Conservation of Linear Momentum
Why this matters
Imagine you and your friend Priya are standing on skateboards in a school parking lot, facing each other. You're both at rest. You decide to push off her hands to move backward. What happens? You roll backward, but she also rolls backward in the opposite direction, even though you were the one who did the pushing.
Why didn't she just stay put? It feels like you created motion out of thin air. But you didn't. You simply redistributed it. This is the core idea behind one of the most powerful principles in physics: the conservation of linear momentum. In this lesson, we'll explore how to define a "system," why the total momentum inside it can stay constant, and how to use this rule to predict the outcome of any collision or explosion.
Diagram
Concept map
flowchart TD
A[Analyze an interaction: collision or explosion] --> B{Choose a system of objects};
B --> C{Is there a net external force on the system?};
C -- Yes --> D[Momentum is NOT conserved];
D --> E[Use Impulse-Momentum Theorem: J_ext = Δp_sys];
C -- No --> F[Momentum IS conserved];
F --> G[Use Conservation Law: p_total,initial = p_total,final];
Core explanation
Hello everyone, it's Saavi. Today we're tackling a concept that is fundamental to all of physics: the conservation of linear momentum. It sounds fancy, but the idea is beautifully simple.
What is a "System"?
First, we need to talk about choosing a system. A system is just a collection of objects that we decide to analyze together. It could be two billiard balls colliding, a rocket expelling fuel, or a quarterback and the football they're about to throw.
The magic of conservation of momentum depends entirely on how you define your system. The key is to draw a mental box around the objects that are interacting with each other.
- Internal ForcesForces that objects within the system exert on each other. When you push off Priya's skateboard, the force you exert on her and the force she exerts back on you are internal to the "you + Priya" system.
- External ForcesForces exerted on the system by something outside the box. Gravity from the Earth pulling on you and Priya is an external force. The normal force from the ground pushing up is also external.
The Law of Conservation of Momentum
Here's the big rule:
If the net external force on a system is zero, the total momentum of that system is conserved.
"Conserved" just means it doesn't change. The total momentum right before an interaction (a collision, an explosion) is equal to the total momentum right after.
p⃗_total, initial = p⃗_total, final
Let's break this down. In our skateboard example, the external force of gravity pulling you down is cancelled out by the external normal force from the ground pushing you up. So, the net external force is zero. This means the total momentum of the "you + Priya" system is conserved.
Initially, you were both at rest, so the total initial momentum was zero.
p⃗_initial = m_you * v_you,i + m_Priya * v_Priya,i = m_you*(0) + m_Priya*(0) = 0
After you push off, you move one way and Priya moves the other. Let's say you have momentum p⃗_you,f and she has p⃗_Priya,f. Conservation of momentum tells us:
p⃗_final = p⃗_you,f + p⃗_Priya,f
Since p⃗_initial = p⃗_final, we must have:
0 = p⃗_you,f + p⃗_Priya,f
...which means p⃗_you,f = -p⃗_Priya,f.
Your final momentum is equal in magnitude and opposite in direction to hers! The total momentum of the system is still zero. Momentum was just transferred between you.
Why Does This Happen? Newton's Third Law.
This isn't a new law of physics; it's a direct consequence of Newton's Third Law. For every action, there is an equal and opposite reaction.
The force you exerted on Priya (F⃗_on_Priya) is equal and opposite to the force she exerted on you (F⃗_on_you).
F⃗_on_Priya = -F⃗_on_you
Since you're in contact for the same amount of time (Δt), the impulses are also equal and opposite:
J⃗_on_Priya = F⃗_on_Priya * Δt
J⃗_on_you = F⃗_on_you * Δt
So, J⃗_on_Priya = -J⃗_on_you
And since impulse equals the change in momentum (J⃗ = Δp⃗), we get:
Δp⃗_Priya = -Δp⃗_you
The change in Priya's momentum is the exact opposite of the change in your momentum. If we look at the change in the total momentum of the system, it's Δp⃗_Priya + Δp⃗_you = 0. The total momentum didn't change.
What if There IS a Net External Force?
In this case, the change in the system's momentum is equal to the net external impulse.
J⃗_external = Δp⃗_system
This is just the Impulse-Momentum Theorem, but applied to the whole system instead of just one object.
Center of Mass
A powerful related idea is the center of mass. It's the average position of all the mass in a system. For a system of objects, the velocity of the center of mass is given by:
v⃗_cm = (Σ mᵢv⃗ᵢ) / (Σ mᵢ)The numerator, Σ mᵢv⃗ᵢ, is just the total momentum of the system, p⃗_total. The denominator, Σ mᵢ, is the total mass, M_total. So:
v⃗_cm = p⃗_total / M_total
Now, think about what this means. If the total momentum of the system is conserved (i.e., p⃗_total is constant), and the total mass is obviously constant, then the velocity of the center of mass must also be constant!
