You're staring at a physics problem. Two carts on a track. One has a spring sticking out front. Because of that, they collide. The spring compresses, stores energy, pushes back. The question hits: is momentum conserved here?
Short answer: yes. That said, always. But the why trips people up.
What Is Momentum Conservation Anyway
Momentum conservation isn't a suggestion. In any closed system — no external forces, no friction messing things up — the total momentum before anything happens equals the total momentum after. Which means it's a law. Period.
$p_{initial} = p_{final}$
Mass times velocity. Still, vector quantity. Because of that, direction matters. On the flip side, a 2 kg cart moving right at 3 m/s has +6 kg·m/s. On top of that, same cart moving left? -6 kg·m/s. Add them up for the whole system. That sum doesn't change.
The spring doesn't change this. Ever.
The system definition trap
Here's where students lose points. Consider this: they define the system as "cart A" or "cart B" or "the spring. " Wrong. The system is everything involved in the interaction*. Both carts. In real terms, the spring. Here's the thing — the track (if it's frictionless). The earth (if we're being pedantic about gravity, but normal force cancels that).
If you leave the spring out of your system, you've created an external force. But the spring is part of the collision. Because of that, the spring pushes on the carts. It's not an outside agent. That push looks external if the spring isn't part of your system*. It's the mechanism.
Why People Think Momentum Isn't Conserved
Energy. That's the confusion. Kinetic energy isn't* conserved during the compression phase. In practice, it transforms into elastic potential energy in the spring. Then it transforms back (mostly) during expansion. Total mechanical energy is conserved if the spring is ideal and there's no internal damping. But kinetic energy alone? Also, nope. Still, it dips. Then rises.
Momentum doesn't dip. It doesn't care about energy transformations.
The force pair illusion
Newton's third law: every force has an equal and opposite reaction. The spring pushes left on cart A. Also, it pushes right on cart B. Because of that, same magnitude. Opposite direction. In practice, same duration. The impulses cancel. Day to day, $\Delta p_A = -\Delta p_B$. Total $\Delta p = 0$.
People see the spring compressing and think "something is happening, momentum must be changing." But the something happening* is internal forces reshuffling momentum between the two carts. The total stays put.
How a Spring Collision Actually Works
Let's walk through it frame by frame. Two identical carts, mass $m$. Cart A moves right at $v_0$. Cart B sits at rest. Spring attached to cart A's front.
Phase 1: First contact
Spring touches cart B. That said, no compression yet. Forces zero. Momentum: $mv_0 + 0 = mv_0$. Cart A still at $v_0$. Cart B still at 0.
Phase 2: Compression
Spring compresses. It pushes left on A, right on B. Cart A slows. Cart B speeds up.
$F_{spring \to A} = -F_{spring \to B}$
$m a_A = -m a_B$
$a_A = -a_B$
Integrate acceleration over time: $\Delta v_A = -\Delta v_B$. So $\Delta(mv_A) = -\Delta(mv_B)$. Total momentum change: zero.
At maximum compression, both carts move together* at the same velocity $v_{cm}$. Center of mass velocity. For equal masses:
$v_{cm} = \frac{mv_0 + 0}{2m} = \frac{v_0}{2}$
Momentum check: $2m \cdot \frac{v_0}{2} = mv_0$. Conserved.
Phase 3: Expansion
Spring pushes back. So cart A slows further (maybe stops, maybe reverses). On the flip side, cart B speeds up past $v_{cm}$. Forces still equal and opposite. Now, impulses still cancel. Momentum still conserved.
Phase 4: Separation
Spring returns to natural length. Forces drop to zero. Carts separate. Final velocities depend on spring stiffness and masses.
$m v_{A,final} + m v_{B,final} = mv_0$
Always.
What Changes With Different Masses
Unequal masses? Same physics. Center of mass velocity stays constant throughout.
$v_{cm} = \frac{m_A v_{A,initial} + m_B v_{B,initial}}{m_A + m_B}$
During maximum compression, both carts move at $v_{cm}$. Now, momentum at that instant: $(m_A + m_B) v_{cm}$ = initial total momentum. Also, the spring just mediates the transfer. It doesn't add or subtract.
Spring attached to the "wrong" cart
Doesn't matter. In real terms, if it's massive, include it in the system. Internal forces are internal forces. Spring on A, spring on B, spring between them with no cart attachment — same result. But the spring's mass? Practically speaking, its momentum counts too. If it's massless (ideal spring), it carries no momentum itself but still transmits forces equally.
Common Mistakes / What Most People Get Wrong
Mistake 1: Confusing momentum with kinetic energy
"I calculated final kinetic energy and it's less than initial! Momentum isn't conserved!"
No. Kinetic* energy isn't conserved during compression. That said, total energy is (ideally). Momentum is conserved always*. Consider this: different quantities. Different rules.
