Classical-Mechanics · Unit 12 · Video 5 · Interactive Practice

Before the Algebra: How to Set Up Any Momentum Problem

IKey Formulas

Formula	Name	What it takes
$\vec{F}^{\,\text{ext}} = \dfrac{d\vec{p}_{\text{sys}}}{dt}$	Second Law for a system	A boundary: only forces crossing it count
$\vec{I} = \vec{F}^{\,\text{ext}}_{\text{ave}}\,\Delta t_{\text{int}} = \Delta\vec{p}_{\text{sys}}$	Integral (impulse) form	A product — small if either factor is small
$p_{\text{sys},x,i} = p_{\text{sys},x,f}$	Constancy, one direction	Only where $F^{\,\text{ext}}_x = 0$
$m_1(v_{x,i})_1 + m_2(v_{x,i})_2 = m_1(v_{x,f})_1 + m_2(v_{x,f})_2$	The equation you solve	A momentum flow diagram: before and after

Key Insight: Because $\vec{F}^{\,\text{ext}} = d\vec{p}_{\text{sys}}/dt$ is a vector equation, "is momentum constant?" is never one question — it is one question per direction. And constant is not zero: $p_{\text{sys},x}$ can hold a fixed nonzero value while $p_{\text{sys},y}$ changes every instant.

IIWhere You Draw the Boundary Decides the Physics

The same collision, the same forces — but which ones count as external depends on what you enclose.

Track

IIIThe Momentum Flow Diagram Becomes the Equation

Three of the four velocities are yours to choose; constancy of $p_x$ fixes the fourth.

m₁ 2.0 kg m₂ 4.0 kg

💡 Constancy of $p_x$ supplies exactly one equation, so it can pin down exactly one unknown. A second condition — elastic, perfectly inelastic, or a measured final speed — is what closes a problem with two unknown final velocities.

IVWhen an External Force Still Leaves $p$ Constant

An external force acts throughout the collision, yet the impulse it delivers can still be negligible.

F^ext_ave 12 N Δt_int 1.0 ms

VQuiz Questions

Problem 1 · Choosing the System

Given: Two carts collide on a horizontal, frictionless track. You define the system as cart 1 + cart 2. Which statement correctly classifies the forces and their consequence?

✅ Correct! The contact pair lives entirely inside the boundary, and the two vertical external forces sum to zero — so

d\vec{p}_{\text{sys}}/dt = \vec{0}

in both directions.

❌ Not quite. A force is external only if its source lies outside the boundary. Cart 1 and cart 2 are both inside, so the pair they exert on each other never crosses it.

❌ Not quite. Gravity is external, but the track's normal force is external too, and on a horizontal track they cancel:

\vec{F}^{\,\text{ext}} = \vec{0}

❌ Constant is not zero. Those are different claims. A system gliding along the track at a steady

6\ \text{kg}\,\text{m/s}

has constant, decidedly nonzero,

p_{\text{sys},x}

Show solution

Sort every force by where its source sits relative to the boundary:

Cart 1 on cart 2 and cart 2 on cart 1 — both sources inside → internal. Internal forces cancel in pairs inside $\vec{F}^{\,\text{ext}}$ .
Earth on each cart ( $m_ig$ down) and track on each cart ( $N_i$ up) — sources outside → external.

On a horizontal track the carts have no vertical acceleration, so $\sum N_i = \sum m_ig$ and

F^{\,\text{ext}}_x = 0, \qquad F^{\,\text{ext}}_y = \sum N_i - \sum m_i g = 0 .

With $\vec{F}^{\,\text{ext}} = d\vec{p}_{\text{sys}}/dt = \vec{0}$ , both components of the system momentum are constant — whatever nonzero values they happen to have.

Problem 2 · One Question Per Direction

Given: A $2.0\ \text{kg}$ projectile is launched at $20\ \text{m/s}$ , $30\degree$ above the horizontal; air resistance is negligible and the system is the projectile alone. Which statement about the flight is correct?

