# Tag Info

1

This seems like a list question which aren't very good for SE. Here is a proof of your statement: $m_1v_1 = m_2v_2$ $\dfrac{m_1}{m_2} = \dfrac {v_2}{v_1}$ Let us assume that $m_1$ < $m_2$ $\dfrac{(m_1)^2}{(m_2)^2} = \dfrac {(v_2)^2}{(v_1)^2} < 1$ $v_1$ must be greater than $v_2$ The only case the energy of a higher velocity is beaten is shown by ...

2

Your analysis will need to be a combination of theory and experiment. You are not maximizing impulse alone: you need to think about drag as well. So you need to think about the shape of your rocket as well as the volume of water and initial pressure. The drag on the rocket is probably well modelled by the ram pressure equation ...

4

A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons. Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by: $$E^2 = p^2c^2 + m^2c^4$$ where $p$ is the ...

12

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + ...

6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...

2

Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ...

2

Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ...

0

The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at ...

0

The conservation of momentum is simply a statement of Newton's third law of motion. During a collision the forces on the colliding bodies are always equal and opposite at each instant. These forces cannot be anything but equal and opposite at each instant during collision. Hence the impulses (force multiplied by time) on each body are equal and opposite at ...

1

You didn't cancel out the mass $m$ properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision, $a=b$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...

1

I will bring an example from classical electrodynamics. In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the ...

1

If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear. For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the ...

2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...

2

The thrust coming from a rocket engine is exerted on the engine bell, and it is directed along its axis of symmetry. It's not completely clear how you're modelling your ship but it is probably more realistic to apply the force to the "thruster fire" block, whatever that is. It's important to note, though, that if applying the force to the engine bell and ...

0

Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion $$F = m g$$ with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$. Therefore, as ...

4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...

-2

F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it. a better equation to represent the effects of gravity would be: F=(G*m1*m2)/d^2 where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) ...

0

Kinetic energy is the work required to accelerate a mass from rest to a velocity (KE = 1/2 mv^2). Momentum is a measure of the amount of movement a mass has at a velocity (p = mv). Kinetic energy may be considered a process, and momentum may be considered the result of a process. Momentum is conserved, but kinetic energy seems to come and go, as it ...

0

That's easy. Think in a simple example that this happens. Imagine two particles of equal masses moving at $\vec v_1 = \vec v$ and $\vec v_2 = -\vec v$. Their momentum: $\vec p_1 = m\vec v_1$ and $\vec p_2 = m\vec v_2$. The momentum of the system is therefore: $$\vec p = \vec p_1 + \vec p_2 = m\vec v_1 + m\vec v_2 = m\vec v - m\vec v = \vec 0$$ The ...

0

Since momentum is a vector, the quantity being measured did indeed change. As you state, while the magnitude is a constant $10\ \mathtt{kg\cdot m/s}$, the direction has altered. But what does that mean? The meaning of this change lies in understanding the distinction between momentum and impulse. The change of momentum is called an impulse. The common ...

1

Hint: The center of mass of pieces A and B moves in the same path as the intact shell would. (This arises from the conservation of momentum.) Edit: The gravitational force is only acting along the vertical direction. So there is net acceleration only in the vertical direction. Looking at the horizontal one, if we neglect things like air resistance (which ...

3

Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ...

0

Suppose a box is pushed in a vacuum.Due to this push the box at rest starts to move.But in a vacuum there are no other forces which oppose the motion of the box.So it will continue to move.Or to think in energy terms,once we gave the box energy there is no mechanism for the energy to transfer from the box.The energy has been "trapped" in it.

1

As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...

1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...

4

Momentum: The resistance of an object to a change in its state of motion. That sounds like a fishy definition of momentum to me. A slightly better definition, at least at your level, is that momentum represents the "amount of motion" an object has. Granted, "amount of motion" is a very vague term, but it stands to reason that if "amount of motion" were ...

4

Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object. A massive object can have any momentum (at least as long as its velocity ...

4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...

0

here's a short simple answer: the Hamiltonian of the pencil can be approximated by an inverted harmonic oscillator near the equilibrium (downward parabola) . It's an easy exercise to solve.

0

Yes. Some of the impulse force is used in work which rotates the rod, and some of the impulse force is used in work which changes the bullet's trajectory. The distribution of impulse force between the rod and the bullet is equal.

3

You are right. If $q$ is a generalized coordinate then $\dot{q}$ is the generalized velocity and hence the generalized momentum is $$p = \frac{\partial L}{\partial\dot{q}}$$ Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the $p_{\theta_i}$ are indeed ...

5

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...

5

That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$v = \frac{pc^2}{E}$$ When $m = 0$, \$E ...

Top 50 recent answers are included