I don't have any specific application. I was thinking in taking the percentage of ##\Delta T## from ##T_i##, and if it's ##<10\%## it's acceptable, but I'm not sure if it could be a valid criteria.
How small should ##\Delta T## be in a collision to be considered elastic? In elastic collisions ##\Delta T =0##, but as far as I know, just atomic collisions are considered perfectly elastic. Then, which criterias are used to considere a collision between two objects elastic?
Homework Statement
A particle of a mass ##m## is embedded in a circular rail, (radius: ##R##), without any friction. In a given moment, the particle finds itselfs without velocity at point C, and a force is applied on the rail, which starts moving with an ## \vec A ## constant acceleration. Use...
Thank you so much! I finally get it, all of my questions were between pages 149 and 150. We will see conservation of energy in three classes, that's why I was so confused about that.
Sorry, sorry, I'm really sorry, i just made a lot of typos, I meant
$$\frac 1 2 Mv^2+\frac 1 2 Cx^2 .$$
That's because
$$ \Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here ##\Delta V=0.## And we also know that
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} M \left(...
(5.12) integral. And yes, i just realized that I cropted it in a wrong way, but basically: $$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2.$$
Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$
and that's only true if ##−Mv_0^2=Cx^2.##
I'm just...
$$ -M v^2_0 = C x^2 .$$ Therefore $$ v_0 = \sqrt {\frac {-C} {M}} x .$$
Btw, the text defines ##w_0 = \sqrt {\frac {C} {M}}, v_0 = w_0 A \cos(\phi). ## Then if ##v_0 = \sqrt {\frac {-C} {M}} x, x = A \cos(\phi). ## But we got that ## x = A \sin(\phi),## so ##\cos(\phi)## would be equal to...
I understand the rest of your argument, but this was my original question and i still not getting why this is a possible solution, i mean, i can´t made a proper demostration or something. Is like I'm not taking into account the constants that integrals implies.
Thank you for answering! I don't think so tho. Of course he did the FBD but the sum of the external forces is to check energy conservation, so i haveto take into account the whole system.
The first pic is the FBD. N: normal force; P: mg; T: tension force.
In the secon one, the sum of the...
When my professor sums external forces (to know if the impulse is constant or not) he always includes tension.
I will use as an example pic related. If my system includes the two particles and the thread, woudn't tension being a internal force?
I'm sorry if it's a silly question, but I also...