In relativistic kinematics, the relativistic momentum of a particle with rest mass $m_0$ and velocity $v$ is given by $p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $c$ is the speed of light in vacuum. Solve this equation explicitly for the particle speed $v$ in terms of $p$, $m_0$, and $c$, and describe the asymptotic behavior of $v$ as momentum $p \to \infty$.

In relativistic kinematics, the relativistic momentum of a particle with rest mass $m_0$ and velocity $v$ is given by $p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$ , where $c$ is the speed of light in vacuum. Solve this equation explicitly for the particle speed $v$ in terms of $p$ , $m_0$ , and $c$ , and describe the asymptotic behavior of $v$ as momentum $p \to \infty$ .

Final answer

$v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$ ; as $p \to \infty$ , $v \to c$

Steps

Square both sides to eliminate the radical

To solve the relativistic momentum equation for the velocity $v$ , we manipulate the formula algebraically to isolate $v$ , and then evaluate the limit as $p \to \infty$ .

The given formula is: $p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$

Squaring both sides of the equation yields: $p^2 = \frac{m_0^2 v^2}{1 - \frac{v^2}{c^2}}$

Clear the fraction and group terms with $v^2$

Multiply both sides by the denominator $\left(1 - \frac{v^2}{c^2}\right)$ : $p^2 \left(1 - \frac{v^2}{c^2}\right) = m_0^2 v^2$

Expand the left side: $p^2 - \frac{p^2 v^2}{c^2} = m_0^2 v^2$

Add $\frac{p^2 v^2}{c^2}$ to both sides to gather all terms containing $v^2$ on one side: $p^2 = m_0^2 v^2 + \frac{p^2 v^2}{c^2}$

Factor out $v^2$ and isolate it

Factor out $v^2$ on the right-hand side: $p^2 = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right)$

Combine the terms inside the parentheses over a common denominator $c^2$ : $p^2 = v^2 \left(\frac{m_0^2 c^2 + p^2}{c^2}\right)$

Multiply both sides by $c^2$ and divide by $(m_0^2 c^2 + p^2)$ : $v^2 = \frac{p^2 c^2}{m_0^2 c^2 + p^2}$

Take the square root

Taking the positive square root (since speed is a non-negative magnitude) gives: $v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$

Determine the asymptotic behavior as $p \to \infty$

To find the limiting speed as momentum $p$ becomes arbitrarily large, divide the numerator and the denominator by $p$ (noting that $p = \sqrt{p^2}$ for $p > 0$ ): $v = \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}}$

As $p \to \infty$ , the term $\frac{m_0^2 c^2}{p^2} \to 0$ . Therefore: $\lim_{p \to \infty} v = \frac{c}{\sqrt{0 + 1}} = c$

As momentum $p \to \infty$ , the speed $v$ approaches $c$ , illustrating that the speed of light $c$ serves as an upper horizontal asymptote for massive particles.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to square the speed of light c when squaring the denominator, leading to an incorrect dimensional expression for velocity.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$p^2 = \frac{m_0^2 v^2}{1 - v^2/c}$
Correct
$p^2 = \frac{m_0^2 v^2}{1 - \frac{v^2}{c^2}}$
When squaring the original denominator $\sqrt{1 - \frac{v^2}{c^2}}$ , squaring the square root removes the radical sign directly, leaving the radicand $1 - \frac{v^2}{c^2}$ . You mistakenly wrote $c$ instead of $c^2$ in the denominator.
Step 2
$p^2 \left(1 - \frac{v^2}{c}\right) = m_0^2 v^2 \implies p^2 - p^2 \frac{v^2}{c} = m_0^2 v^2 \implies p^2 = v^2 \left(m_0^2 + \frac{p^2}{c}\right)$
Correct
$p^2 \left(1 - \frac{v^2}{c^2}\right) = m_0^2 v^2 \implies p^2 - \frac{p^2 v^2}{c^2} = m_0^2 v^2 \implies p^2 = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right)$
Your algebraic manipulation—multiplying out the denominator, expanding terms, and factoring out $v^2$ —was executed correctly, but it carried forward the missing square on $c$ .
Step 3
$v = \frac{p}{\sqrt{m_0^2 + \frac{p^2}{c}}}$
Correct
$v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$
You correctly isolated $v$ and took the square root, but the result inherits the missing factor of $c$ from the initial squaring error.
Step 4
As $p \to \infty$ , $v$ approaches $\sqrt{c}$ .
Correct
As $p \to \infty$ , $v \to c$ .
You correctly evaluated the limit of your derived formula as $p \to \infty$ , but because your formula had an extra power of $\sqrt{c}$ missing, your asymptotic limit yielded $\sqrt{c}$ instead of $c$ . Notice also that $\sqrt{c}$ does not have units of velocity.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that squaring both sides eliminates the radical.