Imagine a firework shooting up in a parabolic arc. At its peak, it explodes. The pieces fly out in all directions. The individual momenta of the pieces are all different. But the center of mass of all those pieces combined continues to move along the exact same parabolic path as if the explosion never happened. Why? Because the explosion was caused by internal forces. The only major external force is gravity, which was already acting on the firework before the explosion.
This is an incredibly useful tool for analyzing complex interactions. If you can show momentum is conserved, you know the center of mass will just keep chugging along at a constant velocity.
Worked examples
Let's put this into practice with a couple of classic scenarios.
Inelastic Collision
Problem: A 2.0 kg cart moving at +3.0 m/s on a frictionless track collides with a 4.0 kg cart that is initially at rest. After the collision, the two carts stick together. What is their final velocity?
Solution:
- 1Define the SystemOur system is the two carts (
cart 1 + cart 2). The track is frictionless, and we can assume the collision happens quickly, so we can ignore air resistance. The vertical forces (gravity and normal force) cancel out. Therefore, the net external force on the system is zero. Momentum is conserved! - 2Set up the Conservation Equation
p⃗_total, initial = p⃗_total, final - 3Calculate Initial MomentumThe total initial momentum is the sum of the individual momenta of the carts. Remember that cart 2 is at rest.
p⃗_initial = m₁v⃗₁ᵢ + m₂v⃗₂ᵢp⃗_initial = (2.0 kg)(+3.0 m/s) + (4.0 kg)(0 m/s)p⃗_initial = +6.0 kg·m/s - 4Calculate Final MomentumAfter the collision, the carts stick together, so they become a single object with a combined mass (
m₁ + m₂) and a single final velocity (v⃗_f).p⃗_final = (m₁ + m₂)v⃗_fp⃗_final = (2.0 kg + 4.0 kg)v⃗_f = (6.0 kg)v⃗_f - 5Solve for the UnknownNow, we set the initial and final momentum equal to each other.
+6.0 kg·m/s = (6.0 kg)v⃗_fv⃗_f = (+6.0 kg·m/s) / (6.0 kg)v⃗_f = +1.0 m/s
The combined carts move to the right at 1.0 m/s. This makes sense: the initial motion was to the right, and now that same momentum is shared across a larger mass, so the speed is lower.
Explosion (Recoil)
Problem: A 1500 kg cannon, initially at rest on a frictionless surface, fires a 25 kg cannonball horizontally with a velocity of +400 m/s. What is the recoil velocity of the cannon?
Solution:
- 1Define the SystemThe system is the cannon and the cannonball. The "explosion" is the firing of the cannonball. Since the surface is frictionless, the net external force is zero. Momentum is conserved.
- 2Set up the Conservation Equation
p⃗_total, initial = p⃗_total, final - 3Calculate Initial MomentumEverything is at rest initially.
p⃗_initial = 0 - 4Calculate Final MomentumAfter firing, the cannon (
c) and the ball (b) are moving separately.p⃗_final = m_c * v⃗_cf + m_b * v⃗_bfp⃗_final = (1500 kg)v⃗_cf + (25 kg)(+400 m/s) - 5Solve for the Unknown
0 = (1500 kg)v⃗_cf + 10000 kg·m/s-(1500 kg)v⃗_cf = 10000 kg·m/sv⃗_cf = -10000 / 1500v⃗_cf = -6.67 m/s
The cannon recoils (moves backward) at 6.67 m/s. The negative sign is crucial—it tells us the direction is opposite to the cannonball's motion.
Try it yourself
Ready to try on your own? Sketch these out and apply the principles we just covered.
- 1Astronaut in SpaceAn 80 kg astronaut is floating at rest in space. She throws a 2.0 kg wrench away from her spaceship at a speed of 15 m/s. What is her recoil speed?
- Hint: What is the total momentum of the "astronaut + wrench" system before she throws it? The system is isolated in space, so what does that tell you about the total final momentum?
- 2Football TackleA 110 kg linebacker moving at +8.0 m/s tackles a 90 kg quarterback who is at rest. They move together after the tackle.
- Part AWhat is their velocity immediately after the tackle?
- Part BWhat is the velocity of the center of mass of the linebacker-quarterback system before the tackle? What about after?
- HintThis is a perfectly inelastic collision. Treat them as one combined mass after the collision. For Part B, you don't need to do a new calculation if you understand the center of mass concept!
- Part A
Practice — 8 questions
In simple terms, conservation of momentum means the total "oomph" of motion in a closed system before a collision is the same as the total "oomph" afterward.
v⃗_cm = (Σ mᵢv⃗ᵢ) / (Σ mᵢ)- 4.3.A: Describe the behavior of a system using conservation of linear momentum.
