Also, the method of solution that you used is the long way around. While valid, there is really no need to resort to Laplace transforms for something this basic.
\(
F(t) = I_o \delta (t)
\)
We know that
\(
\int_{-\infty}^{+\infty} \delta (t) dt = \int_{0^-}^{0^+} \delta (t) dt = 1 \;\; \text{(units?)}
\)
Notice that we have a units problem since if δ(t) is dimensionless then when we integrate over all time we end up with a result that has to have units of time (since the units of an integral are equal to the units of the integrand and dt carries units of time. Thus δ(t) is NOT dimensionless and carries units that are the reciprocal of the independent variable (in our case that is inverse-time). But is the unit 1/seconds? 1/hours? 1/millennia? The only way this can be resolved is to require the use of a scaling coefficient that carries the necessary units, namely whatever the units of the equation need to be multiplied by a unit of time.
Since we need F(t) to have units of force, we need Io to have units of force multiplied by units of time, so something like newton·seconds.
\(
\int_{-\infty}^{+\infty} I_0 \delta (t) dx = I_0 \int_{0^-}^{0^+} \delta (t) dt = I_0
\)
Note that this scaling coefficient is NOT optional (except in a purely dimensionless world), no matter how much textbook authors love to be sloppy and ignore it (keep in mind that most textbook authors, particularly math textbook authors, have never worked in the real world).
If we want the result as a function of time, then we need to do this slightly differently and only integrate from negative infinity up to the time of interest.
\(
\int_{-\infty}^{t} I_0 \delta (\tau) d\tau = I_0 \int_{-\infty}^{t} \delta (\tau) d\tau = I_0 u(t)
\)
where u(t) is the unit step function.
The change of variable in the integral from t to τ is merely to avoid confusion with the variable t in the integration limit. Since the variable of integral is a dummy variable, we can replace it with any variable we want.
We can apply this to our example through direct application of Newton's Second Law:
\(
F(t) \; = \; m \cdot a(t) \; = \; m \cdot \frac{dv(t)}{dt}
\;
\frac{dv(t)}{dt} \; = \; \frac{F(t)}{m} \; = \; \frac{I_0}{m} \delta (t)
\;
dv(t) \; = \; \frac{I_0}{m} \delta (t) dt
\;
v(t) \; = \; \frac{I_0}{m} \int_{-\infty}^{t} \delta (\tau) d\tau
\;
v(t) \; = \; \frac{I_0}{m} u(t)
\)
Notice how the units work out. Io has units of newton·seconds and a newton has units of mass over acceleration which work out to be (mass·distance)/(time²). Thus I0 divided by mass has units of distance/time, which is a unit of velocity.
So given this form for v(t), what are the expressions for x(t), p(t), and E(t)?
\(
F(t) = I_o \delta (t)
\)
We know that
\(
\int_{-\infty}^{+\infty} \delta (t) dt = \int_{0^-}^{0^+} \delta (t) dt = 1 \;\; \text{(units?)}
\)
Notice that we have a units problem since if δ(t) is dimensionless then when we integrate over all time we end up with a result that has to have units of time (since the units of an integral are equal to the units of the integrand and dt carries units of time. Thus δ(t) is NOT dimensionless and carries units that are the reciprocal of the independent variable (in our case that is inverse-time). But is the unit 1/seconds? 1/hours? 1/millennia? The only way this can be resolved is to require the use of a scaling coefficient that carries the necessary units, namely whatever the units of the equation need to be multiplied by a unit of time.
Since we need F(t) to have units of force, we need Io to have units of force multiplied by units of time, so something like newton·seconds.
\(
\int_{-\infty}^{+\infty} I_0 \delta (t) dx = I_0 \int_{0^-}^{0^+} \delta (t) dt = I_0
\)
Note that this scaling coefficient is NOT optional (except in a purely dimensionless world), no matter how much textbook authors love to be sloppy and ignore it (keep in mind that most textbook authors, particularly math textbook authors, have never worked in the real world).
If we want the result as a function of time, then we need to do this slightly differently and only integrate from negative infinity up to the time of interest.
\(
\int_{-\infty}^{t} I_0 \delta (\tau) d\tau = I_0 \int_{-\infty}^{t} \delta (\tau) d\tau = I_0 u(t)
\)
where u(t) is the unit step function.
The change of variable in the integral from t to τ is merely to avoid confusion with the variable t in the integration limit. Since the variable of integral is a dummy variable, we can replace it with any variable we want.
We can apply this to our example through direct application of Newton's Second Law:
\(
F(t) \; = \; m \cdot a(t) \; = \; m \cdot \frac{dv(t)}{dt}
\;
\frac{dv(t)}{dt} \; = \; \frac{F(t)}{m} \; = \; \frac{I_0}{m} \delta (t)
\;
dv(t) \; = \; \frac{I_0}{m} \delta (t) dt
\;
v(t) \; = \; \frac{I_0}{m} \int_{-\infty}^{t} \delta (\tau) d\tau
\;
v(t) \; = \; \frac{I_0}{m} u(t)
\)
Notice how the units work out. Io has units of newton·seconds and a newton has units of mass over acceleration which work out to be (mass·distance)/(time²). Thus I0 divided by mass has units of distance/time, which is a unit of velocity.
So given this form for v(t), what are the expressions for x(t), p(t), and E(t)?