# Tag Info

11

If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds. There is no way out. ...

10

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 ...

8

. After the collision, according to the conservation of energy, the speed of both balls should be 35 m/s. ... Now, obvisouly, 35 m/s in R2 is 85 m/s in R1, not 79m/s. Where does the discrepancy come from? The collision is an artificial example: it is considered inelastic, but there is no loss of energy. If it followed the ordinary rules, in ...

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 ...

6

Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any. A body does not have angular momentum wrt to a point C only when it is circling around it, you know that planets have elliptical orbits and do have L If a body H has linear momentum p it has also and angular ...

4

Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required 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 ...

3

Good question. The main difference is that we cannot freely move in time or in other words, we and everything else are moving together. Other than that, I think nobody can say for sure yet if the answer is 1, 2, 3 or something else. The important thing to realize is that time as a 4th dimension is used to make models or theories of reality. Compare that to a ...

3

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 ...

3

Consider a particle in a nice Keplerian orbit around a star, taken to be a circle for simplicity. If it is at a radius $r$ and the star has mass $M$, then its specific angular momentum will be $$l = vr = \sqrt{GMr}$$ and its total specific energy will be $$e = -\frac{GM}{2r}.$$ If you want to decrease $r$, you are going to have to decrease $l$ as well as ...

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

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

In order to get a clear picture consider the outcome of an impact between the head (m = 1Kg) and a blunt rod (plus arm M = 3 Kg, L = 1 m) rotating with angular velocity 1 r/s: If the sword hits with its tip the effective mass is 1 Kg at speed 1 m/s the head would get momentum 1 Kg*m/s If it hits at the hilt (r = .5 m) the effective mass is the same but ...

2

Now, after the hammer is released, the thrower still has her same angular momentum (and has to slow herself down), but the hammer no longer seems to have any. Even though the hammer isn't rotating around the axis, it still has the same angular momentum it had at release with respect to the original axis. So the formula $$L = mvd$$ is correct both for ...

2

the speed of both should be according to the conservation of energy If both balls have the same speed after the collision, the collision is inelastic, i.e., kinetic energy is not conserved. If the balls are identical, then conservation of momentum requires that $$\mathbf v'_1 + \mathbf v'_2 = \mathbf v_1 + \mathbf v_2 = 150 \mathrm{\frac{m}{s}}$$ If ...

1

First off, when two values are directly proportional, it not only means that they are related, but also that they are related by a constant of proportionality (i.e. as one changes, the other changes proportionally). For example, in your second equation, the centripetal force is directly proportional to the radial distance to the mass and proportional to the ...

1

Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate. In fact they have (as one system) the same angular momentum all the time. When you have a projectile that ...

1

The answer to your question is yes, the existence of $n$ conserved quantities with $n$ degrees of freedom implies separability of HJ. The massless HJ equation is $$g^{MN}\frac{\partial S}{\partial x^M}\frac{\partial S}{\partial x^N}=E.$$ It separates if there exists a new set of coordinates $Y^M$ such that $$S(Y_1,...,Y_n)=\sum_{i=1}^n S_i(Y_i),$$ which ...

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

Angular momentum is conserved in this example! As you already stated, the angular momentum of the thrower doesn't change after the hammer is released. Consider the hammer being in rotation around the origin of our coordinate system for $t < 0$: $$\vec{r}(t) = r_0 \ \ (cos(\omega t), sin(\omega t), 0)^T$$. Its momentum is therefore given by:  ...

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