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Next: Conservation of Momentum Up: The Emergence of Mechanics Previous: Antiderivatives

Impulse and Momentum

Multiplying a scalar times a vector is easy, it just changes its dimensions and length - i.e. it is transformed into a new kind of vector with new units but which is still in the same direction. For instance, when we multiply the vector velocity $\mbox{\boldmath$\vec{v}$\unboldmath }$ by the scalar mass m we get the vector momentum $\mbox{\boldmath$\vec{p}$\unboldmath } \equiv m \,
\mbox{\boldmath$\vec{v}$\unboldmath }$. Let's play a little game with differentials and the SECOND LAW:

\begin{displaymath}\mbox{\boldmath $\vec{F}$\unboldmath } =
{d\mbox{\boldmath $\vec{p}$\unboldmath } \over dt} . \end{displaymath}

Multiplying both sides by  dt  and integrating gives

 \begin{displaymath}\mbox{\boldmath$\vec{F}$\unboldmath } \,
dt = d\mbox{\boldm . . . 
 . . . p}$\unboldmath }
- \mbox{\boldmath$\vec{p}$\unboldmath }_0 .
\end{displaymath} (11.4)

The left hand side of the final equation is the time integral of the net externally applied force $\mbox{\boldmath$\vec{F}$\unboldmath }$. This quantity is encountered so often in Mechanics problems [especially when $\mbox{\boldmath$\vec{F}$\unboldmath }$ is known to be an explicit function of time, $\mbox{\boldmath$\vec{F}$\unboldmath }(t)$] that we give it a name:

 \begin{displaymath}\int_{t_0}^t \mbox{\boldmath$\vec{F}$\unboldmath }(t) \, dt \ . . . 
 . . .  due to applied force }
\mbox{\boldmath$\vec{F}$\unboldmath }
\end{displaymath} (11.5)

Our equation can then be read as a sentence:
$\textstyle \parbox{8cm}{\noindent
\lq\lq The {\sl impulse\/} created by the net ext . . . 
 . . . pplied to
a system is equal to the {\sl momentum change\/} of the system.



 
next up previous
Next: Conservation of Momentum Up: The Emergence of Mechanics Previous: Antiderivatives
Jess H. Brewer - Last modified: Sat Nov 14 12:39:41 PST 2015