You cleared the denominator and successfully gathered all terms involving $v^2$ on one side.

You accurately factored out $v^2$ and took the principal square root to isolate $v$ .

You evaluated the limit as $p \to \infty$ using consistent asymptotic reasoning.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Radicand transcription / squaring error

You accidentally dropped the exponent on $c^2$ when removing the square root, copying the radicand denominator as $c$ instead of $c^2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Inattention to maintaining literal exponents when clearing radicals: squaring an expression of the form $\sqrt{A - B}$ yields $(A - B)$ unchanged, so $\left(\sqrt{1 - v^2/c^2}\right)^2 = 1 - v^2/c^2$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Squaring a Square Root: For any non-negative expression $E$ , $\left(\sqrt{E}\right)^2 = E$ .
Always copy the entire radicand exactly as it is inside the radical without altering any internal exponents.
Dimensional Analysis Sanity Check: Every term added to or equated with a quantity must have matching physical dimensions.
Notice that $v/c$ is dimensionless, while $v^2/c$ has units of $\text{m}^2/(\text{s}\cdot\text{m})$ , which is dimensionally inconsistent.

CORRECTED PATH

ACEpath Teacher's Demonstration

Square both sides properly: $p^2 = \frac{m_0^2 v^2}{1 - \frac{v^2}{c^2}}$ .
Multiply both sides by $\left(1 - \frac{v^2}{c^2}\right)$ : $p^2 - \frac{p^2 v^2}{c^2} = m_0^2 v^2$ .
Group terms containing $v^2$ : $p^2 = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right) = v^2 \left(\frac{m_0^2 c^2 + p^2}{c^2}\right)$ .
Multiply by $c^2$ and divide by $(m_0^2 c^2 + p^2)$ : $v^2 = \frac{p^2 c^2}{m_0^2 c^2 + p^2}$ .
Take the positive square root: $v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$ .
Evaluate the limit as $p \to \infty$ : $\lim_{p \to \infty} \frac{p c}{\sqrt{m_0^2 c^2 + p^2}} = \lim_{p \to \infty} \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}} = c$ .

Final answer

$v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$ ; as $p \to \infty$ , $v \to c$

NEXT STEPS

Final Takeaways from Your Teacher

Double-check that you transcribe every internal exponent faithfully when removing radicals.
Perform a quick unit or dimensional check whenever solving physics literal equations: speed $v$ must have the same dimension as $c$ , not $\sqrt{c}$ .

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when moving the v² term across the equality sign, yielding a difference instead of a sum in the denominator.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$p^2 \left(1 - \frac{v^2}{c^2}\right) = m_0^2 v^2 \implies p^2 - \frac{p^2 v^2}{c^2} = m_0^2 v^2$
Step 2
$p^2 = v^2 \left(m_0^2 - \frac{p^2}{c^2}\right)$
Correct
$p^2 = m_0^2 v^2 + \frac{p^2 v^2}{c^2} = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right)$
When moving $-\frac{p^2 v^2}{c^2}$ to the right-hand side, you should add it to both sides rather than subtract it, which yields a positive sign between the terms.
Step 3
$v^2 = \frac{p^2}{m_0^2 - p^2/c^2} \implies v = \frac{c p}{\sqrt{m_0^2 c^2 - p^2}}$
Correct
$v^2 = \frac{p^2 c^2}{m_0^2 c^2 + p^2} \implies v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$
You followed algebraically sound steps to isolate $v$ and simplify the complex fraction, but the expression carries the sign error from the previous step.
Step 4
As $p \to \infty$ , the quantity under the square root becomes negative, so the speed becomes undefined rather than approaching a finite limit.
Correct
As $p \to \infty$ , $v = \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}} \to \frac{c}{\sqrt{0 + 1}} = c$ .
Because of the negative sign under the radical, the expression became undefined for large $p$ , whereas the correct formula approaches the speed of light $c$ asymptotically.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly squared both sides and cleared the fraction to remove the radical.

You expanded the terms and recognized the need to collect all terms containing the target variable $v^2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error during variable transposition

You changed $-\frac{p^2 v^2}{c^2}$ into $-\frac{p^2}{c^2}$ when gathering $v^2$ terms on the right-hand side instead of adding it.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Losing track of the additive inverse when transposing terms across the equals sign, especially when simultaneously factoring out the target variable.

REASONING PATH

Your ACEpath teacher’s suggested approach

Additive Inverse Rule: Adding or subtracting a term across the equals sign always reverses its sign: $A - B = C \implies A = C + B$ .
Write the explicit addition step $+ \frac{p^2 v^2}{c^2}$ on both sides before factoring out $v^2$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Start from $p^2 - \frac{p^2 v^2}{c^2} = m_0^2 v^2$ .
Add $\frac{p^2 v^2}{c^2}$ to both sides: $p^2 = m_0^2 v^2 + \frac{p^2 v^2}{c^2}$ .
Factor out $v^2$ : $p^2 = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right) = v^2 \left(\frac{m_0^2 c^2 + p^2}{c^2}\right)$ .
Multiply by $c^2$ and divide by $(m_0^2 c^2 + p^2)$ : $v^2 = \frac{p^2 c^2}{m_0^2 c^2 + p^2}$ .
Take the positive square root: $v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$ .
Divide numerator and denominator by $p$ to evaluate the limit: $v = \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}} \to c$ as $p \to \infty$ .

