Introduction
If you've ever tried to revise this topic the night before an exam, you probably know the frustration: one site explains distance and displacement, another explains momentum, a third talks about "fundamental forces" that turn out to be about gravity and nuclear physics instead of the force you actually need for your syllabus. None of them tell you how these ideas fit together.
They do fit together — tightly. Displacement gives you velocity. Velocity and mass give you momentum. Force is what changes momentum. And impulse is force acting over time, which is exactly what causes that change. Once you see the chain, none of these five terms feel like isolated definitions to memorize anymore — they're one continuous story about motion.
This guide walks through that story in order, with the formulas you need and worked examples close to what you'll actually see in an exam.
What Is Motion? Understanding Distance and Displacement First
Before force or momentum make sense, you need a clear handle on how we describe
how far something has moved — because that's where velocity comes from,
and velocity is the seed that momentum grows out of.
Distance is the total length of the path an object actually travels. It doesn't care about direction — it just adds up. If you walk to the shop and back, your distance is however long that round trip measured on the ground, full stop.
Displacement is different. It's the straight-line change in position from where you started to where you ended up, and it includes direction. Walk to the shop and back to your starting point, and your displacement is zero — even though you were clearly moving the whole time.
Here's a concrete example. Say you walk 3 km east, then 4 km north. Your distance is simple addition: 3 + 4 = 7 km. Your displacement is the straight line from your starting point to where you now stand, which you get using the Pythagorean theorem: √(3² + 4²) = 5 km, in a direction somewhere between east and north. Same walk, two different answers, because the two quantities are measuring two different things.
The classic exam trap uses this gap directly: a runner completes one full lap of a 400 m track and ends up exactly where they started. Distance covered: 400 m. Displacement: 0 m. If a question asks for "distance" and you answer with displacement (or vice versa), you'll get it wrong even though your arithmetic was fine — so read the question word carefully.
The only time distance and displacement come out equal is when the motion is a straight line in one direction, with no doubling back. Walk 5 m north in a straight line, and both your distance and displacement are 5 m north. The moment there's a turn or a return trip, the two numbers split apart.
Distance vs Displacement — Key Differences
Once you've got the concept, this table is what you'll actually use to answer direct comparison questions:
|
Property |
Distance |
Displacement |
|
Definition |
Total path length travelled |
Shortest straight-line change in position |
|
Quantity type |
Scalar (magnitude only) |
Vector (magnitude + direction) |
|
Symbol |
d (or s, depending on the text) |
s or Δx |
|
Can it be negative? |
No — always positive or zero |
Yes — sign shows direction |
|
Depends on path taken? |
Yes |
No — only on start and end points |
|
SI Unit |
metre (m) |
metre (m) |
Two quick examples to lock this in:
· Walk 4 m east, then 3 m north → distance = 7 m, displacement = 5 m (Pythagoras again, using the 3-4-5 triangle).
· Drive 10 km to a destination, then drive the same 10 km back home → distance = 20 km, displacement = 0 km.
If you remember nothing else from this
section, remember this: distance
only ever grows, displacement can shrink back to zero. That
single idea resolves most confusion.
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What Is Force?
Quick clarification before we go further, because it trips a lot of people up: if you've searched around this topic, you may have run into articles about the "four fundamental forces of nature" — gravity, electromagnetism, the strong nuclear force, and the weak nuclear force. That's real physics, but it's a different topic (particle physics and cosmology). It has nothing to do with the force that connects to momentum and impulse in your syllabus.
The force we care about here is the everyday, Newtonian kind: a push or a pull that changes an object's state of motion. Newton's First Law tells you an object stays at rest or keeps moving at constant velocity unless a force acts on it. Newton's Second Law tells you exactly what a force does when it does act — and this is the law that becomes the hinge for everything else in this article:
F = ma
Force equals mass times acceleration. But there's a second, more general way to write the same law, and it's the version that actually connects force to momentum:
F = dp/dt
Force is the rate at which momentum changes over time. When mass is constant, this simplifies right back down to F = ma — but writing it in terms of momentum is what lets you handle situations where mass isn't constant (like a rocket burning fuel), and it's the version you'll need for the impulse section coming up.
Force is a vector — it has both size and direction — measured in newtons (N), where 1 N is the force needed to accelerate a 1 kg mass at 1 m/s².
A
simple way to feel this: pushing a stalled car takes real, sustained force
because you're trying to change its momentum from zero to something. Kicking a
football is a force too, but a much larger one applied for a much shorter time
— which is a preview of the impulse idea below.
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Linear Momentum
Momentum is the physics term for "how much motion something has," and it depends on two things: how heavy the object is, and how fast it's going.
p = mv
Momentum equals mass times velocity. It's a vector (it points in the same direction as velocity), and its SI unit is kg·m/s.
