Relativistic momentum p = γmv: Exercises — Physics deep dive

Definition Symbols we use on this page (built here, not assumed)

$p$ — the momentum vector. $p = γ m v$ points along the direction of motion. In these one-dimensional problems all motion is along a single chosen axis (call it $x$ ), so we can describe $p$ by one signed number $p$ , its component along $+ x$ . This is our convention for the whole page: whenever you see the symbol $p$ , it is the signed component $p = γ m v$ , where $v$ is itself signed ( $v > 0$ means " $+ x$ direction," $v < 0$ means "opposite"). Thus a rightward particle has $p > 0$ and a leftward one has $p < 0$ . The signs only ever matter when two particles move toward each other and we add their momenta (L5.1). Its magnitude — the size ignoring direction — is $∣ p ∣ = γ m ∣ v ∣ \geq 0$ ; we write $∣ p ∣$ (with bars) whenever we mean magnitude, and plain $p$ (signed) everywhere else. In L2–L4 every particle moves in the $+ x$ direction, so there $p = ∣ p ∣$ and the distinction is silent.
$E$ — total energy. Total energy is $E = γ m c^{2}$ : rest energy $m c^{2}$ plus kinetic energy. It ties to momentum through $E^{2} = (p c)^{2} + (m c^{2})^{2}$ , the energy–momentum relation. Note this relation contains $p^{2}$ , so it only ever sees the magnitude — signs drop out when squared. We introduce $E$ now so Level 4 has it ready.
$K$ — kinetic energy. The energy of motion only: $K = E - m c^{2} = (γ - 1) m c^{2}$ . It vanishes when the particle is at rest ( $γ = 1$ ).

Intuition Figure — read this before the exercises

What the figure shows. The horizontal axis is

β = v / c

(speed as a fraction of light-speed, running

0 \to 1

). The vertical axis is momentum in natural units,

p / (m c) = γ β

. The blue curve is the true relativistic momentum; the dashed yellow line is what Newton predicts,

p / (m c) = β

(a straight

4 5^{\circ}

line). The dotted red vertical line at

β = 1

is the light-speed wall. The green dots mark two exercises solved below (L2.1 at

β = 0.6

and L2.2 at

β \approx 0.95

). What to notice. At small

β

the blue curve hugs the yellow line — relativity and Newton agree, matching L1.3 and L3.3. As

β

climbs past

\sim 0.5

the blue curve peels upward and rockets toward infinity at the red wall: momentum grows faster than linearly (the point of L3.1) and diverges as

β \to 1

(L3.2). Nearly every exercise on this page is a single point read off this one curve. For motion reversed in direction (

β < 0

) the whole picture reflects through the origin — identical shape, negative

p

Level 1 — Recognition

Goal: plug in and read the formula correctly.

L1.1 — State the relativistic momentum formula and the definition of $γ$ . Then say what $p$ becomes as $v \to c$ and as $v \to 0$ .

Recall Solution L1.1

$p = γ m v = \frac{m v}{1 - v ^{2} / c ^{2}}, γ = \frac{1}{1 - v ^{2} / c ^{2}} \geq 1.$ The direction of $p$ is the direction of motion; its magnitude is $∣ p ∣ = γ m ∣ v ∣$ .

As $∣ v ∣ \to c$ : $β \to \pm 1$ , so $1 - β^{2} \to 0$ , so $γ \to \infty$ , so $∣ p ∣ \to \infty$ . Nothing with mass can reach $c$ — see Why nothing exceeds the speed of light.
As $v \to 0$ : $γ \to 1$ , so $p \to m v$ , the Newtonian value. ✅

L1.2 — Compute $γ$ for $β = 0.6$ .

Recall Solution L1.2

$γ = \frac{1}{1 - 0. 6 ^{2}} = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25.$ What we did: squared $β$ , subtracted from 1, took the reciprocal square root. Why reciprocal: $γ$ must be $\geq 1$ so momentum grows; if you got $0.8$ you took $1 - β^{2}$ by mistake. Note $γ$ depends on $β^{2}$ , so it is the same for $β = + 0.6$ and $β = - 0.6$ — direction of travel doesn't change $γ$ .

L1.3 — A ball of mass $2 kg$ rolls at $3 m/s$ . Is the relativistic correction to its momentum measurable? Estimate $γ$ .

