Internal vs External Forces: The Difference

A force is internal if both objects in the interaction are inside the system you chose, and external if one of them is outside it. Internal forces cancel in third law pairs, so they cannot change the system's total momentum or move its center of mass. Only external forces can.

AP Physics: Unit 2 (topics 2.1 Systems and Center of Mass, 2.2 Forces and Free-Body Diagrams, 2.3 Newton's Third Law, 4.3 Conservation of Linear Momentum). AP Physics 1 introduces the phrase internal forces once, parenthetically, in EK 2.3.A.2: interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass. The framework otherwise carries the idea through the phrase net external force, in EK 2.5.A.3, EK 4.2.A.1, EK 4.2.A.4, EK 4.2.A.5, EK 4.3.A.1.ii, EK 4.3.B.2 and EK 4.3.B.3, and through net external torque in EK 6.4.B.2 and EK 6.4.B.3. EK 2.2.A.1.ii states that an object or system cannot exert a net force on itself, and EK 2.2.B.2 restricts a free-body diagram to the forces exerted on the object by the environment. EK 2.3.A.1 supplies the third law pair, EK 4.3.A.3.i its impulse form, and EK 4.1.A.3.i and 4.1.A.3.iii the collision and explosion models. EK 2.3.A.3 and its subpoints define tension as the net result of forces that segments of a string exert on each other. Unit 2 carries 18 to 23 percent of the AP Physics 1 multiple-choice section and Unit 4 carries 10 to 15 percent.

The distinction, and why it is not a property of the force

Draw a boundary. Every force acting anywhere in the picture now falls into one of two bins.

Internal. Both objects in the interaction are inside the boundary. The push, the pull, the friction, the tension: both the thing exerting it and the thing feeling it are yours.

External. One of the two objects is outside the boundary. The interaction crosses the line.

Nothing in that classification is about the force. It is about the line. The same tension in the same rope between the same two blocks is internal if you enclosed both blocks and external if you enclosed one, and the rope did not notice. This is the point people skip, and it is why the question "is friction an internal or external force?" has no answer until you say what the system is.

AP Physics 1 introduces the term once, parenthetically, and then draws the only consequence it needs. EK 2.3.A.2: interactions between objects within a system, the internal forces, do not influence the motion of a system's center of mass. That single line is what the rest of this page unpacks, because it is doing more work than its length suggests.

The boundary itself is chosen the way object vs system describes, and the glossary covers each side at external force and system. This page is about the difference the classification makes to your answer.

Internal vs external, side by side

	Internal force	External force
Where the two interacting objects are	Both inside your system	One inside, one outside
Is it a property of the force	No. It is a property of your boundary	No. Same
Appears on the system's free-body diagram	Never. EK 2.2.B.2 says the diagram shows forces exerted on the object by the environment	Yes, every one
Both members of the third law pair are inside	Yes, which is why they cancel	No, one partner is outside your sum
Can change the system's total momentum	No	Yes, and it is the only thing that can
Can move the system's center of mass	No, per EK 2.3.A.2	Yes, per EK 2.5.A.3
Can change the system's total kinetic energy	Yes, and this is the exception people miss	Yes
Enters $\sum \vec F$ for the system	No	Yes
Enters $\sum \vec F$ for one part of the system	Yes, if the other part is outside the part's own boundary	Yes
Example, system = two carts	The spring shove between them	Friction from the track
Example, system = one cart	Nothing. A single object has no interior	The spring shove and the friction both

The two rows about momentum and kinetic energy contradict each other in a way that is not a mistake. Internal forces cannot change one and can change the other, and section five is entirely about why.

The case that separates them: you cannot jump by pushing yourself

Stand up and jump. What lifted you?

The honest answer is not your legs. Take yourself as the system, and your muscles, tendons and bones are all inside the boundary, so every force they exert is internal. EK 2.2.A.1.ii forbids the shortcut in one line: an object or system cannot exert a net force on itself. If muscles were enough, a person sealed in a box in deep space could launch themselves across it, and they cannot.

What lifted you is the ground. Your feet push down on it, and by Newton's third law it pushes up on you with an equal force, and that upward normal force is external to you, because the ground is not part of you. The muscles are still necessary. Their job is to press your feet into the floor hard enough that the floor's answering push exceeds your weight. They set up the external force rather than supplying it.

