Momentum, impulse, and collisions

Start with two carts and a boundary

A 2 kg cart A moves right at 3 m/s toward a stationary 1 kg cart B. A is initially behind B on the same straight track. If they meet and stick together, how fast does the pair move afterwards? And can something stay conserved even though A slows down?

Choose the system to be both carts together. We use an exact teaching model: their masses stay constant, no material enters or leaves the system, and only straight-line translation is retained. Choose rightward as positive x. The reference frame is assumed inertial, meaning that a body with zero resultant force would keep a constant velocity in it.

Compare the states immediately before and after one collision. The track and Earth are outside the boundary. Their vertical forces balance, and the model supplies zero net external horizontal force throughout the collision interval. Rotation and the detailed contact force are not specified. The numbers describe the model, rather than measured performance of real carts.

Add momentum and keep energy separate

Velocity includes direction: here v_x is positive for rightward motion and negative for leftward motion, in m/s. Speed is its magnitude |v_x|. In this Newtonian particle model, linear momentum is the vector p = m v. Along x, p_x = m v_x. A positive mass keeps momentum in the velocity’s direction; the unit is kg m/s.

For the pair, add matching components: P_x = p_A,x + p_B,x. Initially p_A,x = (2 kg)(3 m/s) = 6 kg m/s and p_B,x = (1 kg)(0 m/s) = 0 kg m/s. Thus P_x = 6 kg m/s. Capital P denotes this system total, rather than a third object’s momentum.

Kinetic energy describes a different quantity. For a translating particle, K = (1/2)m|v|². The square uses the velocity’s magnitude, so K is a nonnegative scalar and contains no direction sign. In our one-dimensional case |v|² = v_x². Its unit is the joule: 1 J = 1 kg m²/s².

Initially K_A = (1/2)(2 kg)(3 m/s)² = 9 J and K_B = 0 J, giving K_total = 9 J. Squaring a negative velocity would also give a positive energy. Two moving bodies can have opposite momenta that cancel while both contribute positive kinetic energy.

Use impulse for a momentum change

A force changes momentum over time. Write the impulse as I and use Δ for a change: Δp = p_f − p_i. In this model the impulse of the net force on the chosen body equals its momentum change, I_net = Δp. Subscripts i and f mean the beginning and end of the same interval.

For a constant net force over a positive duration Δt, I_net = F_net Δt. The units agree: N s = (kg m/s²)s = kg m/s. Impulse is a vector change, rather than a force or the final momentum itself. Along x, retain the signs in I_net,x = p_f,x − p_i,x.

If a force has consecutive constant pieces, add each piece’s signed force-times-duration contribution. For a general time-varying force, I_net = F_net,ave Δt uses its true time average over that interval. An average of just the first and last force values generally does not supply this time average.

Conversely, F_net,ave = Δp/Δt gives an average force when the positive duration is known. It does not determine the peak force or its detailed history. These finite balances need only multiplication and subtraction; a calculus description uses the time integral of force.

Conserve the pair and solve sticking

In the retained Newtonian interaction model, the force of A on B is opposite to the force of B on A at each instant. Their impulses over the common collision interval are likewise opposite. These forces change each cart’s momentum, but cancel from the combined system’s momentum balance.

For the fixed collection of bodies, ΔP = I_ext, where I_ext is the impulse of the net external force. Zero net external force throughout an interval keeps P constant throughout it. Zero net external impulse is enough for equal endpoint momenta, but does not by itself establish that P stayed constant at every intervening instant.

The main model supplies zero external horizontal force, so P_f,x = P_i,x = 6 kg m/s. Sticking adds the condition that both carts have the same final velocity v_f. Therefore (2 kg + 1 kg)v_f = 6 kg m/s and v_f = 2 m/s rightward.

A then has p_A,f,x = 4 kg m/s and B has p_B,f,x = 2 kg m/s. Their changes are −2 kg m/s and +2 kg m/s: the collision impulses on A and B are −2 N s and +2 N s. Neither cart conserves its own momentum. The conserved quantity is their total.