Recall Solution L1.3

$β = 3/ (3 \times 1 0^{8}) = 1 0^{- 8}$ . Then $γ = \frac{1}{1 - 1 0 ^{- 16}} \approx 1 + \frac{1}{2} (1 0^{- 16}) = 1 + 5 \times 1 0^{- 17} .$ The correction is one part in $1 0^{16}$ — utterly unmeasurable. Everyday speeds are Newtonian: the $γ$ boost only matters when $β ≳ 0.1$ .

Level 2 — Application

Goal: one clean numeric plug-in per problem. (Every particle here moves in $+ x$ , so $p = ∣ p ∣ > 0$ .)

L2.1 — Find the momentum of an electron at $v = 0.6 c$ .

Recall Solution L2.1

$γ = 1.25$ (from L1.2). Lab velocity $v = 0.6 \times 3 \times 1 0^{8} = 1.8 \times 1 0^{8} m/s$ (taken as the $+ x$ direction). $p = γ m v = 1.25 \times (9.11 \times 1 0^{- 31}) \times (1.8 \times 1 0^{8}) .$ $p = 1.25 \times 1.6398 \times 1 0^{- 22} = 2.0498 \times 1 0^{- 22} kg\cdotm/s \approx 2.05 \times 1 0^{- 22} kg\cdotm/s .$ The vector $p$ points in $+ x$ , so $p > 0$ ; if the electron moved the other way, $v = - 0.6 c$ and $p = - 2.05 \times 1 0^{- 22}$ (same magnitude $∣ p ∣$ , opposite sign). Check magnitude: Newton would give $m v = 1.64 \times 1 0^{- 22}$ ; the $\times 1.25$ boost lifts it by $25%$ .

L2.2 — A proton has momentum magnitude $∣ p ∣ = 3 m c$ . Find $γ$ and $β$ .

Recall Solution L2.2

Write $∣ p ∣ = γ m ∣ v ∣ = γ β m c$ (using $β > 0$ for the magnitude), so $γ β = ∣ p ∣/ (m c) = 3$ . Why $γ β$ : it packages both unknowns into the one quantity the data gives us. Square: $γ^{2} β^{2} = 9$ . Use $γ^{2} = 1/ (1 - β^{2})$ : $\frac{β ^{2}}{1 - β ^{2}} = 9 \Rightarrow β^{2} = 9 - 9 β^{2} \Rightarrow 10 β^{2} = 9 \Rightarrow β = 0.9 = 0.9487.$ Then $γ = \frac{1}{1 - 0.9} = \frac{1}{0.1} = 3.162$ . Sanity: $γ β = 3.162 \times 0.9487 = 3.00$ . ✅ (We took the positive root; a proton moving the other way would give signed $p = - 3 m c$ , same $γ$ and same speed $∣ v ∣$ .)

L2.3 — By what factor does an electron's momentum exceed the Newtonian estimate at $β = 0.9$ ?

Recall Solution L2.3

The relativistic-to-Newtonian ratio is exactly $γ$ , because $p_{rel} / p_{Newton} = γ m v / (m v) = γ$ . $γ = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294.$ So the true momentum is $2.29 \times$ the naive value — Newton undershoots by more than half.

Level 3 — Analysis

Goal: reason about how $p$ changes, not just its value.

L3.1 — If $v$ increases from $0.1 c$ to $0.2 c$ , by what factor does $p$ change? Is it exactly $2$ ?

Recall Solution L3.1

$p = γ m v$ , so the factor is $\frac{γ ( 0.2 c ) \cdot 0.2 c}{γ ( 0.1 c ) \cdot 0.1 c}$ .

$γ (0.1 c) = 1/ 1 - 0.01 = 1/ 0.99 = 1.00504.$
$γ (0.2 c) = 1/ 1 - 0.04 = 1/ 0.96 = 1.02062.$ $ratio = \frac{1.02062 \times 0.2}{1.00504 \times 0.1} = \frac{0.204124}{0.100504} = 2.031.$ Interpretation: $p$ grows slightly more than double, because the extra $γ$ itself climbs with $v$ . Doubling speed does not double momentum in relativity.

L3.2 — Show that as $β \to 1$ , $p$ grows without bound, and find $β$ for which $p = 10 m c$ .

Recall Solution L3.2

As $β \to 1$ , $γ β \to \frac{β}{1 - β ^{2}}$ : the numerator tends to $1$ but the denominator tends to $0$ , so the fraction $\to \infty$ . Hence $p = γ β m c \to \infty$ . This is the wall of light-speed. For $p = 10 m c$ : $γ β = 10$ , so $\frac{β ^{2}}{1 - β ^{2}} = 100$ , giving $β^{2} = 100/101$ , so $β = 100/101 = 0.99504.$ Even at $99.5%$ of light-speed, $p$ is "only" $10 m c$ — the last sliver of speed costs enormous momentum.

