Classical-Mechanics · Unit 19 · Video 1 · Interactive Practice

Collision Theory: Relative Velocity, the Center-of-Mass Frame, and Reduced Mass

IKey Formulas

Formula	Name	What it says
$\vec{v}_{1,2} = \vec{v}_1 - \vec{v}_2 = \vec{v}_{1,2}'$	Relative velocity	The boost cancels in a difference
$\mu = \dfrac{m_1 m_2}{m_1 + m_2}, \qquad \dfrac{1}{\mu} = \dfrac{1}{m_1} + \dfrac{1}{m_2}$	Reduced mass	Smaller than either mass
$\vec{v}_1' = \dfrac{\mu}{m_1}\vec{v}_{1,2}, \qquad \vec{v}_2' = -\dfrac{\mu}{m_2}\vec{v}_{1,2}$	Velocities in the CM frame	Momenta $\pm\mu\vec{v}_{1,2}$ sum to zero
$K = \tfrac{1}{2}\mu v_{1,2}^{2} + \tfrac{1}{2}(m_1 + m_2)v_{\mathrm{cm}}^{2}$	Kinetic energy split	Internal motion plus motion of the whole

Key Insight: With no external force, $v_{\mathrm{cm}}$ cannot change, so the second term is the same before and after and drops out of the difference: $\Delta K = \tfrac{1}{2}\mu\,\Delta\!\left(v_{1,2}^{2}\right)$ , one number every relatively inertial observer agrees on.

IIThe Boost Cancels in a Difference

Every velocity changes when the observer starts moving — ask whether their difference does.

Boost

V

(m/s) 0.0 v₁ (m/s) 5.0 v₂ (m/s) -3.0

💡 Challenge: choose the boost that makes $v_2' = 0$ — even riding on particle 2, the relative velocity reads the same.

IIIInside the Center-of-Mass Frame

In the frame riding with the center of mass, each velocity is one fixed multiple of $\vec{v}_{1,2}$ .

m₁ (kg) 2.0 m₂ (kg) 6.0

IVKinetic Energy Splits, and Only One Part Changes

Total kinetic energy depends on the observer; the change across the collision need not.

Observer frame Relative speed after (m/s) 4.0

Sorting collisions by the sign and size of $\Delta K$ — elastic, inelastic, totally inelastic — is the subject of the next video.

VQuiz Questions

Problem 1 · Relative Velocity from a Moving Camera

Given: in the lab frame $S$ , cart 1 moves at $v_1 = +6\ \mathrm{m/s}$ and cart 2 at $v_2 = -2\ \mathrm{m/s}$ . A camera rides past on a trolley at $V = +4\ \mathrm{m/s}$ , defining frame $S'$ . Find the relative velocity $v_{1,2}'$ that the trolley camera measures.

✅ Correct! Both velocities drop by

4\ \mathrm{m/s}

, so the gap between them is untouched:

v_{1,2}' = v_{1,2} = 8\ \mathrm{m/s}

❌ That is $v_1'$ , not $v_{1,2}'$ . The relative velocity is the difference of the two velocities measured in the same frame.

❌ The boost does not survive the subtraction. It is added to

v_1

and to

v_2

identically, so it appears once with each sign and cancels.

Show solution

Transform each velocity into $S'$ :

v_1' = v_1 - V = 6 - 4 = +2\ \mathrm{m/s}, \qquad v_2' = v_2 - V = -2 - 4 = -6\ \mathrm{m/s}

Now take the difference:

v_{1,2}' = v_1' - v_2' = 2 - (-6) = +8\ \mathrm{m/s}

In the lab frame, $v_{1,2} = 6 - (-2) = +8\ \mathrm{m/s}$ — identical. Algebraically,

v_{1,2}' = (v_1 - V) - (v_2 - V) = v_1 - v_2 = v_{1,2}

The relative velocity of two particles is the same in every relatively inertial frame.

Problem 2 · Reduced Mass (Watch the Reciprocals)

Given: $m_1 = 3\ \mathrm{kg}$ and $m_2 = 6\ \mathrm{kg}$ . Find the reduced mass $\mu$ .