Now move the boundary and watch the accounting change without the physics changing.

System	The muscular effort is	The normal force from the ground is	Total momentum during the push
The person	Internal	External	Increases upward
The person and Earth	Internal	Internal	Constant, at zero

In the second row nothing accelerates the system as a whole, and it is still true that you leave the ground: you go up and Earth goes down, by an amount too small to measure and not by an amount equal to zero. The center of mass of the person and Earth together does not move, exactly as EK 2.3.A.2 requires, because the only forces in play are now internal.

Both rows are correct descriptions of the same jump. The first is useful, because it has one unknown and you can solve it. The second is true and unhelpful. That is usually how the choice goes.

Why internal forces cannot change the total momentum

The argument is three lines long and worth being able to reproduce, because free-response questions ask for it.

Every internal force is one half of a Newton's third law pair, and EK 2.3.A.1 gives that pair as $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$ .
If both A and B are inside the system, both members of that pair appear in the system's force sum, and they add to zero. Not approximately: exactly, at every instant, whatever their size.
So $\sum \vec{F}$ over the whole system contains only external contributions, and EK 4.2.A.1 turns that into the momentum statement: $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t}$ , where the net force is the net external force exerted on the object or system.

Set the net external force to zero and the total momentum cannot change, which is EK 4.3.B.2. Make it nonzero and momentum is transferred between the system and the environment, which is EK 4.3.B.3.

The impulse version is worth having alongside it, because it is the version that survives inside a collision. EK 4.3.A.3.i: the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and this is a direct result of Newton's third law. Equal and opposite impulses on two members of the same system means the two momentum changes cancel, which is EK 4.3.A.3 restated: any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system.

This is also what licenses the collision model on a real bench. EK 4.1.A.3.i defines a collision as an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction. Gravity and bench friction are external and they are still there; they are simply too small, over a few milliseconds, to deliver a measurable impulse. And EK 4.1.A.3.iii names the reverse case: an explosion is a model for an interaction in which forces internal to the system move objects within that system apart. An explosion moves the pieces without moving the whole.

Internal forces cannot change momentum, but they can change kinetic energy

This is the asymmetry that catches good students, and the reason for it is precise rather than hand-waving.

Momentum change is force multiplied by time. The two members of a third law pair act for the same duration by definition, and they are equal and opposite, so their impulses cancel exactly. Nothing about the geometry can rescue them.

Energy change is force multiplied by displacement. The two members of the pair are still equal and opposite, but they act at two different places on two different objects, and those two places do not have to move the same way. When they do not, the two amounts of work do not cancel.

So the test for an internal force pair is: do the two points of application have the same displacement along the line of the force?

Internal pair	Do the two ends move the same way	Net work on the system	Effect on total kinetic energy
Ideal rope between two blocks	Yes, the rope cannot stretch, so both ends move the same distance	Zero. Positive work on one block, equal negative work on the other	None
Rigid contact between two blocks pushed as a unit	Yes, they move together	Zero	None
Compressed spring pushing two carts apart	No, the ends move in opposite directions and the spring pushes each one outward	Positive on both carts	Increases it
Friction between two surfaces sliding across each other	No, they slide relative to one another	Negative overall	Decreases it, into thermal energy

The two middle rows are why a spring explosion is possible at all. Total momentum stays at zero while total kinetic energy climbs from zero to whatever the spring had stored, and no external force did any of it. The energy did not come from nowhere: it came from elastic potential energy that was already inside the system, so energy is still conserved, exactly as EK 3.4.C.1 requires. See constant vs conserved for the distinction that keeps that sentence honest.

The fourth row is the reason mechanical energy conservation has an extra condition that momentum conservation does not. EK 4.3.B.2 asks only about external forces. EK 3.4.C.2 asks about external work and about nonconservative interactions inside the system, and internal friction is exactly the case it is guarding against.

Redraw the boundary and you have a different problem, not a different physics

Take one situation and three boundaries. A 4.0 kg block sits on a frictionless table, joined by an ideal rope over an ideal pulley to a 2.0 kg block hanging off the edge. Earth is outside every system below.