L3.3 — For small $β$ , expand $γ$ to show the first correction to Newtonian momentum. What fractional error does $p = m v$ make at $β = 0.1$ ?

Recall Solution L3.3

Use the binomial expansion $(1 - x)^{- 1/2} \approx 1 + \frac{1}{2} x$ for small $x = β^{2}$ : $γ \approx 1 + \frac{1}{2} β^{2} \Rightarrow p \approx m v (1 + \frac{1}{2} β^{2}) .$ Why binomial: it isolates the leading relativistic term without a calculator. The fractional excess of $p$ over $m v$ is $γ - 1 \approx \frac{1}{2} β^{2}$ . At $β = 0.1$ : fractional error $\approx \frac{1}{2} (0.01) = 0.005 = 0.5%$ . (Exact: $γ - 1 = 1.00504 - 1 = 0.504%$ .) So Newton is off by half a percent already at one-tenth light-speed.

Level 4 — Synthesis

Goal: combine momentum with energy and the four-vector. Recall from the top box: $E = γ m c^{2}$ (total energy), $K = (γ - 1) m c^{2}$ (kinetic energy), linked by $E^{2} = (p c)^{2} + (m c^{2})^{2}$ . All particles here move in $+ x$ , so $p = ∣ p ∣ > 0$ .

L4.1 — An electron has total energy $E = 2 m c^{2}$ . Find its momentum $p$ (in units of $m c$ ) using the energy–momentum relation.

Recall Solution L4.1

Use $E^{2} = (p c)^{2} + (m c^{2})^{2}$ from Energy-momentum four-vector. Rearranged: $(p c)^{2} = E^{2} - (m c^{2})^{2} = (2 m c^{2})^{2} - (m c^{2})^{2} = 4 m^{2} c^{4} - m^{2} c^{4} = 3 m^{2} c^{4} .$ $p c = 3 m c^{2} \Rightarrow p = 3 m c = 1.732 m c .$ Cross-check via $γ$ : $E = γ m c^{2} = 2 m c^{2}$ means $γ = 2$ , so $β = 1 - 1/ γ^{2} = 1 - 1/4 = 3 /2$ , and $p = γ β m c = 2 \cdot \frac{3}{2} m c = 3 m c$ . ✅ Both roads agree.

L4.2 — A photon and an electron each have momentum $p = m c$ (where $m$ is the electron mass). Compare their total energies.

Recall Solution L4.2

Photon ( $m_{γ} = 0$ , see Photon momentum and radiation pressure): $E = p c = m c \cdot c = m c^{2}$ .
Electron ( $m \neq = 0$ ): $E = (p c)^{2} + (m c^{2})^{2} = (m c^{2})^{2} + (m c^{2})^{2} = 2 m c^{2} = 1.414 m c^{2}$ . Why the difference: the electron carries extra rest energy $m c^{2}$ that the massless photon lacks. Same momentum, but the massive particle always has the larger total energy. See Mass-energy equivalence E = γmc².

L4.3 — Kinetic energy is $K = E - m c^{2} = (γ - 1) m c^{2}$ . For an electron with $K = m c^{2}$ (i.e. $K$ equals its rest energy), find $γ$ , $β$ , and $p$ .

Recall Solution L4.3

$K = (γ - 1) m c^{2} = m c^{2} \Rightarrow γ - 1 = 1 \Rightarrow γ = 2$ . $β = 1 - 1/ γ^{2} = 1 - 1/4 = 3 /2 = 0.8660$ . $p = γ β m c = 2 \cdot \frac{3}{2} m c = 3 m c = 1.732 m c .$ Note: this is the same physical state as L4.1 — total energy $2 m c^{2}$ means kinetic energy $m c^{2}$ . Consistency confirms our web of relations holds together.

Level 5 — Mastery

Goal: multi-step problems that fuse conservation, frames, and limits. Here the signed convention for $p$ finally does real work.

L5.1 — A particle of rest mass $m$ moving at $β = 0.8$ (in the $+ x$ direction) collides head-on and sticks to an identical particle moving at $β = 0.8$ in the $- x$ direction. Find the rest mass $M$ of the combined object. (Hint: total energy and total momentum are conserved — mind the signs.)

