is positive, so the final combined object must move right.

For any single object:

\[ p=\gamma MV \]

and:

\[ E=\gamma Mc^2 \]

Dividing these relationships gives:

\[ V=\frac{pc^2}{E} \]

Using the conserved total momentum and energy:

\[ \begin{aligned} V_B &=\frac{\left(\frac{\sqrt5}{2}mc\right)c^2} {\frac52mc^2}\\ &=\frac{\sqrt5}{5}c\\ &\approx0.447c \end{aligned} \]

Experiment B’s combined object travels right at approximately \(0.447c\).

Part c: Find the invariant rest masses.

In Experiment A, the final object is stationary.

Its total energy is therefore entirely rest energy:

\[ E_A=M_Ac^2 \]

So:

\[ M_A=2.50m \]

For Experiment B, the final object moves. Its total laboratory energy is not entirely rest energy.

Use:

\[ E^2=(pc)^2+(Mc^2)^2 \]

Rearrange:

\[ M=\frac{\sqrt{E^2-(pc)^2}}{c^2} \]

For Experiment B:

\[ E_B=\frac52mc^2 \]

and:

\[ p_B=\frac{\sqrt5}{2}mc \]

Therefore:

\[ \begin{aligned} M_Bc^2 &=\sqrt{\frac{25}{4}m^2c^4 -\frac54m^2c^4}\\ &=\sqrt5\,mc^2 \end{aligned} \]

Hence:

\[ M_B=\sqrt5m\approx2.236m \]

Compare the results:

Why the equal-energy argument fails

The student assumes that all the energy in each experiment becomes rest energy.

That is true for Experiment A because its final momentum is zero in the laboratory frame.

It is false for Experiment B.

Experiment B begins with non-zero total momentum, and that momentum cannot disappear in an isolated collision.

The final combined object must keep moving, so some of its total energy is associated with that overall motion rather than its invariant rest mass.

Both combined objects have rest masses greater than \(2m\), the sum of their constituents’ original rest masses, because kinetic energy has been retained as internal energy in the collision.

This does not mean the individual particles’ invariant masses increased while they were moving. It means the newly formed composite object has an invariant mass that includes its stored internal energy.

The deeper result is that the invariant mass of a system depends on both its total energy and its total momentum.

Equal total energies in one reference frame do not guarantee equal invariant masses.

06What to carry into the next problem

When a question gives rest mass and speed, the calculation is direct:

\[ p=\gamma m_0v \]

The most important decisions come before and after substituting.

First, identify the reference frame and direction. Then use the invariant rest mass, calculate the Lorentz factor from the measured speed, and keep the velocity’s sign when calculating one-dimensional momentum.

If a result is smaller than \(m_0v\), something has gone wrong: for a massive object moving at non-zero speed, \(\gamma>1\).

If the calculated momentum is large, that doesn’t mean the particle must be travelling faster than light. Relativistic momentum can continue growing while speed remains below \(c\).

The next useful step is to consider what happens when momentum changes. Once you know the initial and final relativistic momenta, their vector difference gives the impulse:

\[ \Delta p=p_f-p_i \]

That connects directly to Impulse and Change in Momentum for HSC Physics, where momentum becomes a tool for analysing forces, interactions, and collisions.