Final answer

$v = \frac{p c}{\sqrt{m_0^2 c^2 + p^2}}$ ; as $p \to \infty$ , $v \to c$

NEXT STEPS

Final Takeaways from Your Teacher

Practice separating transposition and factoring into two distinct lines of algebra to prevent combined mental arithmetic slips.
Perform a sanity check on asymptotic limits: physical speeds in special relativity must remain real and bounded by $c$ .

OverviewACEpath Teacher's Diagnosis

03 The student misinterprets the limit as p → ∞ by treating the denominator as a constant, concluding incorrectly that the speed grows without bound to infinity.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$p^2 \left(1 - \frac{v^2}{c^2}\right) = m_0^2 v^2 \implies p^2 = v^2 \left(m_0^2 + \frac{p^2}{c^2}\right) \implies v = \frac{c p}{\sqrt{m_0^2 c^2 + p^2}}$
Step 2
$\text{As }$ p $\to$ $\infty$ , $\text{ the numerator }$ c p $\text{ grows without bound while the mass term remains fixed, so }$ v $\to$ $\infty$ .
Correct
$\lim_{p $\to$ $\infty$ } $v = \$ \lim_{p \to \infty}\frac{c p}{\sqrt{m_0^2 c^2 + p^2}} $= \$ \lim_{p $\to$ $\infty$ }$ \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}} = c
The variable $p$ appears in both the numerator and the denominator. As $p \to \infty$ , the denominator is dominated by $\sqrt{p^2} = p$ , so dividing both numerator and denominator by $p$ shows that $v \to c$ , not $\infty$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared fractions and isolated $v^2$ algebraically.

You correctly solved for the explicit expression for speed as $v = \frac{c p}{\sqrt{m_0^2 c^2 + p^2}}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Ignoring Variable in Denominator During Limit Evaluation

You observed the numerator $c p$ growing without bound and noted the constant $m_0^2 c^2$ , but overlooked that $p^2$ in the denominator also grows without bound.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a rational or radical expression of the form $\frac{\infty}{\infty}$ as $\frac{\infty}{\text{constant}}$ by failing to identify the leading order term of the denominator before taking the limit.

REASONING PATH

Your ACEpath teacher’s suggested approach

Identify Variable Locations: Before concluding that a fraction grows without bound because its numerator tends to infinity, check whether the denominator also depends on that same variable.
Circle all occurrences of the limiting variable $p$ in both the numerator and the denominator.
Divide by the Dominant Power: For an indeterminate form $\frac{\infty}{\infty}$ , divide the numerator and denominator by the highest power of the variable.
Divide numerator and denominator by $p = \sqrt{p^2}$ (for $p > 0$ ) to obtain $\frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}}$ before evaluating $\lim_{p \to \infty}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor $p^2$ out of the radical in the denominator: $\sqrt{m_0^2 c^2 + p^2} = p \sqrt{\frac{m_0^2 c^2}{p^2} + 1}$ for $p > 0$ .
Simplify the fraction by canceling $p$ : $v = \frac{c}{\sqrt{\frac{m_0^2 c^2}{p^2} + 1}}$ .
Take the limit as $p \to \infty$ : since $\frac{m_0^2 c^2}{p^2} \to 0$ , $v \to \frac{c}{\sqrt{0 + 1}} = c$ .

Final answer

$v = \frac{c p}{\sqrt{m_0^2 c^2 + p^2}}$ ; as $p \to \infty$ , $v \to c$

NEXT STEPS

Final Takeaways from Your Teacher

Always factor out the highest power of the variable from both numerator and denominator when evaluating limits at infinity.
Remember the physical principle of special relativity: massive particles can approach the speed of light $c$ as momentum increases, but can never exceed or reach infinite speed.

Lock it in

Question

What is the exact result of squaring the expression $\sqrt{1 - \frac{x^2}{a^2}}$ ?

Answer

$1 - \frac{x^2}{a^2}$ . The radical is removed and the entire radicand remains completely unchanged.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to square the speed of light c when squaring the denominator, leading to an incorrect dimensional expression for velocity.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student makes a sign error when moving the v² term across the equality sign, yielding a difference instead of a sum in the denominator.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student misinterprets the limit as p → ∞ by treating the denominator as a constant, concluding incorrectly that the speed grows without bound to infinity.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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