System	External horizontal or driving forces	Internal forces	What you can get out of it
Both blocks and the rope	Gravity on the hanging block, 19.6 N	The rope tension, both ends	The acceleration, in one line
The 4.0 kg block alone	The rope tension	None	The tension, once you know the acceleration
The 2.0 kg block alone	Gravity and the rope tension	None	The tension, as a cross-check

Row one gives $a = (19.6\ \text{N})/(6.0\ \text{kg}) = 3.27\ \text{m/s}^2$ with no tension anywhere in the algebra. Row two gives $T = (4.0\ \text{kg})(3.27\ \text{m/s}^2) = 13.1\ \text{N}$ . Row three checks it. The full example is worked below, and how to find tension owns the general routine.

What moved between the rows is which forces you had to know about. The tension is a real force with a real value of 13.1 N in all three rows. In row one it contributes nothing to the sum, because both of its ends are inside, and that is precisely why row one is short.

One CED detail makes the rope case work. EK 2.3.A.3 defines tension as the macroscopic net result of forces that segments of a string, cable, chain, or similar system exert on each other in response to an external force. Read that as a statement about internal forces: the string's own segments pulling on each other. EK 2.3.A.3.i and 2.3.A.3.ii add the idealization you are allowed to assume, that an ideal string has negligible mass and does not stretch, and that the tension in an ideal string is the same at all points within it. EK 2.3.A.3.iii warns that in a string with nonnegligible mass, tension may not be the same at all points, and the Topic 2.3 boundary statement says AP Physics 1 only expects that treated qualitatively, with a hanging chain having greater tension toward the top as its example.

Where it costs a mark

Putting an internal force in the system's free-body diagram. EK 2.2.B.2 says the free-body diagram of an object or system shows each of the forces exerted on the object by the environment. The environment is what is outside. An arrow for the shove between two objects you enclosed does not belong on it.
Adding the tension to the net force on a two-block system. It cancels. Including it once gives an acceleration that is wrong by the whole tension divided by the total mass.
Explaining a jump, a car accelerating or a person walking with an internal force. The engine, the muscles and the tyres are all inside. What accelerates you is the external friction or normal force from the ground, and a justification that never names an external force has not explained the acceleration. EK 2.2.A.1.ii is the sentence to cite.
Claiming internal forces cannot change the kinetic energy. They can, and a spring explosion is the counterexample. What they cannot change is the total momentum. Two different statements, two different quantities.
Asserting momentum conservation without checking for external forces. Friction on the track, an unbalanced gravitational component along an incline and a hand still touching a cart are all external. EK 4.1.A.3.i lets you neglect small external forces during a collision because the interaction forces are much larger, but that is an argument you have to make rather than an assumption you get for free.
Deciding that a force is internal because it is inside the object. The interatomic forces inside a block are internal to the block and irrelevant. The friction between that block and the floor is not internal to the block, because the floor is a different object.
Forgetting that the classification can change midway through a solution. If part (a) took both carts and part (b) takes one, the same force changes bin. Say so, or the marker sees a contradiction.

What the CED actually says, and how the exam frames it

The phrase internal forces appears in the AP Physics 1 framework once, in EK 2.3.A.2, and it appears there in brackets: interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass. The other three Course and Exam Descriptions do not use the phrase at all. The idea is nevertheless load bearing across three units, because the framework carries it in the mirror-image phrase instead, net external force, which turns up throughout Units 2 and 4:

EK 2.4.A.1: the net force on a system is the vector sum of all forces exerted on the system.
EK 2.5.A.3: the velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.
EK 4.2.A.1: the rate of change of momentum is equal to the net external force exerted on an object or system.
EK 4.2.A.4: the impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
EK 4.2.A.5: the net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.
EK 4.3.A.1.ii: the velocity of a system's center of mass is constant in the absence of a net external force.

Every one of those says external, and none of them would be true with the word removed. Unit 6 extends the same structure to rotation with net external torque, in EK 6.4.B.2 and EK 6.4.B.3.

The equation sheet is silent on the distinction, as it has to be: $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ and $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ both print a sum over forces, and which forces belong in that sum is a decision you make before you write the line. The full sheet is here.