Why does this matter as its own concept, separate from velocity? Because mass changes the picture completely. A cricket ball thrown fast and a truck rolling slowly can have comparable momentum, even though their speeds are wildly different — and that's exactly why a slow-moving truck is so much harder to stop than a fast-moving ball. Momentum, not speed alone, is what determines how hard something is to bring to rest.
Rewriting Newton's Second Law as F = dp/dt tells you something important: momentum only changes when a net external force acts on a system. Flip that around, and you get one of the most useful laws in this entire topic:
Law of Conservation of Linear Momentum
If the net external force on a system is zero, its total momentum stays constant. Momentum can be transferred between objects inside the system, but the total never changes.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(where u₁, u₂ are initial velocities and v₁, v₂ are final velocities of two objects in the system)
Worked example: A 4 kg trolley moving at 3 m/s collides with a stationary 2 kg trolley and they stick together after impact. What's their combined velocity?
Total momentum before collision = (4 × 3) + (2 × 0) = 12 kg·m/s Since they stick together, combined mass = 6 kg 12 = 6 × v → v = 2 m/s
No force outside the two-trolley system acted during the collision, so momentum in equals momentum out — that's the whole trick to solving these.
This is also why a rocket lifts off:
before ignition, total momentum of rocket + fuel is zero. As fuel is expelled
downward at high speed, it carries momentum with it — and for the total to stay
at zero, the rocket must gain equal and opposite momentum upward. Same logic
explains the recoil you feel firing a gun, or why a motorboat pushes water
backward to move itself forward.
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What Is Impulse?
This is the piece most revision material skips entirely — which is strange, because it's often the most directly testable idea in this whole unit, and it's genuinely useful outside an exam hall too.
Impulse is what you get when a force acts over a stretch of time:
J = FΔt
But here's the part that makes impulse worth learning as its own concept rather than just a formula: go back to F = dp/dt, multiply both sides by Δt, and you get:
J = Δp
Impulse equals the change in momentum it produces. That's the impulse-momentum theorem, and it's not a coincidence or an approximation — it falls directly out of Newton's Second Law. Impulse is a vector, measured in either kg·m/s or N·s (they're the same unit written two ways).
Why this matters practically: the same change in momentum can come from a huge force over a tiny time, or a small force spread over a longer time. This single idea explains a surprising number of everyday design choices:
· Airbags and crash mats don't reduce the change in momentum in a collision — that's fixed by how fast you were going and your mass. What they do is stretch out the time over which that change happens, which lowers the force your body experiences. Same impulse, smaller force, less injury.
· A boxer "rolling with the punch" is doing the same thing — pulling back extends the contact time and cuts the peak force.
· Follow-through in cricket or golf extends the time the bat or club is in contact with the ball, increasing the impulse delivered (and therefore the ball's final momentum) for the same swing force.
Worked example: A cricket bat exerts an average force of 500 N on a ball for 0.01 seconds during a shot. What impulse does the ball receive, and if the ball has a mass of 0.16 kg, what's its change in velocity?
J = FΔt = 500 × 0.01 = 5 N·s Since J = Δp = mΔv: 5 = 0.16 × Δv → Δv ≈ 31.25 m/s
That's the kind of number that makes "impulse" feel less abstract — it's the direct reason the ball goes flying.
Reading impulse off a graph: If you're given a force-time graph instead
of a single value, impulse is simply the area under that curve. This comes up
often in exam questions where the force isn't constant — instead of a clean
multiplication, you're calculating (or estimating) the area of the shape the
graph traces out.
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How Force, Momentum, Impulse, Distance and Displacement Connect
Here's the full chain, laid out in one place:
Displacement over time → gives you velocity → velocity with mass → gives you momentum → force is the rate momentum changes → force over time → gives you impulse → which equals the change in momentum.
It's a loop that starts and ends at momentum. Let's run one problem through the entire chain, the way an exam might actually combine these ideas.
Worked example: A 1000 kg car travels 100 m in 5 seconds while accelerating from rest, in a straight line. The brakes are then applied, bringing it to a stop over 2 seconds. Find (a) the car's velocity just before braking, (b) its momentum at that point, (c) the average braking force, and (d) the impulse delivered by the brakes.
(a) Velocity: Since it starts from rest and covers 100 m in 5 s with uniform acceleration, average velocity = displacement/time = 100/5 = 20 m/s. Since it starts at 0 and accelerates uniformly, final velocity just before braking = 2 × average velocity = 40 m/s.
(b) Momentum: p = mv = 1000 × 40 = 40,000 kg·m/s
(c) Braking force: The car goes from 40 m/s to 0 in 2 s, so deceleration a = 40/2 = 20 m/s². Force F = ma = 1000 × 20 = 20,000 N (the negative sign showing it opposes motion is usually implied by context).
