calculus-based physics for scientists and engineers

Impulse and Momentum in 1D


Momentum certainly is simple enough to write down: $$\vec{p} = m\vec{v}$$ Your intuition about momentum is pretty good, too. Something with a lot of momentum is harder to slow down than something with little momentum. Forces acting over time change the momentum of an object. Let me walk you through the particulars.


The impulse is the integral of the force over a time interval. The total impulse, or the impulse of the net force on an object is equal to the change in momentum of the object. $$\int_{t_i}^{t_f}\vec{F}(t)\, dt = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$$ If the force is constant in time, this simplifies to $$\vec{F}\Delta t = \Delta \vec{p}$$

The time-average of a force is the constant force that would give you the same impulse over the same time interval as the original force, and is given by: $$\vec{F}_{avg} = \frac{1}{\Delta t}\int_{t_i}^{t_f}\vec{F}(t)\, dt$$ and by substituting the definition of the impulse, you get $$\vec{F}_{avg}\Delta t = \Delta \vec{p}$$ regardless of the time characteristics of the original force. As one might have guessed, the time average of a constant force is itself.

Impulse and Momentum Examples

This is best explained with some examples.


Do Now!   Do these exercises immediately.



Required Problems


1. A 10-kg ball is at rest. A 0.1-kg ball moving at 50 m/s hits the larger ball and bounces directly back with a speed of 40 m/s. What is the resulting speed of the heavy ball? Assume during the collision the force that one ball exerted on the other rose linearly from zero to a maximum in 5 ms, then fell linearly back to zero in another 5 ms. What was the maximum force one ball exerted on the other?

Optional Problems