Alright, there is this three-part problem I am having some trouble with.
Suppose we have a particle of mass m, initially at rest at the origin, that receives an impulse of p at time $t = 0$.
(a) Find the solution $x_\epsilon (t)$ of the problem $mx'' = pd_{0, \epsilon} (t); x(0) = x'(0) = 0$.
This took me a bit to work out, but I solved it by treating $d_{0, \epsilon} (t)$ as the combination of a few unit-step functions. My reasoning is as follows: $$mx'' = p \frac{1}{\epsilon} [ u(t-a) - u(t-\epsilon)] = p \frac{1}{\epsilon} [ u(t) - u(t-\epsilon)]$$
Taking the Laplace transform of this yields the following $$ms^2 X(s) = p \frac{1}{\epsilon} [ \frac{1}{s} - \frac{e^{-\epsilon s}}{s} ] \rightarrow \frac{m\epsilon X(s)}{p} = \frac{1}{s^3} - \frac{e^{-\epsilon s}}{s^3}$$
Simplifying this equation and taking the inverse Laplace transform yields $$m x_\epsilon (t) = \frac{p(t^2 - u(t - \epsilon)(t - \epsilon)^2)}{2\epsilon}$$
Be aware that this part of the question IS NOT the source of my confusion, but rather required to understand the parts that do lend me some confusion.
(b) Show that $\lim_{\epsilon\to\infty} x_\epsilon (t)$ agrees with the solution of the problem $mx'' = p\delta (t); x(0) = x'(0) = 0$
I realized that $mx_\epsilon (t) = p\delta (t)$ can be rewritten as $mx_\epsilon (t) = p \lim_{\epsilon\to\infty} x_\epsilon (t)$, which thus implies that for $t \gt \epsilon$, then $mx_\epsilon (t) = \frac{p(2\epsilon t - \epsilon ^2)}{2\epsilon}$, and hence $mx_\epsilon (t) \rightarrow pt$ as $\epsilon \rightarrow 0$.
I looked to the back of the book and they offered the same reasoning. My question: HOW does this prove that $\lim_{\epsilon\to\infty} x_\epsilon (t)$ agrees with the solution? So what if $mx_\epsilon (t) \rightarrow pt$ as $\epsilon \rightarrow 0$?
(c) Show that $mv = p$ for $t \gt 0$ ($v =x'$).
For this part of the question, I had no idea. I don't even really know the intuition behind what they're asking. The back of the book just states "$mv = (mx)' = (pt)' = p$".
How do I solve part (c) and what is the intuition behind solving it?
Part (b) is much like your companion question and has a similar answer. Saying that $x^{\prime}(0)=0$ doesn't make a whole lot of sense because there is a jump discontinuity of $x^{\prime}(t)$ of magnitude $\Delta x^{\prime}=\frac pm$ so if $x^{\prime}(0^-)=0$, then $x^{\prime}(0^+)=\frac pm$. As with the companion question, there is no jump discontinuity in $x(t)$, so $x(0^+)=x(0^-)=0$.
We can integrate the second differential equation for $t>0$ since $mx^{\prime\prime}(t)=0$, it follows that $x(t)=c_1+c_2t$ and comparing initial conditions, $x(0^+)=c_1=0$ and $x^{\prime}(0^+)=c_2=\frac pm$, so $x(t)=\frac pmt$.
You had $x_{\epsilon}(t)=\frac pmt-\frac{p\epsilon}{2m}$, so the two solutions agree as $\epsilon\rightarrow0$
Part (c) is just differentiation. $x(t)=\frac pmt$, so $\frac d{dt}(mx)=m\frac d{dt}x=mx^{\prime}=mv=\frac d{dt}\left(\frac pmt\right)=\frac pm$.