✅ Correct!

\mu = 18/9 = 2\ \mathrm{kg}

— smaller than either mass, as

\mu

always is.

❌ You stopped one step early.

\tfrac{1}{m_1} + \tfrac{1}{m_2} = 0.5\ \mathrm{kg^{-1}}

1/\mu

, not

\mu

. Invert it.

❌ Not quite.

\mu

is the product over the sum, and it must come out smaller than

3\ \mathrm{kg}

Show solution

Using the product-over-sum form:

\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2\ \mathrm{kg}

Or with reciprocals:

\frac{1}{\mu} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{1}{2}\ \mathrm{kg^{-1}} \quad \Longrightarrow \quad \mu = 2\ \mathrm{kg}

Sanity check: $\mu$ is always less than the smaller mass, and $2\ \mathrm{kg}$ is indeed less than $3\ \mathrm{kg}$ ✓. The sum $9\ \mathrm{kg}$ and the average $4.5\ \mathrm{kg}$ both fail that test.

Problem 3 · Riding with the Center of Mass

Given: $m_1 = 2\ \mathrm{kg}$ moving at $v_1 = +5\ \mathrm{m/s}$ and $m_2 = 6\ \mathrm{kg}$ moving at $v_2 = -3\ \mathrm{m/s}$ along a track. Find particle 1's velocity in the center-of-mass frame, and the magnitude of each particle's momentum in that frame.

What is $v_1'$ ?

What is the magnitude of each particle's momentum in the center-of-mass frame?

✅ Correct!

v_1' = \tfrac{\mu}{m_1}v_{1,2} = +6\ \mathrm{m/s}

and each momentum has magnitude

\mu v_{1,2} = 12\ \mathrm{kg}\cdot\mathrm{m/s}

, pointing opposite ways.

❌ $v_{\mathrm{cm}}$ is a mass-weighted average, not a plain one. The heavier cart pulls it toward

v_2

v_{\mathrm{cm}} = -1\ \mathrm{m/s}

, not

+1\ \mathrm{m/s}

❌ Check $v_1'$ . Subtract the center-of-mass velocity from

v_1

, or use

v_1' = (\mu/m_1)\,v_{1,2}

❌ The total is zero, not each one. The two momenta are equal in magnitude and opposite in direction, so they cancel — but neither is itself zero.

❌ Check the momentum. Use the center-of-mass velocity,

p_1' = m_1 v_1'

, and compare it with

\mu v_{1,2}

Show solution

Step 1 — Relative velocity and reduced mass:

v_{1,2} = v_1 - v_2 = 5 - (-3) = +8\ \mathrm{m/s}, \qquad \mu = \frac{2 \cdot 6}{2 + 6} = \frac{12}{8} = 1.5\ \mathrm{kg}

Step 2 — Velocity of the center of mass:

v_{\mathrm{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{2(5) + 6(-3)}{8} = \frac{10 - 18}{8} = -1\ \mathrm{m/s}

Step 3 — Velocities in the center-of-mass frame:

v_1' = v_1 - v_{\mathrm{cm}} = 5 - (-1) = +6\ \mathrm{m/s} = \frac{\mu}{m_1}v_{1,2} = \frac{1.5}{2}(8) = +6\ \mathrm{m/s}\ \checkmark

v_2' = v_2 - v_{\mathrm{cm}} = -3 + 1 = -2\ \mathrm{m/s} = -\frac{\mu}{m_2}v_{1,2} = -\frac{1.5}{6}(8) = -2\ \mathrm{m/s}\ \checkmark

Step 4 — Momenta:

p_1' = m_1 v_1' = 2(6) = +12, \qquad p_2' = m_2 v_2' = 6(-2) = -12 \ \ (\mathrm{kg}\cdot\mathrm{m/s})

Both magnitudes equal $\mu v_{1,2} = 1.5(8) = 12$ , and $p_1' + p_2' = 0$ — the center of mass is at rest in its own frame. Notice the speeds are in inverse ratio to the masses: $6/2 = 3 = m_2/m_1$ .

