It seems to me that you understand it. I believe the problem is just attempting to familiarize you with the use. If that is t^2 and t^3, then you are correct.
Ohhh, I see. It seems the moment is just the torque about the center of mass. That make sense. Never seen that terminology, thanks for sharing.
Instead of looking at you coordinate system where x-axis is parallel to the bar and y-axis perpendicular to it try this instead:
Orient your...
Sorry, I am starting to feel like I am interjecting. Lets look at it one more way, when integrating your equation you are actually using definite integrals, so the constant is defined as explained above. Here is an example, without being too explicit
\int_i^f dv = \int_i^f dx,
where i and...
This is correct for the final answer. x is the distance to Earth, plugging in the appropriate numbers returns 64 days. (To the amount of sigfigs I was using, it was calculated at about 64.7 days)
DH has good information as to realize what your constant should be. Attention should be put into...
This is in a 2nd semester elementary physics text?
To your problem, it doesn't necessarily mean there is no net force on your test particle. Imagine this problem in one-dimension. (This will create the greatest difference in force) If your test particle was close to the atom it would...
Nice diagram!
I am slightly confused about your statement of the problem though. You say your finding the moment and you define the moment as the time derivative of angular momentum, which is torque.
Anyways, I think you know where your going with he problem and your just curious how you find...
Ok, after analyzing the problem slightly more I see one flaw in your argument. I did get 80 m/s too. The flaw: you assume that v_y=0 when hitting the target. When, in fact, all the question is asking is to hit the top of the cliff. You could be past the highest point or before the highest point...
They are approaching the problem using
F(x) = -\frac{dU(x)}{dx}
that is why they are giving the answer as negative. So, yes, the derivative is positive but the force does carry a negative.
When taking the derivative you are actually talking about infinitesimal displacements, so you can...
I am curious as to why you attempted this via 4-momentum? Is the text you are working with promote that use?
Anyways, the way I would approach this is a lot simpler, you do not need 4-momentum. Think of this by looking at the conservation of momentum thus,
P1=P2+P3
and it follows that...