# General Definitions of Impulse and Work

1. Nov 19, 2011

### motion_ar

In classical mechanics, if we consider the motion of a particle of mass $m$, then

The mass $m$ is $constant$

The vector $\vec{c}$ can be: $\ldots$ or $\vec{r}$ or $\vec{v}$ or $\vec{a}$ or $\vec{j}$ or $\ldots$

$\vec{c}_1 = d\vec{c} / {dt}$

$\vec{c}_2 = d^2 \, \vec{c} / {dt^2}$

Definition of Impulse $\vec{c}$ $\, ( \vec{I}_{\vec{c}} )$

$$\vec{I}_{\vec{c}} \; = \int_a^b m \, \vec{c}_1 \, dt \; = \Delta \; m \, \vec{c}$$
$$\vec{I}_{\vec{c}} \; = \Delta \; \vec{P}_{\vec{c}}$$
where:

$$\Delta \; \vec{P}_{\vec{c}} \; = \Delta \; m \, \vec{c}$$

If $\; \vec{c}_1 = 0$

$$\rightarrow \; \; \Delta \; \vec{P}_{\vec{c}} \; = 0$$
$$\rightarrow \; \; \vec{P}_{\vec{c}} \; = constant$$

Definition of Work $\vec{c}$ $\, (W_{\vec{c}})$

$$W_{\vec{c}} \; = \int_a^b m \; \vec{c}_2 \cdot d\vec{c} \; = \Delta \; {\textstyle \frac{1}{2}} \, m \; \vec{c}_1^2$$
$$W_{\vec{c}} \; = \Delta \; T_{\vec{c}_1} + \Delta \; V_{\vec{c}} \; = \int_a^b m \; \vec{c}_{2n\vec{c}} \cdot d\vec{c}$$
where:

$$\Delta \; T_{\vec{c}_1} = \Delta \; {\textstyle \frac{1}{2}} \, m \; \vec{c}_1^2$$
$$\Delta \; V_{\vec{c}} = - \int_a^b m \; \vec{c}_{2\vec{c}} \cdot d\vec{c}$$
$$\vec{c}_2 = \vec{c}_{2\vec{c}} + \vec{c}_{2n\vec{c}}$$
$$\vec{c}_{2\vec{c}} \; \; \; is \; \; function \; \; of \; \; \; \vec{c}$$
$$\vec{c}_{2n\vec{c}} \; \; \; is \; \; not \; \; function \; \; of \; \; \; \vec{c}$$

If $\; \vec{c}_{2n\vec{c}} = 0$

$$\rightarrow \; \; \Delta \; T_{\vec{c}_1} + \Delta \; V_{\vec{c}} \; = 0$$
$$\rightarrow \; \; T_{\vec{c}_1} + V_{\vec{c}} \; = constant$$

If $\; \vec{c}_2 = 0$

$$\rightarrow \; \; \Delta \; T_{\vec{c}_1} \; = 0$$
$$\rightarrow \; \; T_{\vec{c}_1} \; = constant$$

Last edited: Nov 19, 2011