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9

A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ...


5

Unless I'm missing an easy way to do this problem it seems a surprisingly hard one. This diagram shows the problem (I've exaggerated the altitude of the satellite to make the diagram clearer): The satellites are in circular orbits (dotted line) at a distance $r$ from the centre of the Earth, so their orbital velocity as (as you say): $$ v = ...


3

...truck will experience larger force. But Newton's Third Law of motion says that 'to every action there is an equal and opposite reaction'. So the force experienced by the truck (M) should be same to that experienced by the car (m), but negative, isn't it? Third law states that momentum must be equal and opposite if both vehicles must come to a ...


2

The glass of the windows you see in movies are made of sugar or acrylic, that's why you can fly through the window (with a punch) without a scrape or get a 'bottle' of whisky smashed on your head without a headacke. That's why , especially with plastic you can see a clean hole. ...the stunt glass you see people jump through all the time in movies is not ...


2

The main factor you should consider is impulse: if the force acts for a longer interval of time, it causes greater fractures in your glass, if it acts for a shorter interval of time, it causes less fractures. Again, it depends on the material of the glass and the projectile you are using.


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 ...


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

They both experience the same force because of the impact, due to the Newton's third law, like you say. I think the question is not clear enough. If you assume there is no friction between the trucks and the ground, then you can use momentum considerations. I know this shouldn't be an answer, but I'm new and I can't post a comment, yet.


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

Introducing a side issue to this question: Using $v$ for the orbital velocity at $1000$ km altitude, the total kinetic energy of the two satellites just before the collision is $$KE_{\text{Before}}=\frac12400v^2+\frac12100v^2=250v^2$$and using the final velocity of John Rennie above, the total kinetic energy of the combined satellites just after the ...


1

if mass is assumed to be constant, the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. However, if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. 1) mass is not constant and velocity is different: in ...


1

... if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. True. ... the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. False. So, how can there be a change in kinetic energy of the ...


1

We have to make a few assumptions here, because your question as posed is a bit incomplete. 1) collision is inelastic 2) satellites don't disintegrate on collision (see user58220's interesting observation) 3) there is no atmosphere (so drag is not an issue - let's see if the satellites end up in an orbit above the planet's surface) 4) you said "crash into ...


1

Emmy Noether discovered a fundamental connection between symmetries and conservation laws, embodied in her famous theorem. In simple terms, Noether's theorem is that For every symmetry in a physical system, there must be a conserved quantity. The proof requires neither Lorentz invariance nor causality. By applying Noether's theorem, we find that ...


1

Remember that an inelastic collision means energy is lost but not momentum. Momentum is always conserved in collisions. So you have to figure out what kind of velocity the two joint masses have and use your knowledge of springs to compute the halt at the given position, velocity and mass.


1

I would proceed as follows: Assuming that the stored energy in the spring, compressed/extended a distance $d$ from equilibrium, is given by$$E_{stored}=kd^2$$ Find stored energy, kinetic energy, and thus total energy at the moment of release; Find stored energy, and thus kinetic energy, just before the collision Step aside for a moment, and use ...


1

I think I've get a better idea of what you are looking for now, thanks. As background for others in the future: classic ion-solid interaction theory dates back to the 1960s and is commonly called LSS theory after Linhard, Scharff, and Schiott who first formulated the concepts. It splits the energy loss mechanisms of the ion into two components, electronic ...



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