✅ Correct! Gravity has no

x

-component, so

dp_{\text{sys},x}/dt = 0

and

p_{\text{sys},x}

keeps its launch value

2.0(20\cos 30\degree) = 34.6 \approx 35\ \text{kg}\,\text{m/s}

❌ Not quite. Gravity is external, and it points along

-\hat{\jmath}

F^{\,\text{ext}}_y = -mg \neq 0

, so

p_{\text{sys},y}

changes every instant.

❌ Axes swapped. The force-free direction is the one with no external component — that is

x

, not

y

❌ Check the top of the flight. Only

v_y

vanishes there; the projectile still moves horizontally, so

\vec{p} = 35\,\hat{\imath}\ \text{kg}\,\text{m/s} \neq \vec{0}

Show solution

Apply the Second Law for the system component by component:

F^{\,\text{ext}}_x = 0 \;\Rightarrow\; \frac{dp_{\text{sys},x}}{dt} = 0, \qquad F^{\,\text{ext}}_y = -mg \;\Rightarrow\; \frac{dp_{\text{sys},y}}{dt} = -mg \neq 0 .

So $x$ is the constant direction. Its value is set once, at launch:

p_{\text{sys},x} = m\,v\cos\theta = (2.0)(20)\cos 30\degree = (2.0)(17.32) = 34.6\ \text{kg}\,\text{m/s}.

Meanwhile $p_{\text{sys},y}$ runs from $+20\ \text{kg}\,\text{m/s}$ at launch, through $0$ at the apex, to $-20\ \text{kg}\,\text{m/s}$ at the same height on the way down. One axis constant, one axis changing, in the very same problem.

Problem 3 · Carry Out the Plan

Given: On a frictionless track, cart 1 ( $m_1 = 2.0\ \text{kg}$ ) moves at $(v_{x,i})_1 = +6.0\ \text{m/s}$ toward cart 2 ( $m_2 = 4.0\ \text{kg}$ ) moving at $(v_{x,i})_2 = -1.5\ \text{m/s}$ . After the collision $(v_{x,f})_1 = -1.0\ \text{m/s}$ — find $(v_{x,f})_2$ . Positive $x$ points right; every symbol is a component.

What is $p_{\text{sys},x,i}$ ?

What is $(v_{x,f})_2$ ?

✅ Correct! Cart 2 is turned around: it arrives at

-1.5\ \text{m/s}

and leaves at

+2.0\ \text{m/s}

, while cart 1 rebounds backwards.

❌ Signs, not sizes.

(v_{x,i})_2

is a component, and it is negative: the term is

4.0(-1.5) = -6.0

, not

+6.0

❌ Momentum, not velocity. Each velocity must be weighted by its own mass;

\sum v

is not conserved,

\sum mv

is.

❌ Check the final term.

(v_{x,f})_1 = -1.0\ \text{m/s}

, so

m_1(v_{x,f})_1 = -2.0\ \text{kg}\,\text{m/s}

, which adds

2.0

when moved across the equals sign.

❌ Check the before-state sum. Add signed terms:

m_1(v_{x,i})_1 + m_2(v_{x,i})_2

❌ Check the after-state sum. Set

m_1(v_{x,f})_1 + m_2(v_{x,f})_2

equal to the before-state value and solve.

Show solution

Step 1 — Is $p_x$ constant? The track is frictionless and horizontal, the contact forces are internal, so $F^{\,\text{ext}}_x = 0$ and $p_{\text{sys},x,i} = p_{\text{sys},x,f}$ .

Step 2 — Before state.

p_{\text{sys},x,i} = m_1(v_{x,i})_1 + m_2(v_{x,i})_2 = (2.0)(+6.0) + (4.0)(-1.5) = 12.0 - 6.0 = +6.0\ \text{kg}\,\text{m/s}

Step 3 — After state, same shape.

p_{\text{sys},x,f} = m_1(v_{x,f})_1 + m_2(v_{x,f})_2 = (2.0)(-1.0) + (4.0)(v_{x,f})_2

Step 4 — Equate and solve.

