New answers tagged

1

I had the same doubt as I was figuring out if a certain coupling of angular momenta in the 3$\gamma$ could be found such that parity would be conserved. Turns out (if I am correct) you cannot have three photons with coupled total angular momentum $J=0$ in the first place. For the coupled photon angular momentum $L(3\gamma)$ and spin $S(3\gamma)$ to be able ...


0

Assume all masses are the same and we have a perfectly symmetrical collision. It seems a logical assumption doesn't it? Then, try to understand the following conservation of energy equation. Use it to solve for $v_{af}$: $$\frac{1}{2}mv_{ai}^2 = \frac{1}{2}mv_{af}^2 + \frac{1}{2}mv_{b}^2 +\frac{1}{2}mv_{c}^2$$ Once you've done that, go back and solve the ...


0

You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A. In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common ...


3

No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since $\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same.


0

You have only counted the energy stored in one spring. There are two springs in the diagram. The springs have elastic potential energy initially as well as finally. Look at the formula you have used : you have calculated the increase in elastic energy stored in each spring: $\frac12 k (x_2^2-x_1^2)$. There was elastic energy stored in the springs before ...


1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for


1

You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...


0

Clearly the first law, as stated in Ján Lalinský's answer implies the conservation of energy. Let us consider a system $S$ and a surrounding $\Omega$. Since energy is additive one can writhe the first law to the universe $T=S+\Omega$ as $$\Delta U_T=\Delta U_S+\Delta U_\Omega=Q_S-W_S+Q_\Omega-W_\Omega.$$ But The system can only exchange heat with the ...


0

The thing is, the book claimed the catapult had no recoil during launch thanks to its fixed counterweight. Is this physically correct? Of course not. The projectile has been given some momentum, so the launcher must recoil by conservation of momentum. You can't dodge that. In this case the problem with the proposed mechanism is that after the (initially ...


0

You have luck, the conversion is very simple. Do you remember $E=mc^2$? $c$ is the speed of the light, it is around 300 000 $\frac{km}{s}$. It is big. Its square is yet bigger. In SI, $\mathrm{1kg}$ of mass is $\mathrm{1kg}*(3*10^8\frac{m}{s})^2$ energy. It is $9*10^{16}J$. It is enough for a small city for a century. Unfortunately there is no way to ...


1

When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ...


0

Straight after the man throws the ball, his velocity is determined by conservation of linear momentum, that is, the momentum of the man recoiling is equal to the momentum of the ball leaving his hand: $Mv_1=mv$ After the ball bounces elastically off the wall, it returns towards the man at opposite velocity (equal in magnitude, opposite direction), so ...


1

When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ...


2

@ChesterMiller: answer is good, but I would just simplify it. What you say is that the total momentum is the momentum of the contents of the control volume plus the sum of that which traverses the boundaries. If you take the change in momentum per unit time, you have momentum flux, which equals force.


3

The forces acting on the fluid are changing its momentum. There are forces acting on the stationary portions of the control volume boundary, and there are also forces being applied to force fluid into, and acting to prevent it from flowing out of the control volume. The total rate of change of momentum in the control volume at any instant of time is equal ...


0

First, the authors positioned their reference frame at the center of mass, if you calculate the momentums that way, the total momentum will be zero, at least instantaneously. But that is not the point. In addition to the effects of other objects in our own solar system, mentioned in the other answer, remember that the Sun travels around the milky way ...


1

I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ...


6

Let me explain @ACuriousMind 's answer with some verbiage. The short, regrettably oracular, answer is that the Fabri-Picasso theorem does not hold in a finite superconductor, since translational invariance fails at its boundaries. Really, I do appreciate this is aggressively obscure: will strive to explain. First of all, if you have a chunk of warm ...


3

For generic initial conditions, the answer is Yes, due to Lagrange equations $$ \frac{dp_i}{dt}~\approx~ \frac{\partial L}{\partial q^i}, \qquad p_i~:=~\frac{\partial L}{\partial \dot{q}^i}. $$ [Here the $\approx$ symbol means equality modulo eom.]


0

The equation is built up from considering considering the mass $M$ moving with velocity $v$ splitting into $M+\delta M$ moving at ${\bf v}~+~\delta\bf v$ and a smaller piece with mass $\delta M$ moving with velocity ${\bf v}-\bf V$, for $\bf V$ the velocity of hot gases or plume from the rocket. The diagram below illustrates this Conservation of momentum ...


2

It's unclear what you mean by "nuclear engine", but the main similar notion is a nuclear thermal rocket. Although it derives its energy from nuclear reactions, it uses this to heat gas (usually hydrogen) to very high velocity for propulsion. There is still matter being expelled. More common in space travel is the radioactive thermal generator, which uses ...


2

A conserved quantity is a quantity whose value remains the same over time. An invariant, or scalar quantity is a quantity whose value is the same in all reference frames. These two properties are completely independent. Energy is conserved but not invariant. Mass (i.e. $E^2 - c^2 \mathbf{p}^2$) is invariant but not conserved. Charge is both, and lots of ...


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A physical constant is a physical quantity that is generally believed to be both universal in nature and constant in time. Well-known examples include the speed of light (c), elementary charge (e), and Planck's constant (h). (More about physical constants here.) A scalar (in the context of physics) is a physical quantity that can be described by a number ...


0

In Elastic collisions the interaction forces are conservative. We can represent the total Energy of the System as : E = U + K When the particles are far away from each other (separation > 2R) their potential energies remain constant which I choose to be U. This is true except when the particles are in contact which other. After collision the colliding ...


2

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of ...


1

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$. With the information given, there are 4 unknowns: $v_1, v_2, ...



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