:46289104 hello, yes just working on your problem.
Oblique Collision between 3 Circles : The difficulty with this problem is that there is a simultaneous collision between 3 objects. Impulse is transferred along the common normal, which is the line joining the centres of the larger and smaller circles, which is at an angle of $\theta$ to the horizontal, where $\sin\theta=1/3$.
The smaller circles are initially at rest, so their final velocities are along the common normal. Let this final velocity be $v$ and the initial velocity of the larger circle be $U$. The larger circle has 2x the radius and therefore its mass is $4m$ where $m$ is the mass of the smaller circles. Conservation of linear momentum in the horizontal direction gives $$4mU=2mv\cos\theta$$ $$v=2U\sec\theta$$.
We can account for the simultaneous collision of the larger circle with both smaller circles by treating the larger circle of mass $4m$ as two circles each of mass $2m$ and each colliding with one of the smaller circles of mass $m$. The circles of mass $2m$ cannot be stationary after this collision : they must rebound vertically, then collide with each other and coalesce.
The vertical momentum with which each rebounds must equal the final vertical momentum of the small circle : $$2mV=mv\sin\theta$$ $$V=\frac12 v\sin\theta$$
Now we have enough information to calculate the Coefficient of Restitution, $e$. The relative speed of approach along the common normal is $U\cos\theta$. The relative speed of separation is $$v+V\sin\theta=v(1+\frac12\sin^2\theta)=2U(1+\frac12\sin^2\theta)/\cos\theta$$
@harambe Chat is not very convenient for such a long question. Would you like to post the question on physics.qandaexchange.com/?qa=questions? Then it will be easier to edit the question and answers.
@harambe Sorry I am not finding your question. Did you post on Physics Q&A Exchange?
@harambe If $u$ is the initial velocity of the larger disk and $v$ is final velocity of each smaller disk then by conservation of momentum in horizontal direction $u=2v\cos\theta=4v\sqrt2/3$ so $v=3u/4\sqrt2$. (All 3 disks have the same mass.)
Yea. I was intislly confused about taking the coeffecient of redtitution but seeing your solution tells me all the velocities have to be taken along line of collision. Thanks for the help
My doubt about this solution is that there are two velocities of approach and separation at the same time for the same body. My solution splits the large disk into two, each half collides with a small disk, and then the halves merge. I get $e=11/16$.
Another way of defining $e$ is as the square root of final to initial kinetic energy. Both methods should give the same answer.
A third method of solution is successive collisions : the large disk collides with the upper small disk, rebounds then collides with the small disk and comes to rest. The same COR should be used in both collisions here, and the two small disks should have symmetrical final velocities.
@AvnishKabaj See the 1st question which I linked. It looks simple, 3 objects collide in 1D. However, the outcome depends which pair-wise collision happens first.
If the symmetry argument is correct then the simultaneous answer should be the same as the two different orders of sequential collision. But it isn't. This means that wethod,,, correct?"
... This means that we are making a hidden assumption, as was the case in "Is my method... correct?"
@Hema A wave function must be a solution to the wave equation $$\frac{\partial^2 y}{\partial t^2}=v^2\frac{\partial^2 y}{\partial x^2}$$ Any well-behaved function of the form $y=f(x\pm vt)$ is such a solution. The function $y=(x-vt)^2$ is a possible wave function because it satisfies the wave equation.
@Hema Transverse waves in solids : No I don't think so. Transverse waves in bulk materials require a non-zero shear modulus $G$. But you can have transverse waves on a string which depend on Young's Modulus $E$ rather than shear modulus. Torsional waves on the wire depend on $G$.
@Hema No. For example $y=A\sin(\omega t-kx)$ is a solution to the wave equation. Putting $x=0$ and $t-0$ at the same time also gives $y=0$. There is nothing wrong with this. In any wave the displacement $y$ is zero at places $x$ which are one half wavelength apart and at times $t$ which are one half period apart. This is perfectly natural behaviour for waves.
Perhaps you would care to upload an image of the solution so that we can read what the author is saying?
@Hema I think the author does not make sense. There is no requirement for a solution to the wave equation to be non-zero or finite at any point. It depends on the boundary conditions : the wave can exist and be well-behaved within a restricted region.
All of the functions given contain $x, t$ in the combination $x \pm vt$ so by the definition given in the question they represent possible travelling waves.
The question is a good one, but I suspect that the answers have been supplied by a student and nobody has checked them.
You can tell from the wave function if the wave is moving to the left or right by looking at the sign of the phase argument $x \pm vt$. If this sign is -ve the wave is moving in the +x direction, if it is +ve it is moving in the -x direction.
That tells you that the wave profile (the wave shape) is travelling left or right, but not up or down. In this case the wave profile is travelling to right. Suppose we look at the particle at $x=0$. This particle does not actually move left or right, it only moves up or down.
When the wave moves right the particle at $x=0$ moves down below the x axis. But if we look at the next point to the right for which $y=0$ just now, we can see that as the wave profile moves right the particle at this point moves up.
Yes, that is correct. At any instant in time some particles are moving up and some are moving down. You can also decide from the wave function $y(x,t)$, by looking at the sign of $\frac{dy}{dt}$ for the particular values of $x, t$ which you are interested in. $\frac{dy}{dt} \gt 0$ means the particle at that location & time is moving up ($y$ increasing), while $\frac{dy}{dt} \lt 0$ means it is moving down ($y$ decreasing).
And $\frac{dy}{dt}=0$ means it has stopped moving - it is at a peak or trough.
How do I solve this problem? The period of the free oscillations of the system shown here if mass M1 is pulled down a little and force constant of the spring is k and masses of the fixed pulleys are negligible, is
"The answer is 2π √(M2+4M1/k) Here M2 is mass of movable pulley and M1 is mass of the block. I deduced that when M1 moves down by x, M2 moves up by x/2 and therefore spring stretches by x/2,but I'm not sure what to do next."
Hello! Consider a cable whose mass is $m$. What is the tension in the cable caused by its own weight? $mg$ or $\dfrac{1}{2}mg$? I lean towards the latter but I am not sure how to prove it
I mean, at the topmost point the tension is $mg$ and at the bottommost point it is $0$. But is it fine to just take the average and claim that's the total tension? Or do we have to integrate to solve this?
If the mass of the cable is distributed uniformly along its length then the tension in the cable also varies along its length. (I assume the cable is hanging vertically, rather than resting on a horizontal surface.)
If I were given this problem during a contest, I'd have said that due to the uniform distribution of mass along the cable, the force that causes elongation is half of the cable's weight.
Thanks for the references!
Ok, so I guess this part:
> If the cross-sectional area A of the wire were constant then F (= weight of wire below any layer) would increase linearly with y, the height from bottom to top. Because the only variation (that in F(y)) is linear, the total extension x for the whole wire can be found by using the average value of F at the ends. This is just the same as finding the area under a straight-line graph from the range and the average height at the ends.
I mean, at the topmost point the tension is $mg$ and at the bottommost point it is $0$. But is it fine to just take the average and claim that's the total tension? Or do we have to integrate to solve this?
Yes, I think you are correct. Assuming that the extension is small, the tension increases linearly from bottom to top. So the average tension is $\frac12 mg$.