- 4.3.B: Describe how the selection of a system determines whether the momentum of that system changes.
- 4.3.A.1
- A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.
- 4.3.A.1.i
- For a collection of objects, the velocity of a system's center of mass can be calculated using the equation v⃗_cm = (Σp⃗ᵢ)/(Σmᵢ) = (Σmᵢv⃗ᵢ)/(Σmᵢ)
- 4.3.A.1.ii
- The velocity of a system's center of mass is constant in the absence of a net external force.
- 4.3.A.2
- The total momentum of a system is the sum of the momenta of the system's constituent parts.
- 4.3.A.3
- In the absence of net external forces, 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. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.
- 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. This is a direct result of Newton's third law.
- 4.3.A.3.ii
- A system may be selected so that the total momentum of that system is constant.
- 4.3.A.3.iii
- If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system. Relevant equation: J⃗ = Δp⃗
- 4.3.A.4
- Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.
- 4.3.B.1
- Momentum is conserved in all interactions.
- 4.3.B.2
- If the net external force on the selected system is zero, the total momentum of the system is constant.
- 4.3.B.3
- If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.
flowchart TD
A[Analyze an interaction: collision or explosion] --> B{Choose a system of objects};
B --> C{Is there a net external force on the system?};
C -- Yes --> D[Momentum is NOT conserved];
D --> E[Use Impulse-Momentum Theorem: J_ext = Δp_sys];
C -- No --> F[Momentum IS conserved];
F --> G[Use Conservation Law: p_total,initial = p_total,final];
Read what Saavi narrates
Hello everyone, it's Saavi. Let's talk about one of the most powerful ideas in physics.
Imagine you and your friend Priya are on skateboards in a parking lot. You're both just standing still. Then, you push off her hands to move backward. What happens? You roll backward, of course. But... so does she, in the opposite direction. You created motion, but where did it come from?
This is the heart of the conservation of linear momentum. The big idea is that for any group of interacting objects... what we call a "system"... the total amount of momentum stays constant, as long as no outside forces mess with it. The momentum you gained is perfectly balanced by the momentum Priya gained in the other direction. The total momentum of the "you plus Priya" system started at zero, and it ended at zero.
Let's apply this to a classic problem. Imagine a 2 kilogram cart moving at 3 meters per second. It's about to hit a 4 kilogram cart that's just sitting still. After they collide, they stick together. What's their final speed?
First, we define our system as the two carts. Since there's no friction, momentum is conserved. The total momentum before the collision equals the total momentum after.
Before the collision, the total momentum is just the momentum of the first cart, since the second one is at rest. That's 2 kilograms times 3 meters per second, which gives us 6 kilogram meters per second.
After the collision, the two carts are stuck together. They've become one object with a combined mass of 6 kilograms. Their final momentum is this combined mass... 6 kilograms... times their final velocity, which is what we want to find.
So, we set the initial momentum equal to the final momentum. Six equals six times the final velocity. A little algebra tells us the final velocity is 1 meter per second. It makes sense... the same amount of momentum is now moving a much heavier object, so it has to go slower.
Now, a really common mistake I see all the time is students forgetting that momentum has a direction. It's a vector. If an object is moving to the left, you have to give its velocity a negative sign. If you just add up the speeds, your answer will be wrong. Always, always define a positive direction before you start a problem.
Keep this principle in mind. If you can isolate a system and show there are no net external forces, you have a powerful tool to predict what happens next. You've got this.
You might add magnitudes instead of accounting for direction. If a 5 kg·m/s object moving right hits a 5 kg·m/s object moving left, the total initial momentum is 0, not 10.
Always establish a positive direction (e.g., right is positive). Any velocity or momentum in the opposite direction must have a negative sign.
You might try to apply conservation of momentum when it's not valid. If you're pushing a car, your push is an external force on the car, and its momentum will change. You can only conserve momentum if you include *yourself* in the system, and even then, the friction from the road on your feet is an external force on the "you + car" system.
Before any calculation, explicitly state your system and ask: "Are there any net forces from *outside* this system acting on it?" If the answer is yes (like friction or a continuous push), momentum is not conserved.
Momentum is conserved in *all* collisions (if the system is isolated). Kinetic energy is only conserved in *elastic* collisions. In most real-world collisions, like the carts sticking together, energy is lost to heat, sound, and deformation.
Treat momentum conservation and energy conservation as two separate tools. Never assume kinetic energy is conserved unless the problem explicitly states the collision is "perfectly elastic."
In an inelastic collision where objects stick together, the final mass is the *sum* of the initial masses. In an explosion, the objects move separately, so you use their individual masses for the final momentum calculation.
Draw a "before" and "after" picture. Clearly label the masses and velocities in each state. This will help you correctly write the `p_final` term.