# Tag Info

43

The assumption that you are making here is that with the same motion of a punch, that you are applying $\text{100 N}$ of force to both a wall and to the air. However, you should think about the most fundamental equation of Newton's laws, namely, $F=ma$ The most important part of this in relation to what you are talking about is that the force applied, $F$,...

42

The notion of soft or hard object depends on the velocity of interaction. Water can be soft or hard as rock depending on how fast you fall in (or surf upon). For a shock, the main thing that matter is momentum. In space, where relative speeds can be very high, a simple bolt can cause serious damage to the ISS, and simple flakes of paint cause deep ...

33

At high speeds the structure of the material becomes far less important than it is at low speeds. At high enough speeds, the issue is not whether the tomato can retain structure during the impact (it wont), but rather the issue becomes one of sheer mass. The issue is easiest to see in the tomato's reference frame, where one treats the tomato as holding ...

12

As Anubhav mentioned in his comment, tomato would break into pieces before it hits the plate. However, to answer the logic of the question such event is possible. A ping pong ball can rip a huge hole on the ping pong racket if it is fast enough. https://www.youtube.com/watch?v=acRnKnsddwc Update: It would make a hole, if you make enough assumptions. As ...

11

The frequency of a sound wave cannot change as it crosses the water-air boundary. The wavelength can, and does, change but the frequency cannot because if it did there would be no way to match the two waves at the interface. This means that the higher frequency is not some quirk of the sound propagation, but that the colliding stones emit a higher frequency ...

9

Photons don't directly interact with each other, but if one photon pair produced an e+/e- then the second photon could interact with that pair. The interaction has to conserve the energy of the two photons and conserve their momentum as well of course. But yes they could (and most probably depending on their energy) just pass right "through" each other.

6

The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses. If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed. You ...

6

Assume that after each bounce the velocity decreases in a factor $\xi\in(0,1)$. This means: if the velocity before hitting the floor is $\dot y$, then the velocity after hitting it will be $\xi \dot y$. Let $y(0)=0$ be the initial height and $\dot y(0)=v_0$ be the initial velocity. We choose the units so that $g=1$ and $v_0=1/2$, where $g$ is the ...

5

There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...

5

This is the same problem as the famous question states: which is heavier, 1k of feathers or 1kg of iron? It requires many more feathers to get 1kg of them, so it confuses most people. In this case, it requires you to have a big wing to be able to apply a force of 100N on air.

5

Yes, because even just a water (a big drop of water) would do this. It has been written, in many sources (here for instance), that at high impact speeds the water (or even gas) is as hard as a concrete or glass. Mostly it is about crashing into water at high velocity, but water crashing into something would probably not make any difference.

4

No, Newton's third law is not violated. According to Newton's Second Law, we have that force is the rate of change of momentum with time, i.e. $$F=\frac{\Delta p}{\Delta t}$$ where $p$ is momentum and $\Delta t$ is time elapsed. When Punch Strikes Wall Initial momentum of fist = $mv$ Final momentum of fist = $0$ Force applied = $\frac{mv-o}{t} = \frac{... 4 When two objects collide, they transfer momentum because they exert an equal and opposit force on each other (Newton's third law), and$\Delta p = \int F dt$. In order to know how fast an object moves after a collision, we need to know the velocities and mass of the objects before the collision and how elastic the collision is (conservation of kinetic ... 4 If you fire your projectile on a rigid target it will move, no matter what the diameter of your bullet is. All that counts in this scenario is the momentum and kinetic energy. If the target is a soft target you do also move the part of the target that is directly hit by the bullet, but since this part has a smaller area when you use a smaller caliber the ... 4 The important point here, which I think has been missed so far, is that momentum goes like velocity while energy goes like the square of velocity: $$p = mv$$ but $$E = \frac{mv^2}{2}$$ So let's consider a small object with mass$m$hitting, and sticking to, a much larger object with mass$M$, if the initial velocity of the small object (bullet or baseball) ... 4 Your gut feel is correct. Both are exactly the same. Look at the acceleration in both scenarios. 35 mph to 0 in the time it takes for the cars to fold up and stop. Everybody gets this wrong. Good question. 3 Usually absolutely nothing. Electromagnetism is linear, which means that the result of doing something with two photons is the superposition of the results of doing something with each one individually. By that reasoning, since one photon by itself just goes on its own merry way, then two photons, even if they go near each other, just go along on their ... 3 Q: When we write that, do we suppose a collisionless or collisional nature of the fluids? A: It's the energy-momentum tensor for a perfect fluid Chapter 2.26 Q: If this description corresponds to collisional fluids, why cosmological simulations are N-body simulations (collisionless) and are not simply based on hydrodynamics? A: Cosmological simulations are ... 3 The second solution will mathematically satisfy the conservation equations, but corresponds the objects not actually colliding. Or they ghost'' and fly right through each other. :) 3 There is a general definition of a center of mass in General Relativity, which was proposed by Dixon (W.G. Dixon, Il Nuovo Cimento 34, 317 (1964)) and whose existence and uniqueness were proved by Beiglböck (W. Beiglbock, Commun. Math. Phys. 5, 106 (1967)). These are very old papers, so that people have been working since long ago on this topic. In ... 3 None of these answers really address the question; mostly they just reiterate physics principles that I suspect the poster already understands. The question is saying, 'If kinetic energy changes in different types of collision, then the final velocities must be changing. If the final velocities are changing, the final momentum must be changing, but momentum ... 3 Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ... 3 Liquids and gases are both fluids, meaning they flow(duh) and will take the shape of their container. Gases will expand to fill the container, while liquids will not. Solids are not fluids and do not conform to their container's shape. Yes, I know you can squeeze things to fit (e.g. a sponge), but only at the cost of distorting the structure and storing ... 3 Let's revisit "force on impact" for a moment. I will consider a "sticky" ball of mass$m$traveling at velocity$v$at a stationary wall. When it hits the wall (and sticks), what is the maximum force felt by the ball / wall? That is actually not a trivial question to answer. The ball has momentum$p=mv$, and if the impact time (time for the ball to come to ... 3 I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the$dt$or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ... 3 The assumptions that are being made here are perfectly elastic collisions with the walls where the atom will rebound with the same speed as it came it with. Its trying to calculate the average force based on the number of interactions per unit time with the wall, and is therefore fundamentally related to how long it takes for successive collisions between ... 3 When we describe an inelastic collision of macroscopic objects we normally say the missing energy ends up as heat. That is, the temperature of the colliding objects is higher after the collision than before it. However this is a simplified description because heat is just the kinetic energy of the atoms and molecules making up the objects. Suppose you take ... 3 The way to approach this problem is to use the vector momenta and conserve them. If we orient our coordinate axis such that our$x$-coordinate is pointing North and our$y\$-coordinate is pointing East, then the momentum of the first car is: $$\vec{p}_{1}=m_{1}\vec{v}_{1} = 1400\begin{pmatrix}-11 \\ 0\end{pmatrix}\:\text{kg m s}^{-1}$$ Similarly, the ...

3

Hadronization still doesn't break conservation of energy and momentum. So getting the total energy, or the total momentum, of the quark-antiquark pair is easy: just add up the total energy and momentum of all the reaction products. To get the individual energy or momentum of one particle, i.e. just the quark (or antiquark), we rely on the fact that they ...

2

Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...

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