Momentum and Impulse: Choose a Direction before Comparing Before and After

Choose a direction before you subtract velocities

When an object rebounds, it changes direction as well as speed. Treating both velocities as positive loses that information and usually gives the wrong momentum change. A small arrow labelled “positive” is therefore part of the solution, not decoration.

This lesson uses one-dimensional motion and fictional low-speed cart data. It separates the momentum of one object from the total momentum of a chosen system, then uses that distinction to decide when conservation is justified. No real collision experiment is needed.

1. Write the quantities with their signs

For an object of constant mass, momentum p = mv. Use kilograms for mass and metres per second for velocity, giving momentum in kg·m/s. Momentum points in the same direction as velocity. If right is positive, rightward velocities are positive and leftward velocities are negative; mass remains positive.

Momentum change is final minus initial: Δp = p_final − p_initial = m(v − u). Here u is the initial velocity and v the final velocity. Put negative values in brackets before subtracting. The magnitude of a momentum change is its positive size; its signed value also records direction.

Impulse J is the change in momentum produced by the net force over an interval. For a constant net force, J = FΔt. For a changing force, J = F_averageΔt, where the average is taken over that interval. The unit N·s equals kg·m/s because one newton is one kg·m/s².

Worked case 1: a rebound

A 0.25 kg cart approaches a bumper at 1.8 m/s to the right and leaves at 1.2 m/s to the left. The contact interval is 0.15 s. Choose right as positive and find the cart’s momentum change and average net force.

Native diagram alternative: Right is positive. Before contact, the 0.25 kg cart moves right with u = +1.8 m/s. After contact, it moves left with v = −1.2 m/s. The two drawings represent different moments in time, not two simultaneous carts. Contact lasts 0.15 s.

Solution: Δp = −0.30 − (+0.45) = −0.75 kg·m/s. The impulse on the cart is therefore −0.75 N·s, or 0.75 N·s leftward. Average net force = −0.75 ÷ 0.15 = −5.0 N, or 5.0 N leftward.

The cart first loses its initial rightward momentum and then acquires leftward momentum. This explains why the magnitude of the change involves adding the two momentum magnitudes. Simply calculating 0.25(1.8 − 1.2) = 0.15 gives the wrong result because it uses speeds without their directions.

The calculated force is the average net force on the cart. It is not necessarily the maximum contact force. On a level track where other horizontal impulses are negligible, it also approximates the average horizontal bumper force; that extra interpretation needs the stated assumption.

Worked case 2: force acting for a stated duration

A 0.60 kg cart initially moves left at 0.40 m/s. A constant net horizontal force of 1.5 N acts rightward for 0.80 s. Find the final velocity.

Solution: With right positive, u = −0.40 m/s and F = +1.5 N. Impulse J = 1.5 × 0.80 = +1.2 N·s. Initial momentum = 0.60 × (−0.40) = −0.24 kg·m/s. Final momentum = initial momentum + impulse = −0.24 + 1.2 = +0.96 kg·m/s.

Final velocity = 0.96 ÷ 0.60 = +1.6 m/s, meaning rightward. The force has removed the initial leftward momentum and then produced rightward momentum. An answer that adds 0.40 to an unsigned velocity change would miss the reversal.

As an independent check, acceleration = F/m = 1.5/0.60 = 2.5 m/s². Hence v = −0.40 + 2.5(0.80) = +1.6 m/s. Agreement between the impulse route and this constant-acceleration check confirms the arithmetic under the same force assumption.

2. Define the system before conserving momentum

A system is the set of objects you choose to analyse together. For two colliding carts, the contact forces between them are internal. Their impulses are equal and opposite, so they cancel in the total momentum balance. Each cart’s own momentum can still change substantially.

The total momentum remains constant only when the net external impulse on that system is zero or negligible over the interval. A short collision on a nearly frictionless level track often permits that approximation. A cart hitting a fixed bumper is not automatically an isolated one-cart system: the bumper exerts an external impulse on it.

Write the condition in words before the equation. For two carts with negligible external impulse, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. All velocities must use the same chosen direction. Conservation is a statement about the total, not a claim that each momentum stays unchanged.

Worked case 3: carts that stick together

Cart A has mass 0.80 kg and moves right at 1.5 m/s. Cart B has mass 1.20 kg and is initially at rest. They collide and stick. The net external horizontal impulse during contact is negligible. Find their common final velocity.

