foundation26 min studyLesson 5 in Physics

Momentum, impulses, and collisions

Track momentum across a system boundary and distinguish momentum conservation from kinetic-energy conservation.

Before you begin

On this page

The question for today

What you’ll understand

Relate impulse to momentum change over a finite interval.
Choose a collision system with negligible external impulse.
Test separately whether kinetic energy is conserved.

A collision can involve huge, rapidly changing forces that are hard to measure moment by moment. Momentum offers another route. By tracking how much momentum crosses a chosen system boundary, we can constrain the outcome without resolving every deformation. The crucial move is to conserve a vector for the right system, rather than memorizing one collision formula.

Impulse is an accumulated interaction

Momentum combines mass and velocity, so its direction matters. A heavy slow cart and a light fast cart can carry equal momenta even though their kinetic energies differ. The change in momentum over an interval equals the impulse, the area under a force–time curve. A short sharp force and a gentler longer force can produce the same impulse.

Average force is impulse divided by elapsed time, but it does not reveal the peak force or the detailed time profile. This distinction explains why extending a stopping time can reduce typical loading without altering the momentum that must be removed. An impulse calculation describes the motion change; material damage requires additional information about stresses, deformation, and how the force is distributed.

\mathbf J=\int_{t_i}^{t_f}\mathbf F\,dt=\Delta\mathbf p,\qquad \mathbf p=m\mathbf v

Impulse has units N·s, equivalent to kg·m/s. The momentum formula assumes a Newtonian constant-mass particle; the impulse relation is the integrated force law.↔ Scroll wide equations horizontally.

Internal violence can coexist with total conservation

Two colliding carts exert large forces on each other. For the combined system, their internal forces cancel in the total momentum balance under the Newtonian model. Momentum is conserved when the external impulse is zero, or approximately conserved when that impulse is negligible during the short collision. Gravity may still act; its horizontal component on a level track is zero.

A system does not need zero external force at every instant to have equal initial and final momentum: the net external impulse over the interval is the condition. Conversely, two objects being in contact does not automatically make their total momentum conserved. A wall fixed to Earth can transfer substantial momentum outside a system consisting only of the moving object and a small wall patch.

\mathbf P_f-\mathbf P_i=\mathbf J_{\mathrm{ext}},\qquad \mathbf P=\sum_i m_i\mathbf v_i

Total momentum is conserved across the chosen interval if the net external impulse vanishes. In several dimensions, this is one equation per Cartesian component; conserving only the magnitude loses directional information.↔ Scroll wide equations horizontally.

Collision labels concern kinetic energy

An elastic collision conserves the total kinetic energy of the colliding system as well as momentum when external impulse is negligible. In an inelastic collision, some initial kinetic energy can become internal energy, sound, or deformation. Perfectly inelastic means the bodies share a final velocity because they stick. It does not mean all kinetic energy necessarily disappears.

Momentum conservation alone usually leaves several possible outgoing velocities. Additional information, such as elasticity, sticking, or a measured outgoing speed, closes the problem. The center-of-mass velocity is a useful organizing quantity because internal interactions cannot change it without external impulse. Later scattering theory keeps this conservation structure while replacing definite classical trajectories with probabilities for possible outcomes.

\mathbf V_{\mathrm{CM}}=\frac{\mathbf P}{M},\qquad v_f=\frac{m_1v_{1i}+m_2v_{2i}}{m_1+m_2}

The second equation is for a one-dimensional sticking collision with negligible external impulse. M is the total mass. The shared final velocity equals the system's center-of-mass velocity; it does not imply kinetic-energy conservation.↔ Scroll wide equations horizontally.

Work it through

Two carts lock together

A $2.0\,\mathrm{kg}$ cart moving right at $3.0\,\mathrm{m\,s^{-1}}$ collides with a stationary $1.0\,\mathrm{kg}$ cart. They stick on a nearly frictionless level track. Find their velocity and the kinetic-energy change.

Choose both carts as the system and right as positive. Initial momentum is $P_i=(2.0)(3.0)+(1.0)(0)=6.0\,\mathrm{kg\,m\,s^{-1}}$ . Neglect the horizontal external impulse during collision.
Sticking means a shared velocity, so $P_f=(2.0+1.0)v_f=P_i$ . Thus $v_f=2.0\,\mathrm{m\,s^{-1}}$ to the right. The interaction forces did not need to be known separately.
Initial kinetic energy is $K_i=(2.0)(3.0)^2/2=9.0\,\mathrm{J}$ . Final kinetic energy is $K_f=(3.0)(2.0)^2/2=6.0\,\mathrm{J}$ , so $\Delta K=-3.0\,\mathrm{J}$ .

The result

The combined carts move right at $2.0\,\mathrm{m\,s^{-1}}$ . Momentum is conserved while $3.0\,\mathrm{J}$ of kinetic energy becomes other forms of energy.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Read signed momentum components

Change one vector's angle and magnitude and inspect its signed components. Interpret it as a normalized momentum, then sketch a second momentum to reason about cancellation. The lab is not a collision solver and does not enforce masses, restitution, or conservation.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOpposite components can cancel even when both objects have positive kinetic energy.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Internal forces change the separate cart momenta, but their sum changes only through external impulse. Sticking generally converts some kinetic energy into internal energy, so kinetic-energy conservation is not an additional valid assumption.

Your turn

Try it yourself

A $0.20\,\mathrm{kg}$ ball moving at $+5.0\,\mathrm{m\,s^{-1}}$ rebounds at $-3.0\,\mathrm{m\,s^{-1}}$ . The contact lasts $0.040\,\mathrm{s}$ . Find the impulse on the ball and average contact force, neglecting other impulses.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use final minus initial momentum. Reversal makes the change larger than the change in speed alone.

Compare with the solution

The impulse is $J=m(v_f-v_i)=0.20(-3.0-5.0)=-1.6\,\mathrm{N\,s}$ . Dividing by the contact duration gives $F_{\mathrm{avg}}=-40\,\mathrm{N}$ . Both signs indicate the negative direction. The peak force need not equal this average, and the wall–Earth system receives the opposite impulse in the larger momentum account.

Keep the idea

Take it with you

Impulse measures momentum transfer over time, not peak force.
Momentum conservation is a statement about a chosen system's external impulse.
Conserving momentum does not automatically conserve kinetic energy.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.