Account for the kinetic-energy change

After sticking, the translational kinetic energies are K_A,f = (1/2)(2 kg)(2 m/s)² = 4 J and K_B,f = (1/2)(1 kg)(2 m/s)² = 2 J. Their total is 6 J, down from 9 J. Keeping total momentum constant therefore does not keep total kinetic energy constant.

For this energy comparison, additionally assume no energy is transferred to the surroundings during the collision. The 3 J decrease leaves the modeled translational motion and enters unresolved internal forms, such as deformation, vibration or heating. The balance does not establish how much goes into each form, and it does not mean that energy is destroyed.

An elastic collision preserves the system’s total translational kinetic energy between its before-and-after states. An inelastic collision does not preserve that quantity; sticking is the perfectly inelastic case in this two-body translational model. Bouncing apart alone does not establish an elastic collision.

Even in an elastic model, kinetic energy need not stay constant during contact: some can temporarily be stored in the interaction, such as in the deformation of an ideal spring, and return to kinetic energy afterwards. Our comparison is between the declared endpoint states, not a supplied energy history during contact.

Compare with an elastic outcome

Use the same masses and initial velocities, but replace sticking by a separate elastic collision model. A proposed outcome is v_A,f,x = 1 m/s and v_B,f,x = 4 m/s. Its momenta are 2 kg m/s and 4 kg m/s, again totaling 6 kg m/s. Its kinetic energies are 1 J and 8 J, totaling the original 9 J.

This outcome also has B moving faster than A after their encounter, so they separate with B ahead. The unchanged initial velocities, 3 m/s for A and 0 m/s for B, would satisfy the same momentum and energy totals but would keep the initially rear cart approaching the front one. Those unchanged values do not describe a completed collision followed by separation in this setup.

The figure compares the two different outcome models with their common initial state. Circle and square tails identify A and B; the dotted diamond row is the separately calculated system total. Arrow lengths share one momentum scale, and a zero glyph retains B’s initial zero value. They are momentum arrows, not positions, velocities or forces.

The impulse values under the outcome columns are differences from the common initial momenta. The energy bars use their own common scale and a zero-energy baseline. Reading the 6 kg m/s total arrows alongside the 9 J, 6 J and 9 J bars makes the two conservation questions visible.

Retain an external force and its duration

A separate force-pulse example shows why a chosen body’s momentum need not stay constant. A 2 kg cart initially has v_i,x = +1 m/s. Supply a constant net external force F_net,x = −8 N for 0 ≤ t ≤ 0.75 s. Other components balance. No force after that interval is specified.

Its impulse is I_net,x = (−8 N)(0.75 s) = −6 N s. Initial momentum is +2 kg m/s, so final momentum is +2 kg m/s − 6 kg m/s = −4 kg m/s. Dividing by 2 kg gives v_f,x = −2 m/s. The impulse has reversed the cart’s direction.

Here K changes from 1 J to 4 J. A negative momentum component is compatible with positive kinetic energy. This supplied external-force case is not one of the isolated two-cart collision outcomes and is not classified by their elastic or sticking assumptions.

For comparison, if the original two-cart system instead received an external impulse of −3 N s along x and still stuck together, its final P_x would be 3 kg m/s and its common velocity would be 1 m/s. Internal contact impulses would still cancel in the pair balance; they would not cancel this external contribution.

Check the system, interval and approximation

A useful order is: declare the body or collection, choose a frame and signs, calculate initial momentum, retain the external impulse, and then add a suitable final-state condition. Momentum conservation alone generally does not determine two independent final velocities. Sticking or an elastic assumption supplies more information; encounter geometry must also be respected.

A short collision time alone does not prove that external impulse is negligible. If the true time-average external horizontal force is bounded by 0.2 N in magnitude over 0.3 s, then |I_ext,x| ≤ 0.06 N s. For the original sticking pair, that stipulated bound gives 1.98 m/s ≤ v_f ≤ 2.02 m/s. This is a model bound around the zero-impulse result, not a statistical confidence interval or evidence from measured cart data.

Our p = m v and K = (1/2)m|v|² are Newtonian translational formulas in an assumed inertial frame, with constant positive masses. They do not cover variable-mass flow, field momentum, relativistic motion or quantum collision calculations. A real experiment must justify its retained forces, boundary, energy transfers and required accuracy.

