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47

Amazingly this actually happened to a Russian scientist called Anatoli Bugorski (WARNING: this is pretty gruesome). The beam basically just killed all the tissue it passed through. The symptoms were the relatively mundane ones expected from tissue death. The LHC has a much, much greater energy than the one that struck Bugorski, so it would cause a lot more ...


33

A charged particle will create charge separation (ionization) along its path. This will cause harmful chemical reactions to occur in the body, including DNA damage. The effects of these chemical reactions depend on their amount. The body can heal from a low amount on its own, while a high amount will cause radiation sickness and probably death. This can be ...


13

Nothing happens obviously, when one high energy particle penetrates flesh as cosmic rays continuously impinge on us and some have the energies of the LHC. The cosmic rays reaching us are mainly muons and the damage they do is with electromagnetic scatters/ionisations in their path. The mean energy of muons reaching sea level is about 4 GeV. Muons, being ...


5

While a single LHC particle wouldn't be doing much harm, being hit by the LHC beam would be certainly deadly and it would damage the machine badly. Any dense matter that comes into the LHC beam will instantly act as a beam dump. We have a very good idea about what happens in the LHC beam dump, see e.g. ...


5

Let's turn this around. In an inelastic collision, some of the energy, instead of remaining with the center of mass of the objects colliding, is dissipated as heat - an increase in the random motion of the atoms and molecules of the objects colliding. At the (sub) atomic level, the two particles involved in a simple collision are the same two objects that ...


3

By the way, OA/OB/OC are not force vectors. The only force vector that is acting on the ball before (and after) it hits the ground is its weight, and it is acting vertically downwards. OA/OB/OC are velocity vectors. Try to split OA up into its vertical and horizontal components. When the ball hits the floor, reverse the vertical component due to impulse ...


2

The equation, $v_A-v_B = \frac{4}{5}(u-2u/3)$ is incorrect. The proper equation for the coefficient of restitution is given by, $v_A-v_B$ = $e(u_B-u_A)$, where $u$ and $v$ are velocities along the line of impact. I believe you came across the somewhat incomplete statement that the relative speed of separation after the collision is $e$ times the initial ...


2

What other variables I should know before it can be calculated? : I only know the values before the collision, for example the initial velocities VA,VB, initial temperatures, surface areas, etc. You need to know the $C_R$ coefficient of restitution, CR is the coefficient of restitution if it is 1 we have an elastic collision, if it is 0 we have a ...


2

Yes, the insect does actually travel at 0mph when it reverses direction. The critical aspect that you are missing is that the time that it is not moving is infinitely short. Actually, I would think that the head stops, then the thorax and finally the tail as it is squished up against the train.


2

Both the insect and the train window are deformable. Microscopically so, but deformable nonetheless. Because of that fact, the insect slows continuously to zero, reverses direction, and then speeds up in the direction the train is going. To our human perception this happens imperceptibly fast.


2

Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


1

The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...


1

Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$ Conservation of torque ...


1

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...


1

The force between two colliding bodies is not in the direction of motion. The normal and parallel components have to be treated separately. The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface ...


1

Normally, I think that the surface will react with a force in the OC→ direction Yes, if the ball's force on the surface is in direction A, then by Newton's third law, the surface's force on the wall is in direction C. This is what Newton's third law says. The third law applies to all forces in the same way. However, as pointed out by Jan Hudec in a ...


1

Now if they are both shot at ballistic gelatin, which one is expected to cause more damage if both are stopped by the gelatin ? By damage here I mean more penetration, bigger cavity, heat and any other deformation of the gelatin. In other words, if the velocity of a projectile is doubled, will the amount of damage it causes when it collides with ...



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