Problem 4 · Does the Observer Change $\Delta K$ ?

Given: the same carts ( $m_1 = 2\ \mathrm{kg}$ at $+5\ \mathrm{m/s}$ , $m_2 = 6\ \mathrm{kg}$ at $-3\ \mathrm{m/s}$ , so $\mu = 1.5\ \mathrm{kg}$ and $v_{1,2} = 8\ \mathrm{m/s}$ ) collide, and afterwards their relative speed is $4\ \mathrm{m/s}$ . Find $\Delta K$ in the lab frame, and then $\Delta K$ as measured from a trolley moving at $+5\ \mathrm{m/s}$ .

What is $\Delta K$ in the lab frame?

What is $\Delta K$ measured from the trolley?

✅ Correct!

\Delta K = \tfrac{1}{2}\mu\,\Delta(v_{1,2}^2) = 0.75(16 - 64) = -36\ \mathrm{J}

— and the trolley reads exactly the same number.

❌ You dropped the factor $\tfrac{1}{2}$ .

\mu\,\Delta(v_{1,2}^2) = -72\ \mathrm{J}

; the kinetic energy of relative motion is

\tfrac{1}{2}\mu v_{1,2}^2

❌ That is the total internal energy, not the change.

-48\ \mathrm{J}

would be right only if the relative speed fell all the way to zero.

❌ Check $\Delta K$ . Use

\Delta K = \tfrac{1}{2}\mu\big((v_{1,2}')^2 - v_{1,2}^2\big)

with

\mu = 1.5\ \mathrm{kg}

❌ The center-of-mass term does not enter $\Delta K$ . In the trolley frame it is

144\ \mathrm{J}

— but it is

144\ \mathrm{J}

both before and after, so it cancels in the difference.

❌ It is determined.

\Delta K

depends only on

\mu

and the change in

v_{1,2}^2

, and

v_{1,2}

is the same in every relatively inertial frame.

❌ Check the frame dependence.

K

itself changes with the observer, but the observer-dependent piece

\tfrac{1}{2}(m_1+m_2)v_{\mathrm{cm}}^2

is constant through the collision.

Show solution

Lab frame. Split the kinetic energy:

K = \tfrac{1}{2}\mu v_{1,2}^2 + \tfrac{1}{2}(m_1 + m_2)v_{\mathrm{cm}}^2

With $v_{\mathrm{cm}} = -1\ \mathrm{m/s}$ the second term is $\tfrac{1}{2}(8)(1) = 4\ \mathrm{J}$ , before and after (no external force means the total momentum, hence $v_{\mathrm{cm}}$ , cannot change). So

K_{\text{before}} = \tfrac{1}{2}(1.5)(64) + 4 = 48 + 4 = 52\ \mathrm{J}, \qquad K_{\text{after}} = \tfrac{1}{2}(1.5)(16) + 4 = 12 + 4 = 16\ \mathrm{J}

\Delta K = 16 - 52 = -36\ \mathrm{J}

Equivalently, straight from the formula:

\Delta K = \tfrac{1}{2}\mu\,\Delta\!\left(v_{1,2}^{2}\right) = 0.75\,(16 - 64) = -36\ \mathrm{J}

Trolley frame. Every velocity drops by $5\ \mathrm{m/s}$ , so $v_{\mathrm{cm}}' = -1 - 5 = -6\ \mathrm{m/s}$ and the center-of-mass term becomes $\tfrac{1}{2}(8)(36) = 144\ \mathrm{J}$ :

K_{\text{before}} = 48 + 144 = 192\ \mathrm{J}, \qquad K_{\text{after}} = 12 + 144 = 156\ \mathrm{J}

\Delta K = 156 - 192 = -36\ \mathrm{J}

Both totals more than tripled, but the $144\ \mathrm{J}$ appears in each and cancels. Since $v_{1,2}$ is frame-independent, so is $\Delta K$ — and by the work–energy result of chapter 13, this $-36\ \mathrm{J}$ is exactly the work done by the interaction force.

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