I don't quite understand why they must be perpendicular. :confused: Even if they are perpendicular, in what way should I use the information?
Anyway, the energy can't go into the strings, can it? Furthermore it was assumed that the masses have no internal structure, so the energy can't be loss...
Ah I see...
So the equations would be:
(a) (3√3)/2 v3 + √3 vx = vy
(b) v= ux +2vx +3v3 \Rightarrow ux= v -2vx -3v3
(c) 0 = uy +2vy \Rightarrow uy² = 4vy²
(d) v² = ux² +uy² +2vx² +2vy² + 3v3²
I start by putting (c) into (d) to become (e). Then (a) into (e) to become (f) which is
v² = ux²...
Hi tiny-tim :),
I noticed that since v -v1 is along the string connecting m and 2m, then 2vx +2vy +3v3 is also along the string. So The vertical component divided by the horizontal component will give tan(60 deg)= sqrt(3).
2vy/(2vx +3v3) = sqrt(3) => vy = 3sqrt(3)/2 v3 + sqrt(3) vx which is...
Hmm...
If the velocity of 3m is v3, is it (3v3/2)\sqrt{3} + vx\sqrt{3} = vy ??
And for linear momentum it will be v = v1 +2vx +2vy +3v3
For kinetic energy it would be v² = v1² +2vx² +2vy² +3v3²
And I think v1² = (v -3v3 -2vx)² + (2vy)²
:confused:
Hi tiny -tim.
I'm not sure where the velocities of 3m and 2m might be pointing at, so I can't figure out which distance to multiply to the momentum to give the angular momentum. :frown:
My guess is that the initial velocity for 3m will be horizontal. But for 2m, my intuition tells me it's...
Hi tiny-tim :biggrin:
Here's my attempt:
The tension in string connecting m and 2m both will have no contribution to the moment as the perpendicular distance is zero. The tension in that connecting 2m and 3m both will also give zero net moment because the forces are equal and opposite and the...
Hi everyone, this is a problem I posted here a month ago but it wasn't given any attention from the helpers here. I still cannot solve it. So I will greatly appreciate any help.
1. Homework Statement
Three small balls of masses m, 2m and 3m are placed on a smooth horizontal surface so that...
Homework Statement
A small mass m slides on a hemisphere of mass M and radius R is also free to slide horizontally on a frictionless table. An imaginary vertical line is drawn from the center of the hemisphere to its highest point, where the small mass is originally placed at rest. The angle A...
Hi! Thanks for that tip though I still can't solve it as I'm a rookie. :redface:
These are the steps I took:
Which I believe to have a big mistake somewhere. But never mind my attempt, what would be another (even easier) way to find the time taken for the bubble to rise up?
Homework Statement
The initial volume and depth of the gas bubble is V0 and H respectively. The bubble has a mass M and the gas can be assumed to be ideal. The gravitational field strength is a constant g. And the density of water is Dw.
What then, is the time taken for a bubble initially at...
[SIZE="4"]Sorry! I think I figured the mistake out right after I posted! This thread can be closed. :redface:
[SIZE="1"]I imagine a train powered by an engine of this sort:
A region of hotter temperature evaporates some water and turn the wheels, and the evaporated water is at a lower...
Let me try to 'show' it:
Consider two particles of same mass and one of them approach the other (not necessarily head on) with one of them originally at rest.
The total momentum before and after the collision will thus be:
p1 = p'1 + p'2
The sum of the kinetic energy of the center...
Thanks for the replies, I think I know what being linear means now. :smile:
However, I still do not understand how does "A uniform and rectilinear motion in K must also be uniform and rectilinear in K' " imply that the transformations are linear. I have read the other thread started by facenian...