One sample instructional activity in the Unit 2 guide is built on this exact move. Activity 1, under Topic 2.2, has students consider an accelerating two-object system from everyday life, such as a person pushing a shopping cart or a car pulling a trailer, then draw the forces on one object, then on the other, then the external forces exerted on the two-object system. Three diagrams of one situation, which is the whole skill. Unit 2 carries 18 to 23 percent of the AP Physics 1 multiple-choice section, and Unit 4 carries 10 to 15 percent. Topic 2.3 Newton's Third Law is where the CED states the internal-force consequence.

A spring explosion: internal forces change the kinetic energy and leave the momentum alone

Cart A, 0.50 kg, and cart B, 1.5 kg, are held together on a level frictionless track with a compressed spring between them storing 3.0 J of elastic potential energy. The carts are released from rest and the spring falls away. Find each cart's final speed, then state what happened to the system's total momentum and its total kinetic energy.

Declare the convention: positive is to the right, cart B goes right and cart A goes left. Take both carts and the spring as the system.
Classify the forces. The spring pushes A one way and B the other, and both carts are inside the boundary, so the spring force is internal. The track is frictionless and level, so the only external forces are gravity and the normal force, which are vertical and cancel. The net external horizontal force is zero.
Apply EK 4.3.B.2. Zero net external force means the total momentum is constant, and everything started at rest, so the total momentum is zero before and zero after: $m_A v_A = m_B v_B$ , giving $(0.50)v_A = (1.5)v_B$ and therefore $v_A = 3.0\, v_B$ .
Now use energy. Total energy is conserved and the 3.0 J of stored elastic potential energy ends up as kinetic energy of the two carts: $\frac{1}{2}(0.50)v_A^2 + \frac{1}{2}(1.5)v_B^2 = 3.0\ \text{J}$ .
Substitute $v_A = 3.0 v_B$ : $\frac{1}{2}(0.50)(9.0 v_B^2) + \frac{1}{2}(1.5)v_B^2 = 2.25 v_B^2 + 0.75 v_B^2 = 3.0 v_B^2$ .
Solve: $3.0 v_B^2 = 3.0$ , so $v_B = 1.0\ \text{m/s}$ to the right and $v_A = 3.0\ \text{m/s}$ to the left.
Check the momentum: $(0.50\ \text{kg})(3.0\ \text{m/s}) = 1.5\ \text{kg}\cdot\text{m/s}$ leftward against $(1.5\ \text{kg})(1.0\ \text{m/s}) = 1.5\ \text{kg}\cdot\text{m/s}$ rightward. They cancel, so the total is still zero.
Check the energy: $\frac{1}{2}(0.50)(3.0)^2 = 2.25\ \text{J}$ and $\frac{1}{2}(1.5)(1.0)^2 = 0.75\ \text{J}$ , summing to 3.0 J, which is what the spring held.

Cart A leaves at 3.0 m/s left, cart B at 1.0 m/s right. Total momentum went from zero to zero, because the spring force was internal and EK 2.3.A.2 forbids internal forces from moving the center of mass. Total kinetic energy went from 0 J to 3.0 J, because the two ends of that same spring moved in opposite directions, so the two equal and opposite forces did positive work on both carts instead of canceling. This is EK 4.1.A.3.iii in numbers: an explosion is an interaction in which forces internal to the system move objects within that system apart.

A jump: the muscles are internal, so something outside you has to push

A 65 kg person crouches and then extends, pushing on the ground so that the ground exerts a constant upward normal force of 1150 N on them for 0.25 s. Find the net force on the person during the push, their speed at the moment they leave the ground, and the height they reach. Then explain what would happen if the muscular force were the only force considered.

