Classical-Mechanics · Unit 19 · Video 3 · Interactive Practice

Seen from the Center of Mass, an Elastic Collision Simply Reverses

IKey Formulas

Formula	Name	What it gives you
$v_{x,\text{cm}} = \dfrac{m_1 v_{1x,i} + m_2 v_{2x,i}}{m_1 + m_2}$	Center-of-mass velocity	Total momentum divided by total mass; unchanged by the collision
$v'_{1x,f} = -v'_{1x,i}, \qquad v'_{2x,f} = -v'_{2x,i}$	Elastic collision in the CM frame	Same speeds, both reversed (primes mean $v' = v - v_{x,\text{cm}}$ )
$v_{1x,f} = v_{1x,i}\dfrac{m_1 - m_2}{m_1 + m_2} + v_{2x,i}\dfrac{2 m_2}{m_1 + m_2}$	Lab-frame result, elastic	The reversal with $v_{x,\text{cm}}$ added back; swap $1 \leftrightarrow 2$ for the other cart
$\dfrac{\Delta K}{K_{\text{initial}}} = -\dfrac{m_2}{m_1 + m_2}$	Totally inelastic, target at rest	The share of the mass sitting still is the share of the energy destroyed

Key Insight: Three moves settle any one-dimensional collision — subtract $v_{x,\text{cm}}$ , act in the frame where the total momentum is zero, add $v_{x,\text{cm}}$ back. Elastic means both velocities reverse; sticking drives the relative velocity to zero and destroys exactly $\tfrac{1}{2}\mu\,(v^{\text{rel}}_i)^2$ with $\mu = \dfrac{m_1 m_2}{m_1 + m_2}$ , while the center-of-mass energy $\tfrac{1}{2}(m_1 + m_2)v_{x,\text{cm}}^2$ survives untouched.

IIVisualization 1 — The Only Root That Collides

With the total momentum pinned at zero, only two final velocities leave the kinetic energy unchanged.

Mass ratio m₂/m₁ 2.00 Trial ratio s −0.55

IIIVisualization 2 — Subtract, Reverse, Add Back

One velocity axis, three moves: shift to the zero-momentum frame, flip through zero, shift back.

Step lab Mass ratio m₂/m₁ 2.00 Cart 2

Step 1 — In the laboratory frame

v_{x,\text{cm}} = \frac{m_1 v_{1x,i} + m_2 v_{2x,i}}{m_1 + m_2}

No external force acts along the track, so this one number is fixed for all time.

Step 2 — Subtract $v_{x,\text{cm}}$

v'_{1x,i} = v_{1x,i} - v_{x,\text{cm}} = (v_{1x,i} - v_{2x,i})\,\frac{m_2}{m_1 + m_2}

v'_{2x,i} = v_{2x,i} - v_{x,\text{cm}} = (v_{2x,i} - v_{1x,i})\,\frac{m_1}{m_1 + m_2}

m_1 v'_{1x,i} + m_2 v'_{2x,i} = 0

Step 3 — Reverse

Zero momentum forces

v'_{2x} = -\dfrac{m_1}{m_2} v'_{1x}

before and after, so the elastic condition collapses to

\big(v'_{1x,f}\big)^2 = \big(v'_{1x,i}\big)^2

The

+

root is a miss; the

-

root is the collision:

v'_{1x,f} = -v'_{1x,i}, \qquad v'_{2x,f} = -v'_{2x,i}

Step 4 — Add $v_{x,\text{cm}}$ back

v_{1x,f} = -v'_{1x,i} + v_{x,\text{cm}} = v_{1x,i}\,\frac{m_1 - m_2}{m_1 + m_2} + v_{2x,i}\,\frac{2 m_2}{m_1 + m_2}

v_{2x,f} = -v'_{2x,i} + v_{x,\text{cm}} = v_{2x,i}\,\frac{m_2 - m_1}{m_1 + m_2} + v_{1x,i}\,\frac{2 m_1}{m_1 + m_2}

The lab-frame formulas, with no simultaneous equations solved anywhere.