+6.0 = -2.0 + 4.0\,(v_{x,f})_2 \;\Longrightarrow\; (v_{x,f})_2 = \frac{8.0}{4.0} = +2.0\ \text{m/s}

Look back: the units are $\text{kg}\,\text{m/s}$ on every line; the positive component means cart 2 ends up moving in the $+\hat{\imath}$ direction, which is the reversal a collision from the right should produce. Check: $2.0(-1.0) + 4.0(+2.0) = +6.0\ \checkmark$

Problem 4 · Is the Impulse Negligible?

Given: The same two carts ( $m_{\text{sys}} = 6.0\ \text{kg}$ , $p_{\text{sys},x} = 6.0\ \text{kg}\,\text{m/s}$ ) collide on a track with kinetic friction, $\mu_k = 0.20$ , and the contact lasts $\Delta t_{\text{int}} = 1.0\ \text{ms}$ . Take $g = 9.8\ \text{m/s}^2$ . Estimate the fraction of $p_{\text{sys},x}$ that friction removes during the contact.

✅ Correct! Friction is a genuine external force, but

1.0\ \text{ms}

is short compared with the momentum already in play, so

\Delta p_{\text{sys}} \to 0

and the constancy statement is licensed.

❌ Check the arithmetic of the impulse.

I = f_k\,\Delta t_{\text{int}}

with

\Delta t_{\text{int}} = 1.0 \times 10^{-3}\ \text{s}

, not

10^{-1}\ \text{s}

❌ You compared a force with a momentum.

f_k/p_{\text{sys}} = 11.8/6.0 \approx 2

has units of

\text{s}^{-1}

— you must multiply by

\Delta t_{\text{int}}

first. Impulse is the product, and either factor can make it small.

❌ Not quite. Build the impulse in two moves:

f_k = \mu_k m_{\text{sys}} g

, then

I = f_k\,\Delta t_{\text{int}}

, and only then divide by

p_{\text{sys},x}

Show solution

Step 1 — Size the external force.

f_k = \mu_k m_{\text{sys}} g = (0.20)(6.0)(9.8) = 11.8\ \text{N}

Step 2 — Turn it into an impulse using the integral form $I = F^{\,\text{ext}}_{\text{ave}}\Delta t_{\text{int}}$ :

I = (11.8)(1.0 \times 10^{-3}) = 1.2 \times 10^{-2}\ \text{kg}\,\text{m/s}

Step 3 — Compare with the momentum already in play.

\frac{|F^{\,\text{ext}}_{\text{ave}}|\,\Delta t_{\text{int}}}{|p_{\text{sys},x}|} = \frac{1.2 \times 10^{-2}}{6.0} = 2.0 \times 10^{-3} = 0.20\%

Since $|F^{\,\text{ext}}_{\text{ave}}|\Delta t_{\text{int}} \ll |p_{\text{sys}}|$ , we set $\Delta p_{\text{sys}} \approx 0$ and use $p_{\text{sys},x,i} = p_{\text{sys},x,f}$ during the collision.

Short is never absolute. The same $1.0\ \text{ms}$ contact would be a $20\%$ correction for a system carrying only $0.06\ \text{kg}\,\text{m/s}$ — and friction still degrades the momentum over the long glide before and after the collision, which is a different interval and a different question.

Solved: 0 / 4

Before the Algebra: How to Set Up Any Momentum Problem

IKey Formulas

IIWhere You Draw the Boundary Decides the Physics

IIIThe Momentum Flow Diagram Becomes the Equation

IVWhen an External Force Still Leaves ppp Constant

VQuiz Questions

IVWhen an External Force Still Leaves $p$ Constant