Recall Solution L5.1

Momentum (signed components add): particle 1 moves in $+ x$ so its signed component is $p_{1} = + γ m (0.8 c)$ ; particle 2 moves in $- x$ so its signed component is $p_{2} = - γ m (0.8 c)$ . They cancel: total $p_{tot} = p_{1} + p_{2} = 0$ . The lump is at rest. This is exactly why we keep signs: with magnitudes we'd have wrongly added to $2 γ m (0.8 c)$ . Energy (a scalar — always adds): each has $γ = 1/ 1 - 0.64 = 1/0.6 = 5/3$ , so each $E = γ m c^{2} = \frac{5}{3} m c^{2}$ . Total: $E_{tot} = 2 \cdot \frac{5}{3} m c^{2} = \frac{10}{3} m c^{2} .$ Since the lump is at rest, all its energy is rest energy: $E_{tot} = M c^{2}$ . Therefore $M = \frac{10}{3} m = 3.333 m .$ The lesson: the combined mass is more than $2 m$ ! The kinetic energy of the collision became rest mass — a direct display of Mass-energy equivalence E = γmc². Rest mass is not conserved; total energy and (signed) total momentum are.

L5.2 — Verify the invariant. For the single particle at $β = 0.8$ (mass $m$ ), compute $E^{2} - (p c)^{2}$ and confirm it equals $(m c^{2})^{2}$ regardless of the frame's motion.

Recall Solution L5.2

$γ = 5/3$ , so $E = \frac{5}{3} m c^{2}$ and $∣ p ∣ = γ β m c = \frac{5}{3} \cdot 0.8 m c = \frac{4}{3} m c$ , giving $p c = \frac{4}{3} m c^{2}$ . (The invariant uses $p^{2}$ , so the sign of $p$ is irrelevant here — only $∣ p ∣$ enters.) $E^{2} - (p c)^{2} = (\frac{5}{3})^{2} (m c^{2})^{2} - (\frac{4}{3})^{2} (m c^{2})^{2} = (\frac{25}{9} - \frac{16}{9}) (m c^{2})^{2} = \frac{9}{9} (m c^{2})^{2} = (m c^{2})^{2} .$ Why this is deep: $E^{2} - (p c)^{2}$ is the squared "length" of the energy–momentum four-vector. Different observers measure different $E$ and $p$ , but this combination is invariant — everyone computes $m c^{2}$ . That invariance is why we differentiated by proper time in the first place.

L5.3 — Show that the relativistic momentum can be written $p = \frac{E v}{c ^{2}}$ , and check it for the $β = 0.8$ electron.

Recall Solution L5.3

Derivation. We have two facts built in the parent note: total energy $E = γ m c^{2}$ and momentum $p = γ m v$ . The trick is to notice both share the same factor $γ m$ . Solve the energy equation for that shared factor: $E = γ m c^{2} \Rightarrow γ m = \frac{E}{c ^{2}} .$ What we did / why: we isolated $γ m$ so we can substitute it into the momentum formula and eliminate $γ$ and $m$ in favour of the single quantity $E$ . Now put $γ m = E / c^{2}$ into $p = γ m v$ : $p = (γ m) v = \frac{E}{c ^{2}} v \Rightarrow p = \frac{E v}{c ^{2}} .$ Equivalently, dividing the two original equations component-by-component gives the same thing directly: $\frac{p}{E} = \frac{γ m v}{γ m c ^{2}} = \frac{v}{c ^{2}}$ . Why useful: it says momentum is just energy carried along at velocity $v$ (divided by $c^{2}$ ), and it inherits the direction of $v$ automatically — so the sign of $p$ follows the sign of $v$ with no extra work. For a photon, $∣ v ∣ = c$ gives $∣ p ∣ = E / c$ — the massless case pops straight out. Check ( $β = 0.8$ , so $v = 0.8 c$ ): from L5.2, $E = \frac{5}{3} m c^{2}$ . Then $p = \frac{E v}{c ^{2}} = \frac{( \frac{5}{3} m c ^{2} ) ( 0.8 c )}{c ^{2}} = \frac{5}{3} \cdot 0.8 m c = \frac{4}{3} m c .$ This matches the direct computation $p = γ β m c = \frac{4}{3} m c$ from L5.2. ✅

Connections

Lorentz factor and time dilation — the source of every $γ$ used above.
Mass-energy equivalence E = γmc² — powers the L4 and L5 energy problems.
Energy-momentum four-vector — the invariant $E^{2} - (p c)^{2}$ behind L4.1 and L5.2.
Why nothing exceeds the speed of light — the $p \to \infty$ wall of L3.2.
Newtonian momentum p = mv — the low- $β$ limit checked in L1.3 and L3.3.
Photon momentum and radiation pressure — the $m = 0$ case in L4.2 and L5.3.