Hot answers tagged collision
18
First of all -- it wouldn't be called "the Large Hadron Collider", right?
Looks like one would rather call it something like "Large Electron-Positron Collider".
In that case one definitely would need another abbreviation for it. Something like "LEP" instead of "LHC"...
Now, guess what was there in the same tunnel before?
Edit: since my shenanigan got ...
17
I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...
From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first ...
13
Contrary to what is stated in many textbooks, energy-momentum conservation alone cannot explain the behavior of Newton’s cradle. For N balls we have two equations and N final velocities to calculate. Hence, conservations laws can do the job only for N=2. This means that if we want to give an explanation of the cradle behavior based on conservation laws, we ...
12
You should be able to use energy conservation to write down the velocities of the bodies as a function of time.
$$
\textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right)
$$
And
$$
\frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right)
$$
Momentum ...
12
While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation.
General Remarks
The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer ...
12
There are two points in answering this question:
Design: The design of the collider would have to be different. Electrons/positrons in a cyclotron radiate synchrotron radiation when they are accelerated (which itself is a useful device). To get above a few GeV, researchers use linear accelerators, such as SLAC. The proposed International Linear Collider is ...
11
It conserves both energy and momentum in the collision at the same time.
By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side). This means there are no horizontal forces from the string on the balls so linear momentum in the direction of swing must be conserved in the collision. Energy is ...
9
First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the ...
8
The reason that jumping can make a relatively large difference is that the kinetic energy is proportional to the square of the velocity. Thus relatively small changes to the velocity can result in relatively large changes to the kinetic energy. In addition, the velocity which a human can achieve in jumping is a substantial percentage of the velocity of fatal ...
8
As an addition to already posted answers and while realising that experiments on Mythbusters don't really have the required rigour of physics experiments, the Mythbusters have tested this theory and concluded that:
The jumping power of a human being cannot cancel out the falling velocity of the elevator. The best speculative advice from an elevator ...
7
Without having heard this argument before I would guess that the plan it to reduce the degree to which the head rattles around.
Most of the brain damage (short term and long term) associate with a punch comes from the brain bouncing off the skull a few times as the head whips back and forth. Minimize the motion, minimize the damage. By leaning in you get a ...
7
The anti-particle corresponding to a neutron is an anti neutron!
The neutron is made up of one up quark and two down quarks. The anti-neutron is made up of an anti-up quark and two anti-down quarks. Both have zero charge because the charges of the quarks within them balance out.
You are correct that elementary particles with no charge are often their own ...
5
Head on collision is not required for elastic collision. Or the collision you described above can be an elastic collision.
To be an elastic collision, the momentum and kinetic energy should both be conserved, that is: assume the velocity for sphere 1 and 2 are $\vec{v_1}$ and $\vec{v_2}$ accordingly, and the direction of neither of them is along the ...
5
As a clarification the energy per nucleon is always lower: for example, currently in the LHC the proton top energy is 3.5 TeV. Now the Pb energy is 3.5 TeV times Z so the energy per nucleon is 3.5*Z/A and A is greater than Z for every nucleus (except the proton where it is equal to one).
But the goal of ion-ion collision is not to increase the total energy ...
5
The hypothetical collision has been mathematically modeled, and the results of those models are consistent with what see.
The impacting body (referred to as "Theia") would have hit at a low velocity and relatively shallow angle. It would certainly have affected the Earth's orbit, but not enough to knock it out of the Solar System or into the Sun.
Knocking ...
5
A subtle problem you seem to overlook is that the proton-proton cross section is very small, about 0.07 barns (a barn is $10^{-28}$ square meters) at the LHC energies and not dramatically different at your lower "fusion energies". It means that at the LHC, much like at your dream machine, most of the protons simply don't hit their partners. It is not really ...
5
The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision.
The change of momentum is final ...
5
The rotating surfaces of the spheres would just slide over each other at the instant of contact: no forces perpendicular to the line connecting the centers of the two spheres would exist (i.e. no torque would exist). They would undergo a perfectly elastic collision (no loss of energy, thus no friction), thus conserving angular (they just keep spinning) and ...
4
Mark Eichenlaub's answer is 100% correct, but you can also do it without changing reference frames, and I think it's probably easier that way. Here's how I would set it up.
Suppose that you've determined that two objects are going to collide within the next time step. Determine the positions of the two objects at the moment of collision. Draw a line ...
4
The large-scale structure of the Universe is expanding. However, gravity still works, and it's especially powerful if the distance is small. E.g., the Earth is still pulling your body closer to it, even though the Universe is expanding. The Earth and the Moon still attract each other, even though the Universe is expanding. Our whole galaxy is still held ...
4
For completeness, here's an another solution (though not so elegant as Shanth's solution).
The equations of motion (from the 2nd Newton's Law) of the two point masses $m_1$ and $m_2$, respectively, are:
$$G\frac{m_1 m_2}{(r_2-r_1)^2}=m_1 \ddot{r_1}\Leftrightarrow \ddot{r_1}=G\frac{m_2}{(r_2-r_1)^2}$$
$$-G\frac{m_1 m_2}{(r_2-r_1)^2}=m_2 ...
4
It doesn't look possible, because the stopping power starts growing for very high energies even for non-strongly interacting particles (see Fig. 27.1 in Passage of particles through matter).
To do a best case estimate, let's consider two cubic spaceships colliding, each of them with $1\,{\rm mm^3}$ of volume, a density of $1\,{\rm g\cdot cm^{-3}}$ and a ...
4
Space is full of cosmic rays with vastly higher energies than this, and humans have so far been unaffected (though looking around at my co-workers I wonder sometimes).
Ions travelling towards the Earth get deflected by the Earth's magnetic field. The only ones energetic enough to make it through scatter off the atmosphere and generate a shower of less ...
4
The board did exert an equal and opposite force, but your mass is considerably greater than the mass of the board so all the friction between you and the ground keeps you from accelerating.
If you punched the board while standing on perfectly slippery ice, or in a vacuum, you would accelerate also, but at a much smaller rate than the board due to the mass ...
4
If I punch a wooden board hard enough and it breaks in two, has the board still exerted a force of equal magnitude on my fist?
Yes.
When the board breaks in two due to my force, the halves have a component of acceleration in the direction of my striking fist...that implies the board did not exert an equal and opposite reaction, no?
Incorrect, the ...
4
In the ideal case where the collision is instantaneous and there is only a single point of contact, the forces experienced by each object can only be along the line that connects the centers and passes through the contact point. Actually, the "force" will be infinite, but it will impart a finite impulse (i.e. change in momentum) during the infinitesimal ...
4
As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...
3
You haven't explicitly stated the nature of the contact between the spheres (i.e. is it frictionless?)
If we have ideal frictionless contact between the spheres, then yes the collision can be elastic regardless of whether it is head-on or glancing.
In the real world, with friction, then it does matter whether it is head on or not. For glancing contact, ...
3
In an elastic collision between spheres, we can use conservation of energy and momentum, but this alone is not enough.
Start by looking in a frame where one ball is stationary (the target) and the other one is coming in from the side (the striker). The problem is to find the final momenta of the two balls. Assuming everything stays in a plane, there are ...
3
Also, in addition to all of this said above, galaxies are full of dust. When they collide, they will dump a bunch of dust onto all of the stars in the other galaxy. This then accelerates the rate of fusion in the stars--this means that you will get a bunch of time-correlated supernovae, which will then eject out a bunch of hot gas. Several of these ...
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