Declare the convention: positive is upward. Take the person as the system, so Earth and the ground are both outside it.
Classify the forces on the person. Every muscular force is internal, because both the muscle and the bone it pulls on are inside the boundary. The two external forces are the person's weight and the normal force from the ground.
Weight: $F_g = mg = (65\ \text{kg})(9.8\ \text{m/s}^2) = 637\ \text{N}$ downward.
Net force during the push: $F_{\text{net}} = 1150\ \text{N} - 637\ \text{N} = 513\ \text{N}$ upward.
Impulse and takeoff speed: $J = F_{\text{net}}\Delta t = (513\ \text{N})(0.25\ \text{s}) = 128.25\ \text{N}\cdot\text{s}$ , so $v = J/m = (128.25)/(65) = 1.97\ \text{m/s}$ upward.
Height, using $v^2 = v_0^2 + 2a\Delta y$ with $v = 0$ at the top and $a = -9.8\ \text{m/s}^2$ : $\Delta y = v_0^2/(2g) = (1.97)^2/(19.6) = 3.893/19.6 = 0.199\ \text{m}$ , about 20 cm.
Now the counterfactual. Delete the ground and keep the muscles, and the person is a system exerting forces only on itself. EK 2.2.A.1.ii says an object or system cannot exert a net force on itself, so the net force is zero, the momentum does not change, and the person does not move. The extension of the legs would still happen; the center of mass would not go anywhere.
Cross-check by enlarging the system. Take the person and Earth together and the normal force becomes internal too, so the total momentum stays at zero. The person's 128 kg m/s upward is matched by 128 kg m/s downward given to Earth, whose mass makes the resulting speed far too small to detect. Nothing is inconsistent: momentum is constant for the larger system and increases for the smaller one, because the smaller one has an external force acting on it.

Net force 513 N upward, takeoff speed 1.97 m/s, height 0.199 m. The muscles never appear in any of those lines. They are internal to the person, so the entire 513 N came from the external normal force and the external weight. What the muscles do is set the size of the normal force by pressing the feet into the ground, which is a completely different job from supplying the net force, and confusing the two is the standard error in the justification part of this question.

Two blocks and a rope: the same tension is internal in one system and external in another

A 4.0 kg block rests on a frictionless horizontal table, connected by a light inextensible rope over an ideal pulley to a 2.0 kg block hanging beside the table. The system is released from rest. Find the acceleration and the rope tension, first by treating the two blocks as one system and then by treating each block separately. Take Earth as outside the system.

Declare the convention: positive is the direction of motion, so the 4.0 kg block moves toward the pulley and the 2.0 kg block moves down. The rope is inextensible, so both blocks have the same speed and the same magnitude of acceleration at every instant.
System 1, both blocks. The tension pulls each block, and both blocks are inside, so the tension is internal and cancels. The table is frictionless, so the horizontal external force on the 4.0 kg block is zero. The one external force that drives the motion is gravity on the hanging block: $F_g = (2.0\ \text{kg})(9.8\ \text{m/s}^2) = 19.6\ \text{N}$ .
Apply the second law to the whole system, total mass 6.0 kg: $a = (19.6\ \text{N})/(6.0\ \text{kg}) = 3.2667\ \text{m/s}^2$ , which rounds to $3.27\ \text{m/s}^2$ .
System 2, the 4.0 kg block alone. Now the rope is attached to something outside the boundary, so the tension is external and it is the only horizontal force: $T = m a = (4.0\ \text{kg})(3.2667\ \text{m/s}^2) = 13.07\ \text{N}$ , which rounds to 13.1 N.
System 3, the 2.0 kg block alone, as an independent check. Two external forces: gravity down and tension up. $19.6\ \text{N} - T = (2.0\ \text{kg})(3.2667\ \text{m/s}^2) = 6.53\ \text{N}$ , so $T = 19.6 - 6.53 = 13.07\ \text{N}$ . The two routes agree to three figures.
Confirm the tension does no net work on the whole system. Over a displacement $d$ the rope does $+Td$ on the 4.0 kg block, pulling it forward, and $-Td$ on the hanging block, pulling it up while it moves down. The rope cannot stretch, so the two displacements have the same magnitude and the two amounts of work cancel exactly.
Sanity check the magnitude of the answer. If the table block had zero mass the acceleration would be the full 9.8 m/s squared, and if it were enormously heavy the acceleration would approach zero. 3.27 m/s squared for a two-to-one mass ratio sits sensibly between them, and it equals $g$ times $2.0/6.0$ , which is $9.8/3 = 3.2667$ .

