Impulsive forces and finite forces

Question

I was reading a message on collisions and it stated that during a collision the finite forces like gravitational force are ignored and we only take into account the impulsive forces, In case the impulsive forces act in the horizontal direction we can conserve the momentum in the vertical direction. The meaning of the passage was clear to me but I want to know that why do we ignore finite force. What I think is that the finite forces are very small as compared to the infinite forces so we ignore them, but this sounds more like an approximation. Please tell me what is the correct reason.

Answer 1

It's more or less like you say, if the collision occurs in a "very small" interval of time.

We could treat this "very small" interval $[t^{coll}_1, t^{coll}_2]$ as a discontinuity in a smooth motion. We can model this discontinuity using a Dirac delta for impulsive forces when analyzing the state of the system just before and after the collision. As an example, looking at the momentum

$\dfrac{d \mathbf{Q}}{d t} = \mathbf{R}^{ext}$,

being $\mathbf{Q}$ the momentum of the system, $\mathbf{R}^{ext}$ the external forces acting on the system. We can write the external forces as the sum of the forces due to the impact and the other kind of forces, as

$\mathbf{R}^{ext}(t) = \mathbf{F}^{coll}(t) + \mathbf{F}^{smooth}(t) \\ \qquad \quad = \mathbf{I}^{coll} \delta(t-t_{coll}) + \mathbf{F}^{smooth}(t)$,

assuming that $\mathbf{F}^{smooth}(t)$ is finite. If we integrate over the interval $[t^{coll}_1, t^{coll}_2]$, we get

$\displaystyle\mathbf{Q}_2 - \mathbf{Q}_1 = \int_{t^{coll}_1}^{t^{coll}_2} \dfrac{d \mathbf{Q}}{d t} = \int_{t^{coll}_1}^{t^{coll}_2} \mathbf{R}^{ext}(t) dt = \\ \displaystyle \qquad \qquad =\int_{t^{coll}_1}^{t^{coll}_2} \left[ \mathbf{I}^{coll} \delta(t-t_{coll}) + \mathbf{F}^{smooth}(t) \right] dt = \\ \displaystyle \qquad \qquad = \mathbf{I}^{coll} + \int_{t^{coll}_1}^{t^{coll}_2} \mathbf{F}^{smooth}(t) dt $,

and if $t_2^{coll} - t_1^{coll} \rightarrow 0$, the last integral goes to zero, i.e., $\mathbf{Q}_2 - \mathbf{Q}_1 = \mathbf{I}^{coll}$, being $\mathbf{Q}_1$, $\mathbf{Q}_2$ the momentum of the system just before and after the collision.

Remark. When you define a Dirac delta in time, it has physical dimension of $\frac{1}{\text{time}}$, and thus the force the the collision is defined as $\mathbf{F}^{coll}(t) = \mathbf{I}^{coll} \delta(t-t_{coll})$, being $\mathbf{I}^{coll}$ the impulse during the collision.