Conserve momentum across a two-object collision

Select both objects as the system

Object A is 2 kg and travels right at +3 m/s. Object B is initially to its right and travels left at −1 m/s. They approach each other at a relative speed of 4 m/s. Start with B also at 2 kg. The initial momenta are +6 kg·m/s for A and −2 kg·m/s for B, giving a total of +4 kg·m/s.

During contact, each object pushes on the other. Their interaction forces are equal in magnitude and opposite in direction, but they act on different objects. They therefore do not cancel on A alone or on B alone. Each object’s velocity can change. When the momentum changes are added for the two-object system, the internal impulses cancel.

The model assumes negligible net external horizontal impulse during contact. There is no modeled drive or track friction. With that boundary and assumption, Pbefore = Pafter, where P is total signed momentum in kg·m/s. This statement concerns the sum, not each term separately.

The OpenStax conservation-of-momentum discussion develops the internal-force cancellation and system boundary. This investigation uses original before-and-after numerical cases rather than a measured collision.

Find the velocity of a sticking pair

Set restitution to zero. In this model the objects stick after contact and share a final velocity V. Their combined mass is 2 + 2 = 4 kg. Momentum conservation gives (mA + mB)V = mA uA + mB uB. The symbols mA and mB are the masses; uA and uB are the signed velocities before contact. V is the shared velocity afterward.

Insert the known values: 4V = 2 × 3 + 2 × (−1) = 4. Therefore V = +1 m/s. Both move right after the collision. Opposite initial directions do not imply that they must stop, because the initial momenta were not equal and opposite.

A’s momentum falls from +6 to +2 kg·m/s, a change of −4 kg·m/s. B’s rises from −2 to +2 kg·m/s, a change of +4 kg·m/s. Their collision impulses are −4 and +4 N·s. The changes add to zero, while the total momentum remains +4 kg·m/s. The signed bars make both the redistribution and unchanged sum visible.

Compare with the elastic limit

Change restitution to one while keeping both masses 2 kg. A finishes at −1 m/s and B at +3 m/s. For these equal masses in a one-dimensional elastic collision, they exchange velocities. Their final momenta are −2 and +6 kg·m/s, still totaling +4.

The impulses are now −8 N·s on A and +8 N·s on B. Conservation does not require the individual impulse magnitude to stay the same when collision properties change. It requires the two internal impulses to be equal and opposite within a given trial under the stated assumptions.

Capture the sticking and elastic cases. Explain why both conserve total momentum despite very different individual final velocities. Momentum conservation alone is one equation with two unknown velocities. A second physical condition is required: a shared velocity for sticking, or a suitable energy/relative-speed condition for another collision type.

Build a zero-total-momentum case

Return restitution to zero and increase B mass to 6 kg. Its initial momentum is now −6 kg·m/s, balancing A’s +6. The total is zero. Since the objects stick, their common final velocity is zero.

Before contact they were both moving. A had 9 J of kinetic energy and B had 3 J, totaling 12 J. Zero total momentum did not mean no kinetic energy. The opposite signed momenta canceled; the nonnegative kinetic energies added. After sticking, their translational kinetic energy is zero and the model records 12 J of combined internal-energy increase.

This is a different prepared system from the equal-mass case. Changing B mass changes both initial total momentum and initial kinetic energy. Do not compare totals across those separate trials as though conservation demanded that they be identical. Conservation compares before and after within the same system and initial conditions.

Interpret the center-of-mass velocity

The center-of-mass velocity is Vcm = P/(mA + mB). It is a mass-weighted average of the object velocities. It stays constant here because total mass and total momentum stay constant. For the equal-mass case it is +1 m/s, whether the objects stick or separate.

When they stick, their common velocity equals Vcm. When they separate, individual velocities can differ greatly from Vcm. A zero center-of-mass velocity does not require both objects to be stationary. For B mass 6 kg and an elastic collision, the final velocities are −3 and +1 m/s: both move, but their momenta still sum to zero.

The lab does not provide collision duration or force history. An impulse value by itself cannot give a unique force; average force would also require the contact interval, and peak force would require more information.

Transfer to a new pair

A 1 kg object at +5 m/s collides with a 3 kg object at −1 m/s and sticks. Predict the common velocity and both momentum changes. Then explain how a non-negligible external impulse would alter your equation.

Check the collision-momentum transfer

Initial total momentum is 5 − 3 = +2 kg·m/s. Dividing by combined mass 4 kg gives V = +0.5 m/s. The first object's momentum changes from +5 to +0.5, so its impulse is −4.5 N·s. The second changes from −3 to +1.5, so its impulse is +4.5 N·s. With external impulse Jexternal, the combined equation becomes Pafter = Pbefore + Jexternal. Internal impulses still cancel in the total, but the total need not remain unchanged.