# Tag Info

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You can't always nail in a nail with any object. On the low speed scales, its obvious that a piece of bread cannot nail the nail in because the bread will deform before the nail overcomes friction and starts to move. Your interest is in the faster cases, where the bread doesn't get the opportunity to deform out of the way. At high speeds, what you find is ...

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As you say, momentum is conserved when there is no external resultant force acting on the system. This is a statement of Newton's 2nd law: when the net force acting on a system is zero, there is no change in momentum. $$\vec{F}_\mathrm{net} = \frac{d\vec{p}}{dt}$$ an example It might help to think about an example. Lets consider the head on collision of ...

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As it moves along covering some total travel distance $L$, the particle's motion carves out a tube of length $L$ and cross sectional area $\sigma$, making a volume $L \sigma$. If $n$ is the number density of other particles that it might hit, then there are $N = n L \sigma$ such particles in the tube, so that is the number of collisions it will have in the ...

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The average impact force is $$F_{average}=ma_{average}=m\frac{\Delta v}{\Delta t}$$ Clearly, if $\Delta t=0$ the average impact force would be infinite. If the object is a human body jumping on his legs from a finite height, his bones don't break even if the collision is almost visibly non-elastic and $t$ is apparently too small. What is the reason behind ...

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You are on the right track. IF the collision were to happen in zero time, it would imply an infinite force. In practice no collision is that brief and no force is infinite, but it may be a good approximation in some situations. More generally, there's no reason to restrict the collision to a short time - it might take place over quite a long time, and the ...

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I think I understand your question. How is it possible, that even though the first ball deforms and stores elastic energy during the collision, it suddenly ends up with no motion after. The short answer is that some elastic energy is temporarily stored in the left-most ball during the collision, but somehow during the collision, the combination of forces, ...

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Why? Because that is the only direction where relative motion is not allowed. The centers cannot get any closer to each other. If force was along a direction where motion was allowed, then the contact force would do work and either add or remove energy to the system violating the conservation of energy. The only time the contact force is at an angle to the ...

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Yes, here are some references. There are likely many more. Eugene J. Saletan (1997) "Minkowski diagrams in momentum space" American Journal of Physics 65, 799 (1997); https://doi.org/10.1119/1.18651 Minkowski diagrams in configuration space, with points representing events are often used in undergraduate courses on special relativity. Similar ...

