# Mechanics

calculus-based physics for scientists and engineers

## Impulse and Momentum in 1D

### Momentum

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.

### Impulse

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.

### Solution

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?