Classical-Mechanics · Unit 6 · Video 2 · Interactive Practice

Newton's First Law: Inertia and Inertial Reference Frames

IKey Relationships

Statement	Name	What it means
$\sum \vec{F} = 0 \iff \vec{v} = \text{const}$	Newton's First Law	Zero net force ⟺ rest or uniform straight-line motion
$\Delta \vec{v} \ne 0 \iff \sum \vec{F} \ne 0$	Contrapositive form	Any change in velocity signals a nonzero net force
$\vec{v} = \|\vec{v}\|\,\hat{u}$	Velocity = speed + direction	Changing the speed or the direction changes $\vec{v}$
$\sum \vec{F} = 0 \iff \vec{a} = 0$	Force–acceleration link	A balanced body has zero acceleration

Key Insight: Rest and uniform motion are placed side by side as the same mechanical state. No experiment performed inside a body can tell whether it is at rest or gliding at constant velocity — this equivalence is the basis of inertial reference frames.

IIVisualization 1 — Zero Net Force ⟺ Constant Velocity

A body keeps its velocity exactly when the forces on it cancel: balance the push and the drag and the equal-time snapshots stay evenly spaced; unbalance them and the spacing spreads.

Applied push

F_{\text{push}}

2 Resistance

F_{\text{drag}}

2 Time 0.0s

💡 Both directions of the biconditional: even spacing (constant velocity) forces the arrows to cancel, and cancelling arrows force even spacing. Neither can hold without the other.

IIIVisualization 2 — Every Change in Velocity Needs a Force

Velocity carries a direction as well as a magnitude, so it changes if the speed changes, the direction changes, or both — and each change is a $\Delta\vec{v}$ that only a net force can produce.

💡 Turning counts: drag the tip around the circle $|\vec{v}| = 4$ — the speed never changes, yet $\Delta\vec{v}$ is nonzero. Rounding a corner at steady speed still demands a force.

IVVisualization 3 — Uniform Motion Is Indistinguishable from Rest

The same journey is described equally well from the platform or from the train; because the net force on the passenger is zero either way, no observation can decide who is "really" moving.

Reference frame Time 0.0s

💡 Both frames are inertial: each moves at constant velocity relative to the other, so the laws of motion look identical in both. That is exactly what "physically indistinguishable" means.

VQuiz Questions

Problem 1 · Where Is the Net Force Zero? (Basic)

Given: Newton's First Law says the net force is zero exactly when a body moves with constant velocity. In which case is the net force zero?

✅ Correct! Constant velocity in a straight line means zero net force — rest and uniform motion are the same mechanical state.

❌ Constant speed, not constant velocity. In orbit the direction changes every instant, so gravity is a nonzero net force continually turning the satellite.

❌ Not quite. At the peak the ball is momentarily at rest, but its velocity is still changing — gravity acts the whole time, so the net force is not zero.

❌ Not quite. A body that is speeding up has a changing velocity, so the net force on it cannot be zero.

Show solution

Zero net force is equivalent to constant velocity — unchanging speed and direction.

Speeding up: $|\vec{v}|$ changes ⟹ $\sum\vec{F} \ne 0$ .
Orbiting at constant speed: direction changes ⟹ $\sum\vec{F} \ne 0$ (gravity).
Sliding at constant velocity on frictionless ice: nothing changes ⟹ $\sum\vec{F} = 0$ . ✓
Top of the arc: velocity is passing through zero but its rate of change (gravity) is undiminished ⟹ $\sum\vec{F} \ne 0$ .

Only the hockey puck has $\sum\vec{F} = 0$ .

Problem 2 · Constant Speed Around a Curve (Common Pitfall)

Given: A car drives around a circular track at a constant speed of $20\,\text{m/s}$ . Is there a net force on it?

✅ Correct! Even at constant speed, a changing direction is a changing velocity, so a net (centripetal) force is required.

❌ Speed is constant, but velocity is not. Velocity includes direction, and the direction changes every instant around the curve.

❌ Not quite. Turning the corner changes the direction of

\vec{v}

, so there must be a nonzero net force pointing toward the center.

Show solution

Write velocity as $\vec{v} = |\vec{v}|\,\hat{u}$ . The magnitude $|\vec{v}| = 20\,\text{m/s}$ is fixed, but the direction $\hat{u}$ rotates continuously.

A rotating $\hat{u}$ means $\Delta\vec{v} \ne 0$ , and by the First Law $\Delta\vec{v} \ne 0 \iff \sum\vec{F} \ne 0$ .

The net force points toward the center of the circle (centripetal). This is why the First Law is "really a law about acceleration in disguise."

Problem 3 · Reading Off the Balance (Multi-step)

Given: You push a box across a level floor at constant velocity with a steady horizontal force of $30\,\text{N}$ . What is the force of friction on the box?

✅ Correct! Constant velocity means zero net force, so friction must exactly cancel your

30\,\text{N}

push.

❌ Not quite. With no friction, your

30\,\text{N}

push would be unbalanced and the box would accelerate. Constant velocity requires the forces to cancel.

❌ Right size, wrong direction. Friction opposes the sliding, so it points backward, against your push.

❌ Motion needs no net force to continue. At constant velocity the net force is exactly zero, so friction equals — not exceeds — the push.

❌ Not quite. Set the net force to zero: friction balances the

30\,\text{N}

push exactly.

Show solution

Step 1 — Apply the First Law. Constant velocity ⟹ $\sum\vec{F} = 0$ .

Step 2 — Sum the horizontal forces. Push (forward) plus friction (unknown) must total zero:

F_{\text{push}} + F_{\text{fric}} = 0 \;\Rightarrow\; F_{\text{fric}} = -30\,\text{N}

Step 3 — Interpret. The minus sign means friction is $30\,\text{N}$ pointing opposite to the push. There is no leftover "net force keeping it moving" — the box already has velocity, and nothing is removing it.

Problem 4 · Inside the Moving Train (Transfer)

Given: A train cruises in a straight line at a constant $30\,\text{m/s}$ . A passenger holds a coin at arm's length and releases it. Relative to the train, the coin lands:

✅ Correct! The train is an inertial frame, so inside it physics is identical to rest — the coin shares the train's horizontal velocity and falls straight down.

❌ Not quite. Nothing removes the coin's horizontal velocity when you let go; it keeps moving forward with the train and lands directly below your hand.

❌ No speed is special. Every constant velocity is an inertial frame, so the outcome is the same at

30\,\text{m/s}

as at rest.

❌ Not quite. In the train's frame the coin begins with no horizontal motion relative to you, so it falls straight down.

Show solution

Before release, the coin travels with the passenger at $30\,\text{m/s}$ — this is uniform motion, so the coin's horizontal velocity is unchanged by its own inertia.

Releasing it removes nothing horizontal; only gravity now acts, straight down. In the train's frame the coin starts from rest horizontally and falls straight to the floor, landing beneath the release point.

Because uniform motion is indistinguishable from rest, the result is identical to dropping the coin in a stationary train — and it does not matter whether the speed is $30\,\text{m/s}$ or any other constant value. This is the meaning of an inertial reference frame.

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