Define momentum as $p = mv$, define impulse as $F\Delta t$, and apply the impulse-momentum relationship $F\Delta t = \Delta p$ to calculate force, time or change in momentum.

What this topic is asking

Momentum measures the "quantity of motion" of an object, and impulse measures the effect of a force acting over time. The Physical Setting/Physics course asks you to calculate momentum with $p = mv$, to understand impulse as $F\Delta t$, and to link them with the impulse-momentum relationship $F\Delta t = \Delta p$. The Regents tests straightforward momentum calculations, impulse problems, and the safety applications (airbags, crumple zones, follow-through) that the relationship explains.

Momentum

Momentum captures how hard it is to stop a moving object: it grows with both mass and speed. A slow truck and a fast car can carry the same momentum if the products $mv$ match. Because momentum is a vector, its direction matters, and a sign convention is essential whenever an object reverses direction.

Impulse

Impulse is what a force "delivers" over time. The same change in motion can be produced by a large force for a short time or a small force for a long time, as long as the product $F\Delta t$ is the same. This trade-off is the heart of the topic.

The impulse-momentum relationship

To use it, identify the change in momentum (with signs, treating velocity as a vector) and relate it to the force and time. A ball that bounces back has a larger momentum change than one that stops, because its velocity reverses, so $\Delta p = m(v_f - v_i)$ includes the sign change.

Why extending the time reduces the force

This is a favorite Regents application, tested both as a calculation and as an explanation. The reasoning is always the same: the momentum change is set by the masses and velocities, so the only way to soften the force is to spread that change over more time.

Reference Tables note

Both $p = mv$ and $F\Delta t = \Delta p$ are printed in the Mechanics section of the Reference Tables. The relationship to Newton's second law is not stated explicitly but is worth knowing for explanations. The conservation of momentum, which is the next dot point (conservation of momentum), is not printed as an equation, so you recall it.

Try this

Q1. A $2.0$ kg object moves at $6.0$ m/s. Calculate its momentum. [2 points]

• Cue. $p = mv = (2.0)(6.0) = 12$ kg m/s in the direction of motion.

Q2. Explain why an airbag reduces the force on a passenger in a crash. [2 points]

• Cue. The airbag increases the time over which the passenger's momentum changes; since $F = \Delta p / \Delta t$, a longer time means a smaller force.