IVVisualization 3 — What Sticking Costs

When the carts stick, the relative kinetic energy is destroyed and the center-of-mass energy survives.

Mass ratio m₂/m₁ 1.00

💡 Let the same two carts bounce elastically instead and the ledger closes untouched: the reversal in the center-of-mass frame hands back every joule of relative kinetic energy, so $\Delta K = 0$ no matter what the mass ratio is.

VQuiz Questions

Problem 1 · Elastic, Target at Rest

Given: cart 1 of mass $m_1$ moves along $+x$ with speed $v_{1,i}$ and strikes cart 2 of mass $m_2 = 3m_1$ sitting at rest; the collision is elastic — find $v_{1x,f}$ .

✅ Correct! The mass factor

\frac{m_1 - m_2}{m_1 + m_2} = \frac{-2m_1}{4m_1} = -\tfrac{1}{2}

: cart 1 rebounds at half its incoming speed.

❌ That is the CM-frame behavior. Only in the zero-momentum frame does a cart leave with the speed it brought in. In the lab frame you must still add

v_{x,\text{cm}} = \tfrac{1}{4}v_{1,i}

back, which turns

-\tfrac{3}{4}v_{1,i}

into

-\tfrac{1}{2}v_{1,i}

❌ Check the sign.

m_1 - m_2 = m_1 - 3m_1 = -2m_1

is negative, so the lighter cart comes back the way it came.

❌ Those are the numbers for $m_2 = 2m_1$ . With

m_2 = 3m_1

the factor is

\frac{-2m_1}{4m_1}

, not

\frac{-m_1}{3m_1}

❌ Not quite. Use

v_{1x,f} = v_{1x,i}\frac{m_1-m_2}{m_1+m_2} + v_{2x,i}\frac{2m_2}{m_1+m_2}

with

v_{2x,i} = 0

Show solution

Lab-frame formula. With $v_{2x,i} = 0$ the second term vanishes:

v_{1x,f} = v_{1,i}\,\frac{m_1 - 3m_1}{m_1 + 3m_1} = v_{1,i}\,\frac{-2m_1}{4m_1} = -\tfrac{1}{2}v_{1,i}

The same answer through the center of mass.

v_{x,\text{cm}} = \frac{m_1 v_{1,i}}{4m_1} = \tfrac{1}{4}v_{1,i}, \qquad v'_{1x,i} = v_{1,i} - \tfrac{1}{4}v_{1,i} = \tfrac{3}{4}v_{1,i}

Reverse, then add $v_{x,\text{cm}}$ back:

v_{1x,f} = -\tfrac{3}{4}v_{1,i} + \tfrac{1}{4}v_{1,i} = -\tfrac{1}{2}v_{1,i}

Check. Cart 2 leaves at $v_{2x,f} = \frac{2m_1}{4m_1}v_{1,i} = \tfrac{1}{2}v_{1,i}$ , so the momentum is $m_1(-\tfrac{1}{2}v_{1,i}) + 3m_1(\tfrac{1}{2}v_{1,i}) = m_1 v_{1,i}$ ✓ and the kinetic energy is $\tfrac{1}{2}m_1 v_{1,i}^2(\tfrac{1}{4} + \tfrac{3}{4}) = \tfrac{1}{2}m_1 v_{1,i}^2$ ✓.

Problem 2 · Into the Center-of-Mass Frame

Given: $m_1 = 1.0$ kg with $v_{1x,i} = +4.0$ m/s meets $m_2 = 3.0$ kg with $v_{2x,i} = -2.0$ m/s — find $v_{x,\text{cm}}$ and then $v'_{1x,i}$ .

What is $v_{x,\text{cm}}$ ?

What is $v'_{1x,i}$ ?

✅ Correct! The center of mass drifts backward at

0.50

m/s, and cart 1 closes on it at

4.5

m/s. Cart 2 gives

v'_{2x,i} = -1.5

m/s, so

m_1 v'_{1x,i} + m_2 v'_{2x,i} = 4.5 - 4.5 = 0

❌ That is the plain average. The velocities are weighted by their masses: the

3.0

kg cart counts three times as much as the

1.0

kg cart.