Mistake 2: Forgetting the spring is part of the system
You draw a free-body diagram for cart A. On top of that, you call it "external. On the flip side, put the spring in the system. Spring force appears. " The spring force is only external if the spring isn't in your system*. Here's the thing — " Then you say "external force acts, so momentum not conserved. Problem solved.
For more on this topic, read our article on what is an example of newton's first law or check out ap calc ab exam score calculator.
Mistake 3: Thinking maximum compression means zero momentum
At max compression, both carts move at $v_{cm}$. That said, system momentum is $(m_A + m_B)v_{cm}$. That's not zero (unless total initial momentum was zero). People see "relative velocity zero" and think "momentum zero." Relative velocity zero $\neq$ momentum zero.
Mistake 4: Ignoring the spring's mass
Real springs have mass. If the spring is heavy and oscillates, its center of mass moves. That momentum counts. In intro physics we usually assume massless springs. But if a problem gives you spring mass, include it. And the spring's coils have velocity. They have momentum.
Mistake 5: Assuming the spring force is constant
It's not. That's why $F = -kx$. Here's the thing — changes continuously. But impulse $\int F dt$ still cancels for the two carts. Also, the integral* of the force over time is what matters for momentum change. The forces are equal and opposite at every instant*, so their integrals are equal and opposite. Done.
Practical Tips / What Actually Works
Define your system first
Before writing equations, circle the system. C
Practical Tips / What Actually Works
Define your system first
Before writing any equations, circle the system. If you want to apply momentum conservation, include every object that can exchange momentum with the surroundings. In a typical cart‑spring problem this usually means the two carts plus the spring (and any attached mass). Once the boundary is fixed, any force that crosses that boundary is external; everything inside is governed only by internal forces, which cancel in pairs.
Write the momentum balance explicitly
For a closed system the vector sum of external forces equals the rate of change of total momentum:
[ \sum \mathbf{F}{\text{ext}} = \frac{d\mathbf{p}{\text{total}}}{dt}. ]
If you have chosen a truly isolated system, the right‑hand side is zero, so (\mathbf{p}_{\text{total}}) is constant. Write this statement out; it forces you to check whether any forces you have omitted are truly external.
Keep track of directions
Momentum is a vector. When you set up the conservation equation, write separate components (e.g., (p_x), (p_y)) if the motion is not strictly one‑dimensional. This prevents sign errors that often masquerade as “violations” of momentum conservation.
Use impulse for variable forces
Since the spring force varies with displacement, it is often easier to work with impulse rather than trying to integrate the force over time manually. The impulse delivered to each cart is equal in magnitude and opposite in direction, so the total impulse on the system is zero. If you need the final velocities, you can write:
[ \Delta p_A = -\Delta p_B = \int_{t_i}^{t_f} F_{\text{spring}},dt, ]
and then apply (p = mv) to each cart.
Separate the “compression” and “release” phases
The interaction can be split into two distinct intervals:
- Compression – the spring shortens, storing potential energy while the carts approach each other. Momentum is still conserved, but relative speed decreases.
- De‑compression (or rebound) – the spring expands, pushing the carts apart. Momentum continues to be conserved, and the relative speed reverses sign (for an ideal elastic interaction).
Treating the two phases separately helps avoid confusion about “when” a particular velocity occurs.
Check energy consistency (optional)
While momentum is always conserved, kinetic energy is not necessarily conserved during the compression stage. If the problem states that the spring is ideal (no damping) and the collision is elastic, then the total mechanical energy (kinetic + spring potential) remains constant. Using both conservation laws together can serve as a useful sanity check.
Example: Two carts with a massless spring
Suppose cart A (mass (m_A)) moves rightward with speed (v_{0}) and cart B (mass (m_B)) is initially at rest. The spring is initially uncompressed. After a short interval the spring reaches maximum compression; both carts share a common velocity (v_{c}). By momentum conservation:
[ m_A v_{0}= (m_A+m_B)v_{c};;\Longrightarrow;;v_{c}= \frac{m_A}{m_A+m_B}v_{0}. ]
During the subsequent expansion the spring returns its stored energy, giving each cart a final speed that can be found by combining the momentum result with the energy equation. This illustrates how the common‑velocity condition at maximum compression follows directly from the conservation principle, without invoking any “extra” forces.
Conclusion
The apparent paradox of “missing” momentum in spring‑mediated interactions dissolves as soon as the system is defined correctly and the distinction between internal and external forces is kept clear. So momentum is a vector that is conserved for any isolated system, no matter how complicated the internal forces may be. The spring merely acts as a conduit for internal forces; it does not create or destroy momentum.
- selecting an appropriate system boundary,
- writing the momentum balance explicitly,
- handling variable forces through impulse,
- separating the compression and rebound phases, and
- optionally cross‑checking with energy conservation,
students can predict the motion of connected objects with confidence. The key takeaway is that internal forces always cancel in pairs, and therefore the total momentum of a closed system remains unchanged from start to finish. This principle holds for massless or massive springs, for carts on frictionless tracks, and for any collection of objects that interact solely through forces that originate inside the system.
