Define momentum as p = mv, define impulse as a force acting over a time, and relate impulse to the change in momentum (MA STE Introductory Physics, Motion and Forces).

What this topic is asking

Momentum is how physics measures "amount of motion," and the Massachusetts Introductory Physics MCAS builds the whole collisions module on it. You must define momentum as $p = mv$, define impulse as a force acting over a time, and relate the impulse to the change in momentum. The crosscutting idea is cause and effect: a force applied for a time is the cause, a change in momentum is the effect. This sets up conservation of momentum, the centerpiece of the module.

Momentum

Momentum captures both how much is moving and how fast. A slow-moving truck and a fast-moving bullet can have similar momentum, because the truck's huge mass offsets its low speed. Because velocity is a vector, momentum is too, so direction matters: two objects moving toward each other have momentum in opposite directions, which become opposite signs in a calculation.

The reference-sheet formula is

where $p$ is the momentum (kg m/s), $m$ is the mass (kg), and $v$ is the velocity (m/s). Doubling the mass or the velocity doubles the momentum.

Impulse

Impulse links force to momentum the way Newton's second law links force to acceleration. In fact it is the second law in another form: a force changes velocity (acceleration), and through the mass it changes momentum. The practical message the MCAS tests:

• To produce a given change in momentum, you can use a big force for a short time or a small force for a long time.
• Extending the time over which a momentum change happens reduces the force needed, which is the principle behind airbags, crumple zones, and bending your knees when you land.

Change in momentum and direction

When an object speeds up or slows down in a straight line, the change in momentum is just the final momentum minus the initial momentum. The subtlety the MCAS loves is a direction reversal, as when a ball bounces back. If you call the initial direction positive, the rebound velocity is negative, so the change in momentum is larger than for simply stopping the object. A ball that bounces back experiences about twice the momentum change of a ball that is merely caught.

Reference-sheet note

The reference sheet prints momentum as $p = mv$. It does not print a separate impulse equation, so on this test impulse is handled qualitatively (a force over a time produces a change in momentum) or by computing the change in momentum directly from $p = mv$ before and after. Knowing that impulse equals the change in momentum is something you recall.

Try this

Q1. A $2.0$ kg object moves at $5.0$ m/s. Calculate its momentum. [2]

• Cue. $p = mv = (2.0)(5.0) = 10$ kg m/s in the direction of motion.

Q2. Explain why bending your knees when you land from a jump reduces the force on your legs. [2]

• Cue. Bending your knees extends the time over which your momentum drops to zero; a longer time for the same change in momentum means a smaller force.