Newton's second law — F = ma (net force), impulse-momentum form: Question bank

True or false — justify

A constant velocity means zero net force acts on the object.

True. Constant velocity ⇒

a = 0

⇒

F_{net} = m a = 0

. Individual forces may still exist (they just cancel), but their vector sum is zero.

If the net force on a body is zero, its momentum cannot change.

True.

F_{net} = \frac{d p}{d t} = 0

means

p

is constant. This is exactly Newton's First Law (Inertia) living inside the Second.

A body can have zero velocity yet nonzero acceleration at that instant.

True. At the top of a thrown ball's flight

v = 0

but gravity still acts, so

a = g

downward — velocity is about to change even though it's momentarily zero.

Two objects with equal momentum always have equal kinetic energy.

False.

p = m v

but

K E = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}

. For fixed

p

, a lighter object carries more kinetic energy — see Work-Energy Theorem.

Doubling an object's speed doubles its momentum but also doubles its kinetic energy.

False on the second half. Momentum

m v

doubles, but

K E = \frac{1}{2} m v^{2}

quadruples because it depends on

v^{2}

Impulse and force have the same units.

False. Force is N; impulse is N·s = kg·m/s. Impulse is force accumulated over time, a fundamentally different quantity.

A large force always produces a large impulse.

False. Impulse is

F Δ t

. A large force acting for a vanishingly short time can give a tiny impulse; a gentle force over hours can give a huge one.

In the constant-mass case, the momentum form and

F = ma

give identical predictions.

True. With

\frac{d m}{d t} = 0

the product rule kills the

v \frac{d m}{d t}

term, leaving

F = m a

exactly. They only diverge when mass changes.

Heavier objects fall faster in a vacuum because gravity pulls them harder.

False. Weight

m g

is larger, but so is inertia

m

;

a = m g / m = g

cancels the mass. All masses accelerate at

g

A force pointing perpendicular to the velocity does no work but still changes momentum.

True. Momentum is a vector; a sideways force changes its direction (hence

Δ p \neq = 0

) while doing zero work since it's perpendicular to motion.

Spot the error

"The car keeps moving, so there must be a forward force pushing it."

Error: confuses "moving" with "accelerating". At constant speed the engine's push only cancels drag/friction; net force is zero. Motion needs no force to continue (First Law) — only to change.

F = ma

, so I can compute the thrust on a rocket by mass times acceleration."

Error: a rocket loses mass, so

\frac{d m}{d t} \neq = 0

. You must use

F = \frac{d ( m v )}{d t} = m a + v \frac{d m}{d t}

; the discarded term is the whole point (see Variable Mass Systems & Rocket Equation).

"The ball went from

20

m/s to

30

m/s the other way, so

Δ p = 10 m

Error: momentum is a vector. Taking initial direction as

+

Δ p = m (- 30 - 20) = - 50 m

, not

10 m

. The reversal must carry a sign.

"Friction is a force, so a block with only friction on it must accelerate forward."

Error: friction opposes motion, so it decelerates the block; the net force points backward, giving negative acceleration. See Friction.

"Airbags reduce the driver's change in momentum."

Error:

Δ p

is fixed once initial and final speeds are set. The airbag increases the time

Δ t

, so

F = Δ p /Δ t

shrinks — momentum change is unchanged, force is not.

"Since impulse equals

Δ p

, an object with more momentum has more impulse on it."

Error: impulse is the change in momentum during an interval, not the momentum itself. A fast object cruising at constant velocity has huge

p

but zero impulse.

"Newton's second law says force causes velocity."

Error: force causes acceleration (rate of change of velocity/momentum), not velocity itself. A body already has velocity without any force acting.

Why questions

Why is the momentum form

\frac{d p}{d t}

considered "more fundamental" than

m a

Because it stays valid when mass changes (rockets, raindrops, conveyor belts).

m a

is a special case that silently assumes

\frac{d m}{d t} = 0

Why do we specify "net" force rather than just "force" in

F_{net} = ma

Because acceleration responds only to the vector sum of all forces. Ten forces that cancel produce zero acceleration despite each being real.

Why does bending your knees when landing reduce the force on your legs?

It stretches the stopping time

Δ t

. Since

Δ p

is fixed by your landing speed, a bigger

Δ t

makes

F = Δ p /Δ t

smaller.

Why can two very different force-time graphs produce the identical velocity change?

Velocity change depends only on impulse

= \int F d t

= the area under the force–time curve. Different shapes with equal area give equal

Δ p

Why does Newton's Third Law not appear in the equation

F_{net} = ma

Because

F_{net}

counts only forces acting on this one body. The Third-Law reaction acts on the other body, so it never enters this object's equation — see Newton's Third Law.

Why does conservation of momentum follow directly from the Second Law?

F_{net} = 0

then

\frac{d p}{d t} = 0

, so

p

is constant. For an isolated system internal forces cancel in pairs — see Conservation of Linear Momentum.

Why is impulse the time-integral of force while the work–energy theorem uses the space-integral?

Impulse

\int F d t

tracks how force accumulates over time (changing momentum); work

\int F d x

tracks accumulation over distance (changing energy). Same force, different accountant — see Work-Energy Theorem.

Edge cases

A raindrop grows as it falls with no horizontal force; does its horizontal momentum stay constant even as it slows?

Yes. With

F_{x} = 0

\frac{d ( m v _{x} )}{d t} = 0

, so

m v_{x}

is conserved. As

m

grows,

v_{x}

must drop —

ma

alone would wrongly predict constant

v_{x}

If mass is zero, what does

F = ma

predict, and is it physical?

It would demand infinite

a

for any finite

F

(or

F = 0

). Massless bodies aren't described by this law; it's a degenerate case outside its domain.

For an instantaneous, ideal collision (

Δ t \to 0

) with finite momentum change, what happens to the force?

The average force

F = Δ p /Δ t \to \infty

. Real contacts always take nonzero time, so the "instantaneous" force is an idealisation — see Collisions and Elasticity.

A ball bounces off a wall with the same speed it arrived; is its momentum change zero because speed is unchanged?

No. Momentum is a vector: it reverses direction, so

Δ p = 2 m v

in magnitude. Only the speed is unchanged, not the momentum.

An object experiences a force that is always exactly perpendicular to its velocity; what is its long-term speed and momentum behaviour?

Speed stays constant (zero work), but momentum continually changes direction — e.g. uniform circular motion. Constant

∣ p ∣

, ever-changing

p

Over a full oscillation a spring pushes a block back and forth symmetrically; what is the net impulse over one complete period?

Zero. The block returns to its starting velocity, so

Δ p = 0

, hence total impulse (area under

F

–

t

) is zero — positive and negative halves cancel.

Recall One-line survival kit

Ask three questions and most traps dissolve: (1) What is the net force on this body? (2) Is mass changing — do I need the $v \frac{d m}{d t}$ term? (3) Is momentum a vector here — did I keep the signs and directions?

Connections

Newton's Second Law — F = ma (net force), impulse-momentum form — parent topic.
Newton's First Law (Inertia) — the $F_{net} = 0$ traps.
Newton's Third Law — why the reaction force never enters this body's equation.
Conservation of Linear Momentum — the zero-net-force consequence.
Collisions and Elasticity — reversal and short- $Δ t$ edge cases.
Friction — the "forward force" and deceleration traps.
Variable Mass Systems & Rocket Equation — the changing-mass edge cases.
Work-Energy Theorem — the time- vs space-integral distinction.