Solution: Choose the two carts as the system and right as positive. Initial total momentum = 0.80(1.5) + 1.20(0) = 1.20 kg·m/s. After contact they share one velocity v and have combined mass 2.00 kg. Thus 2.00v = 1.20 and v = +0.60 m/s.

Check individual changes: A ends with momentum 0.80(0.60) = 0.48 kg·m/s, a change of −0.72 kg·m/s. B ends with 1.20(0.60) = 0.72 kg·m/s, a change of +0.72 kg·m/s. The changes cancel, as expected for the internal interaction.

Do not also assume kinetic energy is conserved. Initial kinetic energy = ½(0.80)(1.5)² = 0.90 J. Final kinetic energy = ½(2.00)(0.60)² = 0.36 J. The decrease is 0.54 J. Energy has been transferred into other forms during the sticking interaction; momentum conservation does not require unchanged kinetic energy.

3. Diagnose signs and missing information

A negative answer is not automatically wrong. Interpret it against the direction you chose. If you reverse the positive direction consistently, the signs reverse but the physical prediction does not. Avoid changing the sign convention halfway through a calculation.

Also check whether the available equation is enough. If two carts separate after a collision and both final velocities are unknown, momentum conservation supplies only one equation for two unknowns. You need an additional condition or measurement, such as one final velocity. Saying “they collide” alone does not establish an elastic collision.

Finally, distinguish a force pair from a net force. Cart A’s force on B and cart B’s force on A act on different carts. They cancel in the combined-system momentum balance, but cannot be cancelled when drawing the forces acting on cart A alone.

Three tasks with complete solutions

Task 1: convert mass and handle reversal

A 150 g model object changes velocity from +2.4 m/s to −1.6 m/s in 0.12 s. Find signed impulse and average net force.

Solution: Mass = 0.150 kg. J = 0.150[−1.6 − (+2.4)] = −0.600 N·s. Average net force = −0.600/0.12 = −5.0 N. The force and impulse point in the negative direction; the impulse magnitude is 0.600 N·s.

Task 2: opposite-moving carts stick

A 0.90 kg cart moves right at 1.4 m/s and a 0.60 kg cart moves left at 0.60 m/s. They stick with negligible external horizontal impulse. Find the final velocity.

Solution: Taking right positive, initial momentum = 0.90(1.4) + 0.60(−0.60) = 1.26 − 0.36 = 0.90 kg·m/s. Total mass = 1.50 kg, so v = 0.90/1.50 = +0.60 m/s, rightward. Subtracting the opposing momentum is essential.

Task 3: include an external impulse

A chosen system has initial momentum +0.70 kg·m/s. It receives a net external impulse of −0.20 N·s. What is its final momentum, and can you set initial equal to final momentum?

Solution: Final momentum = +0.70 − 0.20 = +0.50 kg·m/s. You cannot set the two totals equal because the external impulse is explicitly nonzero. Use p_final = p_initial + J_external instead.

Frequently asked questions

Why use velocity rather than speed?

Momentum has direction. Signed velocity preserves that direction in one dimension, while speed alone cannot distinguish a rebound from continued motion in the same direction.

What does a negative impulse mean?

It points in the negative direction you chose. Its magnitude remains positive. A negative sign does not by itself mean that the calculation is wrong.

Are N·s and kg·m/s interchangeable?

Yes. Since one newton equals one kg·m/s², multiplying by seconds gives kg·m/s. Impulse and momentum change therefore have equivalent units.

When may I conserve total momentum?

When the net external impulse on the chosen system is zero or negligible during the interval. Internal collision impulses alone do not change the system total.

Do sticking carts conserve kinetic energy?

Not generally. They can conserve total momentum while kinetic energy is transferred to deformation, heating and other forms. Use their common final velocity in the momentum balance.

Does equal and opposite force mean zero force on each cart?

No. The two forces act on different carts. They cancel when summing the combined system interaction, not when calculating the net force on one cart.

Can momentum conservation find two unknown final velocities?

Not by itself in a general one-dimensional two-cart collision. It gives one equation, so another condition or measurement is needed if both final velocities are unknown.

Continue revision

Explore the Physics topical revision collection; access to individual resources may vary. Reference checks: OpenStax on impulse and system momentum conservation. The cart cases and solutions are original.