❌ A sign was dropped.

v_{2x,i} = -2.0

m/s, so the numerator is

4.0 - 6.0 = -2.0

, not

4.0 + 6.0

❌ Check the weighted sum.

v_{x,\text{cm}}

is the total momentum

m_1v_{1x,i} + m_2v_{2x,i}

divided by the total mass.

❌ That is the relative velocity.

v_{1x,i} - v_{2x,i} = 6.0

m/s still has to be scaled by cart 2's share of the mass,

\frac{m_2}{m_1+m_2} = \tfrac{3}{4}

❌ The shift was added, not subtracted.

v' = v - v_{x,\text{cm}} = 4.0 - (-0.50) = +4.5

m/s: subtracting a negative drift makes the primed velocity larger.

❌ Not quite. Every primed velocity is the lab velocity minus

v_{x,\text{cm}}

Show solution

Center-of-mass velocity.

v_{x,\text{cm}} = \frac{(1.0)(+4.0) + (3.0)(-2.0)}{1.0 + 3.0} = \frac{4.0 - 6.0}{4.0} = -0.50\ \text{m/s}

Cart 1 in that frame.

v'_{1x,i} = v_{1x,i} - v_{x,\text{cm}} = 4.0 - (-0.50) = +4.5\ \text{m/s}

The compact form gives the same thing:

v'_{1x,i} = (v_{1x,i} - v_{2x,i})\frac{m_2}{m_1 + m_2} = (6.0)\left(\tfrac{3}{4}\right) = +4.5\ \text{m/s}

Check the frame. $v'_{2x,i} = -2.0 - (-0.50) = -1.5$ m/s, and

m_1 v'_{1x,i} + m_2 v'_{2x,i} = (1.0)(4.5) + (3.0)(-1.5) = 0 \ \checkmark

Problem 3 · Head-On, Both Moving

Given: $m_1 = 3.0$ kg with $v_{1x,i} = +2.0$ m/s meets $m_2 = 1.0$ kg with $v_{2x,i} = -2.0$ m/s in an elastic collision — find both final laboratory velocities.

What is $v_{1x,f}$ ?

What is $v_{2x,f}$ ?

✅ Correct! The heavy cart stops dead and the light one leaves at

+4.0

m/s, carrying all

8.0

J and all

4.0\ \text{kg}\cdot\text{m/s}

of the momentum.

❌ Velocities only reverse in the CM frame. Here

v_{x,\text{cm}} = +1.0

m/s is not zero, so the lab velocities are the reversed primed values shifted by

+1.0

m/s.

❌ That is $v'_{1x,f}$ , still in the CM frame. Cart 1 arrives at

v'_{1x,i} = +1.0

m/s and leaves at

-1.0

m/s; adding

v_{x,\text{cm}} = +1.0

m/s back gives

v_{1x,f} = 0

❌ Not quite. Work in the CM frame: subtract

v_{x,\text{cm}}

, reverse, add it back.

❌ Velocities only reverse in the CM frame. Cart 2's primed velocity is

-3.0

m/s, not

-2.0

m/s, because the center of mass is itself moving at

+1.0

m/s.

❌ That is $v'_{2x,f}$ , still in the CM frame. Add

v_{x,\text{cm}} = +1.0

m/s back to reach the laboratory value.

❌ Not quite. Cart 2 approaches the center of mass at

-3.0

m/s, so it leaves at

+3.0

m/s in that frame.

Show solution

Step 1 — the center-of-mass velocity.

v_{x,\text{cm}} = \frac{(3.0)(+2.0) + (1.0)(-2.0)}{4.0} = \frac{6.0 - 2.0}{4.0} = +1.0\ \text{m/s}

Step 2 — subtract it.

v'_{1x,i} = 2.0 - 1.0 = +1.0\ \text{m/s}, \qquad v'_{2x,i} = -2.0 - 1.0 = -3.0\ \text{m/s}

Momentum in that frame: $(3.0)(1.0) + (1.0)(-3.0) = 0$ ✓

Step 3 — reverse. $v'_{1x,f} = -1.0$ m/s and $v'_{2x,f} = +3.0$ m/s.