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While formulating the theory, we assume that the particles obey Newton's laws (third one) and thus, the forces exerted by them on each other gets cancelled out. This is sometimes called as the weak law of action and reaction. Mathematically speaking, $$\mathbf{F} = m\mathbf{a}$$ $$\frac{d^2}{dt^2} \sum_i{m_i \mathbf{r}_i} = \sum_i{\mathbf {F}^{(ext)}_{i}} + ... 2 Consider a monoenergetic beam of neutrons all traveling in the x direction through a slab target. Let I_0 be the intensity (neutrons per square cm per sec) incident on the slab, denoted as the flux. Let \Sigma represent the macroscopic cross section of the target (units are 1/cm). See any physics/nuclear engineering test for a discussion of the ... 1 the answer is simple. we divide the target thickness into discrete slabs of equal thickness. Imagine that after traversing the first slab element, 50% of the neutrons have been absorbed. Then, after the second slab, we have 50% of 50% or 25% of the original beam strength remaining. After three slabs we have 50% of 25% or 12.5%; four slabs is 50% of 12.5 or 6.... 1 I think the key is for you to take account that energy is a scalar quantity and momentum is a vector quantity. So let's say two particles are approaching each other and their momenta are 7 to the right and 5 to the left (omitting units here), so net momentum to the right of magnitude 2. If they have an inelastic collision, after the collision let's say their ... 1 Your equation should be$$h=s^2A^2t^2/2gm^2$$Your observations about ideal and human bodies do not contradict your theory. By definition ideal bodies are not real, so there are no observations of their behaviour. It is not contradictory for ideal bodies which do not deform (hence t=0) to break when h=0. For real bodies if t is almost zero then the ... 1 A ball with a higher pressure gas inside it (or made of a material with a high elasticity) will have a higher coefficient of restitution given by$$e = \frac{h_b}{h_i}$$where h_b is the height the ball bounces from an initial dropped height of h_i. The reason why e is higher for a ball with a higher gas pressure is because the more pressurised such a ... 1 Having calculated v3 (velocity of conjoined bodies after the crash), we can now find the kinetic energy of the metorite before the crash and subtract kinetic energy of conjoined bodies after the crash to find out how many heat was converted from kinetic heat to internal heat. Strictly thermodynamically speaking, the lost macroscopic kinetic energy of the ... 1 The figure is correct. It shows the 4-velocity of the particles after collisions. Note that the spatial components of the 4-velocity are not the same as the spatial components of the velocity in x and y directions. x-directional 4-velocity is, using the matrix equation given in the question:$$u^1=-\beta \gamma_v \gamma_u c + \gamma_v \gamma_u \vec{u}^1$$... 1 As you have said, Coulomb's force is an internal force. Hence momentum is always conserved and both methods would give the same answer(first one). You must realize three things: Conservation of momentum is just another way of representing Newton's Laws. This is a situation of completely elastic collision with e=1. Here repulsion between the protons act as ... 1 Your object consists of atoms composed of charges (electrons and protons). Likewise, the ground is as well. As the object hits the ground, the atoms in the ground are displaced and pushed closer together. The electrons in the atoms begin to repel each other and therefore resist this compression and as a consequence your object is then pushed by the atoms in ... 1 Start with the equation for the trajectory of a projectile launched from the origin with speed u at angle \theta to the horizontal, with \tan\theta=m :$$y=mx-\frac{g}{2u^2}(1+m^2)x^2$$The deflection point is the origin, at height h above ground. Launch speed is given by u^2=2g(H-h). The projectile hits the ground where y=-h. Let \frac{u^2}{g}=... 1 The time of collision is zero here therefore force must be infinite . Is this correct ? Something doesn't feel right here. Can Infinite force act ? You have been tricked by an idealization, and took it for reality. In order to make explanations simpler the colliding bodies are often treated as being infinitely hard. An infinitely hard body would not deform. ... 1 An easy way to understand this is that the total kinetic energy could be divided into two part:$$ E_k=\frac{1}{2}(m_1+m_2)v_c^2+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}v_r^2 the first term is the kinetic energy of the mass center and the second term is the relative kinetic energy, where $v_r=v_1-v_2$ is the relative velocity and the relative kinetic energy is ...

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then why does the second body ( the body which was at rest earlier) only moves after the collison and why not both of them move with some kinetic energy since both had some stored potential energy ? Here's the thing, energy is a scalar quantity. Conservation of energy is only dependent on the initial final state. Energy does not care which direction a body ...

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The crux of your question seems to be attempting to reconcile the apparent symmetry of two balls compressing under each other's equal but opposite applied force, but moving asymmetrically after rebound. Others answers have detailed how, in the reference frame where one ball is at rest before collision and the other ball is moving, this asymmetrical result ...

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NEWTON'S LAWS OF MOTION: It can be shown for a perfectly elastic head on collision involving identical objects, strictly using Newton's laws of motion, Object 1 will come to a stop and Object 2 will continue with the initial velocity of Object 1. We can consider the two colliding objects as a system. Given no friction forces, and the fact that the ...

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Let us make it easy. Now suppose you take two identical balls that are not rigid (but they also do not lose energy on changing shape). So now as they strike, you assume of it as if one ball striked the other that was at rest and both got maximum deformed (They attain equal velocity i.e. a common velocity) Now the ball wants to restore its shape so it will ...

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A perfectly elastic head on collision between two objects (say spheres) of equal masses will result in the first object coming to a stop and the second moving on with the same momentum and kinetic energy possessed by the first object just prior to collision. Now I believe you may be thinking: Why doesn't the first ball continue to move after the collision ...

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If you solve this problem in the center of mass, each object is colliding with $\pm v$ and rebounding with $\mp v$. That is the only way to conserve energy and momentum. Now if you (Galilean) boost it by $\pm v$, it's pretty clear one ball starts off at rest, and the other ball ends up at rest. Now in our experience with bouncing balls: harder balls bounce ...

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A perfectly elastic collision is defined by the fact, that no energy goes into the deformation of the bodies. So the misunderstanding lies in the sentence Now if we consider a perfectly elastic collison between two identical bodies [...], we know that the two bodies are going to deform. Of course, no collision process is perfectly elastic. More or less ...

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