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When solving a problem like this, it is always a good idea to draw a diagram: For an elastic collision, there are three conserved quantities: Energy Linear momentum Angular momentum And in this case there are three unknowns - $v_1, v_1', v_2$. We are only interested in $v_1$ so we should organize our equations to eliminate the other two. The set of ...

8

The correct answer has been given by linus since what causes more damages in an impact is momentum, but the 'soft spot' has been vaguely suggested as a better solution in a comment and in an answer and probably it is opportune to give some concrete and precise information about that issue. The center of percussion is the point on an object where a ...

7

The stroke of genius The laws of Nature are wonderful: extremely simple - make a tremendously complex world, a few symbols can be extremely powerful and beautiful all roads lead to Rome, you may go from London to Rome following many different routes, only a fool would get there via Moscow. But an eagle can fly over the Alps in a straight-line. I'll try to ...

4

Since the collision is elastic, the kinetic energy of the system is the same before and after the collision: $$0.5m_1v_1^2=0.5J_2 \omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$ This kind of problem has usually 3 equations: conservation of: 1. Ke, 2. p, 3. L, and 3 unknowns: $y= v_3, z =v_2, x = \omega$, when the initial velocity $v_1$ is known. But in ...

4

Because the equations for linear and angular motion are very symmetrical In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object. In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor. So, the ...

3

This is the right way to represent the collision: Since the masses are equal the resulting vectors will form a right angle and linear momenta on the y-axis are equal and the sum of momenta on the x-axis is equal to the original momentum of the cueball $mv'+m_ov' = mv$

3

Even classically, forces arise from field being propagated at the speed of light. A physically relevant object is the energy-stress tensor, whose components represent energy density and momentum current density, so indeed momentum can be interpreted as a current that is conserved over time (as a consequence of symmetries). This point of view is also ...

3

.... can I say that at the exact moment the force is removed, impulse equals momentum? Yes, but it is more correct to say that "it equals the change (increase/decrease) of momentum" Force is an abstract concept. As defined by wiki: A force is any interaction which tends to change the motion of an object.[1] In other words, a force can cause an ...

2

There are two objects with mass $m$ and $M$, respectively. The object with mass $m$ has a velocity of $\sqrt{2gl}$ and collides with the other object that is initially at rest. If the collision is elastic, what is the velocity of the object with mass $M$ right after the collision? Friction is negligible. This is an old question but it may still interest ...

2

Center of mass before the disintegration is in the initial particle. This means that the center of mass moves with velocity $\vec V$ in lab frame. Thus, to switch from center of mass frame to lab frame you just use the Galilean transformation $$\vec v=\vec v_0+\vec V.$$ As for conservation of mass, it is certainly implied because the system is considered ...

2

Let's call our two colliding objects $A$ and $B$. So object $A$ and object $B$ come together, collide and ricochet away again. The collision may be elastic or inelastic. You are quite correct that there must be a force acting during the collision, and because the force acts for some time there is an associated change in momentum. Consider just object $A$. ...

2

$F = \frac{dp}{dt}$ means that force is the rate of momentum transfer per unit time. Lets say we have mass $m_1$ moving to the right, and mass $m_2$ is on the left side of $m_1$ with zero velocity. If $m_1$ put a force to pull $m_2$, that force will create the acceleration on $m_2$ and increase its velocity, this also means the change in momentum. At the ...

2

TL;DR: If you have to choose either "near the handle" or "near the tip", the tip will work better. But there's a point in between these two that works even better; exactly where that point is depends on how you swing the sword, and how its weight is distributed. UPDATED now I am near a computer and can draw diagrams etc. If cutting off the zombie's head ...

2

Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass: $$m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1$$ $$\frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2}$$ and ...

2

There are many good answers already. This answer is basically an expanded version of Emilio Pisanty's answer. Let us start by recalling the standard convention to write the position wavefunction $$\tag{1} \psi(x)~=~\langle x | \psi \rangle$$ as an overlap with a position bra state $\langle x |$. The CCR $$\tag{2} [\hat{x},\hat{p}]~=~i\hbar{\bf 1}$$ is ...

2

Yes, in that instance impulse equal momentum, since impulse equals change in momentum. If the momentum then was 0, and the momentum now is p, then impulse = p - 0 = 0.

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When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.

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I think you have the answer for your second question. For your first question let me clarify: $$\langle x\ |\ \hat{x}\ |\ p\rangle \overset{(1)}{=} \int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(2)}{=} \int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(3)}{=} x\langle x\ |\ p\rangle ... 1 It's acutally possible. The phenomenon is called Bremsstrahlung (its direct translations would be something like "stopping radiation"). It can be described purely with classical theories like special relativty and electromagnetism. If a charged particle is accelerating, it "borrows" some of his momentum and energy to the EM-field which is then radiated as ... 1 I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ... 1 It is not a valid quantum state, it is an idealization of very long wave-packets emitted by atom-lasers. These wave-packets are almost coherent waves, very close, by their quantum description, to Fourier components, though they have finite length, e.g. 0.35mm (see arXiv quant-ph/9812.258, "An Atom Laser with a cw Output Coupler", by Bloch-Hänsch-Esslinger, ... 1 You only have a problem if you start without any tangential velocity and have no reflection (or you collide with a celestial body before you have a chance to evade it). Let us say you start in a stable orbit. Your ship has a bow and stern, in the direction of your velocity. If you can reflect the light towards your stern, it will accelerate you and you get ... 1 A spread of momentum is sometimes used literally (as in, the momentum is definitely bigger than a and definitely less than b so the spread of momentum of is b-a), and sometimes it is used colloquially just to say that most of the time the momentum is between an a and a b such that b-a equals your spread of momentum. The reason having a spread of ... 1 I'd like to offer a thought experiment in return: same scenario, but instead of hot gases coming from a rocket, we've got a device firing BB's in the same direction as the exhaust gasses were going. The results should be the same (gasses behave a lot like BB's in a vacuum), but BB's are a bit more tangible and have some heft, so the results will feel a bit ... 1 A collision is always a continious process meaning that one shouldn't think of a force acting for a single instance of time. There's a theorem in physics called "Noether's theorem" (wikipedia) which states that if the underlying physics are invariant under certain continious transformations, there exists a conserved quantity. If that transformation is ... 1 This question (v6) [concerning the overall minus sign in OP's calculation] is essentially a Fourier transformed version of e.g. this Phys.SE post, see Emilio Pisanty's answer and my answer. The main point is again that the derivative in the momentum Schrödinger representation$$\hat{x}~=~i\hbar\frac{\partial}{\partial p}, \qquad \hat{p}~=~p, acts on the ...

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