Step 4 — add $v_{x,\text{cm}}$ back.

v_{1x,f} = -1.0 + 1.0 = 0, \qquad v_{2x,f} = 3.0 + 1.0 = +4.0\ \text{m/s}

Checks. Momentum before $= (3.0)(2.0) + (1.0)(-2.0) = 4.0$ ; after $= 0 + (1.0)(4.0) = 4.0$ ✓. Kinetic energy before $= \tfrac{1}{2}(3.0)(2.0)^2 + \tfrac{1}{2}(1.0)(2.0)^2 = 8.0$ J; after $= 0 + \tfrac{1}{2}(1.0)(4.0)^2 = 8.0$ J ✓. The relative velocity goes from $+4.0$ m/s to $-4.0$ m/s, reversed as it must be.

Problem 4 · Reading the Loss Backwards

Given: cart 1 of mass $m_1$ and speed $v_{1,i}$ hits cart 2 at rest and the two stick together; the collision destroys 80% of the kinetic energy — find the mass ratio and the speed of the pair.

What is $m_2/m_1$ ?

What is the final speed of the pair?

✅ Correct! A target four times heavier carries

\tfrac{4}{5}

of the mass and takes

\tfrac{4}{5}

of the energy with it; the pair crawls off at

v_f = v_{x,\text{cm}} = \tfrac{1}{5}v_{1,i}

❌ That is the ratio for a 20% loss.

m_2/(m_1+m_2)

is the fraction lost;

m_1/(m_1+m_2)

is the fraction kept.

❌ $0.80$ is the fraction lost, not the mass ratio. Solve

\frac{m_2}{m_1 + m_2} = 0.80

for

m_2

: it gives

0.20\,m_2 = 0.80\,m_1

❌ Not quite. Set

\frac{m_2}{m_1+m_2}

equal to

0.80

and solve for

m_2

in terms of

m_1

❌ That is the lost share, used as a speed. The pair moves at

v_f = \frac{m_1}{m_1+m_2}v_{1,i}

, which is the kept share.

❌ No square root is needed here. With a stationary target

K_{\text{final}}/K_{\text{initial}} = \frac{m_1}{m_1+m_2}

and

v_f/v_{1,i} = \frac{m_1}{m_1+m_2}

are the same number,

0.20

❌ Not quite. Momentum alone fixes the pair's speed:

m_1 v_{1,i} = (m_1 + m_2)v_f

Show solution

Step 1 — invert the loss formula. For a stationary target,

\frac{|\Delta K|}{K_{\text{initial}}} = \frac{m_2}{m_1 + m_2} = 0.80 \;\Longrightarrow\; 0.80\,m_1 = 0.20\,m_2 \;\Longrightarrow\; m_2 = 4m_1

Step 2 — momentum gives the speed.

m_1 v_{1,i} = (m_1 + m_2)v_f \;\Longrightarrow\; v_f = \frac{m_1}{5m_1}v_{1,i} = 0.20\,v_{1,i}

Check the energy directly.

K_{\text{final}} = \tfrac{1}{2}(5m_1)(0.20\,v_{1,i})^2 = 0.20\left(\tfrac{1}{2}m_1v_{1,i}^2\right) = 0.20\,K_{\text{initial}}

So 80% is gone ✓. That missing energy is exactly $\tfrac{1}{2}\mu v_{1,i}^2$ with $\mu = \frac{(m_1)(4m_1)}{5m_1} = 0.80\,m_1$ , and the surviving 20% is the center-of-mass energy $\tfrac{1}{2}(5m_1)v_{x,\text{cm}}^2$ , since $v_f = v_{x,\text{cm}}$ .

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