Acceleration 3.27 m/s squared, tension 13.1 N. The same 13.1 N was internal in the first system and external in the second, and the classification changed because the boundary moved, not because anything happened to the rope. This is also why the whole-system route is the fast one: it deliberately throws away the tension, and you pay that back with a second diagram when the tension is what the question wants. The ideal string is what makes it clean, since EK 2.3.A.3.ii gives the tension as the same at all points within it.

Frequently asked questions

What is the difference between an internal and an external force?

A force is internal if both objects in the interaction are inside the system you chose, and external if one of them is outside. The classification belongs to your boundary, not to the force: the tension in a rope joining two blocks is internal when you enclose both blocks and external when you enclose one, and nothing about the rope changed. The reason it matters is that internal forces come in Newton's third law pairs with both members inside, so they add to zero in the system's force sum. AP Physics 1 EK 2.3.A.2 states the consequence directly: interactions between objects within a system, the internal forces, do not influence the motion of a system's center of mass.

Why do internal forces not change a system's momentum?

Because they cancel in pairs. Newton's third law gives every internal force a partner that is equal in magnitude and opposite in direction, and if both objects are inside the system, both members of the pair appear in the sum and add to zero. That leaves only external contributions, so EK 4.2.A.1 can say that the rate of change of momentum equals the net external force. The impulse version is the same argument in the collision setting: EK 4.3.A.3.i says the impulse one object exerts on a second is equal and opposite to the impulse the second exerts on the first, so the two momentum changes inside a system always cancel. One object's momentum changes a great deal; the system's total does not change at all.

Can an internal force change a system's kinetic energy?

Yes, and this is the exception worth remembering. Momentum change depends on force times time, and the two members of a third law pair act for the same time, so their impulses cancel exactly. Energy change depends on force times displacement, and the two members act at different places, so if those places move differently the two amounts of work do not cancel. A compressed spring between two carts pushes each of them outward, so it does positive work on both, and the system's kinetic energy rises from zero while its total momentum stays at zero. An ideal rope is the opposite case: its two ends move the same distance, so the positive work on one block cancels the negative work on the other.

Is friction an internal or external force?

It depends entirely on the system, and the question has no answer until the system is named. Friction between a block and the floor is external if your system is the block alone, and internal if your system is the block, the floor and Earth. That choice changes what you can claim: with friction external, momentum is transferred out of the system as the block slows, and with friction internal, the system's total momentum is constant while its mechanical energy still falls. The energy behaves that way because EK 3.4.C.2 requires both zero external work and no nonconservative interactions inside the system before it will call mechanical energy constant, and internal friction breaks the second condition.

Why can you not push yourself forward with your own muscles?

Because your muscles are internal to you, and AP Physics 1 EK 2.2.A.1.ii states that an object or system cannot exert a net force on itself. Every force a muscle exerts on a bone has an equal and opposite partner inside the same body, so the pair adds to zero and your center of mass cannot accelerate. Walking, running and jumping all work by using muscles to press against something external, usually the ground, so that the ground's answering push does the accelerating. The muscular effort is essential and it is not the net force. This is exactly why a person floating in the middle of a frictionless surface with nothing to push against cannot get moving.

Do internal forces appear on a free-body diagram?

Not on the diagram for the system that contains both of their objects. AP Physics 1 EK 2.2.B.2 says the free-body diagram of an object or system shows each of the forces exerted on the object by the environment, and the environment means whatever lies outside your boundary. So a diagram of two blocks treated as one system carries no arrow for the push between them. Redraw with one block as the system and that same push becomes external and must be drawn. A useful habit is to write the system down beside the diagram, because a diagram without a stated system cannot be checked for missing or surplus arrows.

How does an explosion increase kinetic energy if momentum is conserved?

The two statements are about different quantities and neither one blocks the other. AP Physics 1 EK 4.1.A.3.iii models an explosion as an interaction in which forces internal to the system move objects within that system apart. Those internal forces cancel in pairs, so the total momentum is unchanged and typically stays at zero, with the fragments carrying equal and opposite momenta. Kinetic energy is not restricted the same way, because the internal forces act through displacements in opposite directions and so do positive work on both pieces. The energy comes from a store that was already inside the system, such as a compressed spring or a chemical propellant, so total energy is conserved throughout.