Signed arithmetic and vector components suffice for these cases; the force history and collision material law are not needed to compute the stated endpoint balances. Center of mass and many-particle systems, the next lesson, extends the description of the collection without treating a conserved total as a conserved value for every part.

Worked examples

Try it yourself

1. Two 1 kg bodies move at +2 m/s and −2 m/s on the same line. Calculate their total momentum and kinetic energy. If they meet, stick and receive zero external impulse, find their common final velocity and translational kinetic energy.

   Show the answer and reasoning

   Initially P_x = 0 kg m/s and K_total = 4 J. After sticking, v_f = 0 m/s and K_total = 0 J.

   The signed momenta +2 kg m/s and −2 kg m/s cancel, but the initial kinetic energies 2 J and 2 J add. Zero total momentum did not mean that either body was initially at rest. Sticking gives (2 kg)v_f = 0 kg m/s.

2. In the main sticking case A receives −2 N s and B receives +2 N s. Supply a collision duration of 0.25 s; the mutual interaction is the only unbalanced force on either cart. Find their time-average interaction forces. Do the impulses and duration also determine the peak magnitudes?

   Show the answer and reasoning

   The average force on A is −8 N and on B is +8 N. Their peak magnitudes are not determined.

   Divide each signed impulse by the positive common duration: (−2 N s)/(0.25 s) = −8 N and (+2 N s)/(0.25 s) = +8 N. Several force histories can have the same time average. No constant contact-force history has been supplied.

3. Return to the original 2 kg and 1 kg pair with initial P_x = 6 kg m/s. Change the model: its net external horizontal force is constantly −6 N for a 0.5 s collision interval, and the bodies stick. Find the external impulse and common final velocity.

   Show the answer and reasoning

   I_ext,x = −3 N s and v_f = +1 m/s.

   The external impulse is (−6 N)(0.5 s) = −3 N s, so P_f,x = 6 kg m/s − 3 kg m/s = 3 kg m/s. Divide by the total 3 kg. Equal-and-opposite internal impulses do not cancel an external impulse.

4. In the elastic outcome, A changes from 3 m/s to 1 m/s and B from 0 m/s to 4 m/s. Their masses remain 2 kg and 1 kg. How much kinetic energy does each gain or lose, and which total is conserved?

   Show the answer and reasoning

   A loses 8 J and B gains 8 J. Their combined kinetic energy stays 9 J.

   A changes from 9 J to 1 J and B from 0 J to 8 J. Elastic conservation applies to the system’s total before and after, rather than to each body’s kinetic energy separately.

5. After sticking, the diagram shows p_A,x = 4 kg m/s, p_B,x = 2 kg m/s and a dotted total P_x = 6 kg m/s. A learner adds all three and reports 12 kg m/s for the pair. Correct the calculation and explain what the dotted row represents.

   Show the answer and reasoning

   The pair’s momentum is 4 kg m/s + 2 kg m/s = 6 kg m/s. The dotted row is that calculated sum.

   There are two bodies in the boundary. P is neither a third body nor an additional contribution. Adding it to its own components counts the same momenta twice.

6. For the original sticking pair, suppose |F_ext,ave,x| ≤ 0.2 N over 0.3 s. The initial total is 6 kg m/s and the mass is 3 kg. Bound the final common velocity. Is the bound a statistical confidence interval?

   Show the answer and reasoning

   1.98 m/s ≤ v_f ≤ 2.02 m/s. This is a stipulated model bound, not a statistical confidence interval.

   The impulse magnitude is at most (0.2 N)(0.3 s) = 0.06 N s. Thus P_f,x lies from 5.94 to 6.06 kg m/s. Dividing both inclusive limits by the positive 3 kg gives the velocity limits. No probability law or measurement evidence was supplied.

What to carry forward

• Momentum is a vector; kinetic energy is a nonnegative scalar with different units.
• The net external impulse changes a fixed system’s total momentum across the declared interval.
• Internal impulses can change individual body momenta while canceling in the total.
• An elastic collision preserves total kinetic energy between the before-and-after states; sticking need not.
• State the boundary, frame, force-duration assumptions and final-state condition before applying a conservation balance.