(d) Impulse: J = Δp = 0 − 40,000 = −40,000 kg·m/s (the brakes remove all the car's momentum — and note this matches F × Δt = 20,000 × 2 = 40,000, confirming the two formulas agree).
Notice how the answer to each part fed into the next one. That's not a coincidence built for this example — it's how these five concepts actually behave together in any real motion problem.
One
more useful parallel while we're here: on a velocity-time graph, the area under
the curve gives you displacement. On a force-time graph, the area under the
curve gives you impulse. Same graphical trick, two different physical
quantities — worth remembering as a pair.
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Scalar vs Vector
A fast reference for every quantity covered above, worth bookmarking for last-minute revision:
|
Quantity |
Type |
Formula |
SI unit |
Can Be negative? |
|
Distance |
Scalar |
— |
metre (m) |
No |
|
Displacement |
Vector |
Δx |
metre (m) |
Yes |
|
Speed |
Scalar |
distance/time |
m/s |
No |
|
Velocity |
Vector |
displacement/time |
m/s |
Yes |
|
Force |
Vector |
F = ma |
newton (N) |
Yes |
|
Momentum |
Vector |
p = mv |
kg·m/s |
Yes |
|
Impulse |
Vector |
J = FΔt = Δp |
N·s or kg·m/s |
Yes |
MCQs
1.
Which of the
following is not
a vector quantity?
A. Momentum
B. Displacement
C. Torque
D. Speed
Answer: D — speed is commonly tested as the odd one out among vector quantities like momentum, displacement, and torque, since speed has no direction.
2. If the velocity of a body is doubled
while its mass stays constant, its momentum:
A. Remains the same
B. Doubles
C. Becomes half
D. Becomes four times
Answer: B — momentum doubles when velocity doubles at constant mass, directly from p = mv.
3. In the equation of motion 2as = v² −
u², the term "s" represents:
A. Speed
B. Displacement
C. Velocity
D. Acceleration
Answer: B — "s" in this kinematic equation stands for displacement, not distance (a frequently-tested distinction).
4. If a distance-time graph is a straight
inclined line, it represents:
A. Uniform speed
B. Non-uniform speed
C. Constant displacement
D. Non-uniform velocity
Answer: A — a straight inclined distance-time graph indicates uniform speed.
5. The area under a velocity-time graph,
bounded by the curve and two time ordinates, gives:
A. Acceleration
B. Displacement
C. Force
D. Momentum
Answer: B — this is a standard graph-interpretation question frequently asked in SSC previous papers, testing whether you know displacement = area under v-t graph (the direct parallel to impulse = area under F-t graph from this guide).
6. Newton's Second Law of Motion can be most generally
expressed as the rate of change of which quantity with time?
A. Velocity
B. Momentum
C. Displacement
D. Force
Answer: B — F = dp/dt; UPSC prelims tends to test this as a concept statement rather than a numerical.
7. The SI unit of impulse is the same
as the SI unit of:
A. Force
B. Work
C. Momentum
D. Power
Answer: C — impulse and momentum share the unit kg·m/s (equivalently N·s), since J = Δp.
8. The law of conservation of linear
momentum applies to a system when:
A. The system's mass
is constant
B. The net external force on the system is zero
C. All collisions are elastic
D. The system is at rest
Answer: B — this is the standard condition tested; note it does not require elastic collisions (a common distractor).
9. A body of mass 5 kg moving at 4 m/s
collides with a stationary body of mass 3 kg and they move together after
collision. Find their common velocity.
A. 1.5 m/s
B. 2.5 m/s
C. 3.5 m/s
D. 4 m/s
Answer: B — Total momentum = 5×4 = 20 kg·m/s; combined mass = 8 kg; v = 20/8 = 2.5 m/s.
10. Two bodies of different masses
have the same linear momentum. Which one has greater kinetic energy?
A. The heavier body
B. The lighter body
C. Both have equal kinetic energy
D. Cannot be determined
Answer: B — for the same momentum p,
KE = p²/2m, so the lighter body (smaller m) has greater kinetic energy.
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Conclusion:
None of these five ideas exist in isolation, no matter how separately they're taught. Displacement is what lets you calculate velocity. Velocity, paired with mass, becomes momentum. Force is simply the rate at which momentum changes — and impulse is what you get when you track that force over time, landing you right back at a change in momentum. Once that chain clicks, you stop memorizing five disconnected definitions and start seeing one continuous idea about how motion happens and how it changes.
That's also the mindset worth carrying into the exam hall. When a question mixes these concepts — a braking car, a collision, a bat hitting a ball — don't look for which single formula fits. Look for where you are in the chain, and work outward from there: displacement to velocity, velocity to momentum, force to impulse, impulse back to momentum change. Most "hard" problems in this unit are really just two or three of these easy steps stacked together.
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