# What is an example of an impulse practice problem?

Jun 16, 2015

First of all, using the definitions $a = \frac{\mathrm{dv}}{\mathrm{dt}}$ and $F = m a$, the definition of impulse is:

$I = \int F \mathrm{dt} = \int m a \mathrm{dt} = m \int \frac{\mathrm{dv}}{\cancel{\mathrm{dt}}} \cancel{\mathrm{dt}}$

$I = m \int \mathrm{dv}$

$I = m \Delta v$

...whereas $p = m v$

Thus, an impulse causes an object to change velocity as a result of an impact. Or, it can be said that it is summation of the infinite instances of instantaneous force applied over a small amount of time.

A nice example is right when a golf club hits a golf ball. Let's say there was a constant impulse for $0.05 s$ on a golf ball started at rest. If the golf ball is $45 g$ and its velocity after it leaves contact with the golf club is $50 \frac{m}{s}$, what was the impulse?

$I = m \Delta v$

$I = \left(0.045 k g\right) \left(50 \frac{m}{s} - 0 \frac{m}{s}\right) = 2.25 k g \cdot \frac{m}{s}$

The average force applied onto the golf ball in these $0.05$ seconds would be what?

${F}_{a v g} = \frac{1}{N} {\sum}_{i = 1}^{N} {F}_{i} = \frac{F \Delta t}{\Delta t} = \frac{I}{\Delta t} = \frac{2.25 N \cdot s}{0.05 s} = 45 N$

And I'll leave this one for you:

If the golf ball leaves the tee at a ${45}^{o}$ from the horizontal, how long does it take for it to cross the spot that lines up horizontally with its initial position?