# Tag Info

-1

Your calculation of final momentum after the collision has a sign error in it. The pulley serves to change the direction of the motion. This means that a mass moving upward on the left side of the pulley is given a mathematical sign of "+" for the associated velocity. As the string goes over the pulley, the direction of the motion changes such that a ...

0

The ceiling is applying a total upwards force on the pulley of magnitude $2T$, in order to counteract the force of equal magnitude applied by the masses. So the net force on the system (consisting of three masses and the massless pulley and string) is not zero. Since $\int {Tdt} = mV = mv/3$, $\int 2Tdt = {2 \over 3} mv$ which is the amount of momentum lost. ...

0

In my opinion, the statement regarding the conservation of momentum says that "The momentum of a system remains conserved if no external force acts on it". I think that the mistake you have committed is that you tried to apply momentum conservation principle along the y-direction, along which gravity(an external force for the system) acts. So according ...

2

According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, ...

4

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 ...

0

I think the real reason is because if you change your velocity then everything that used to be at rest should now continue just as they did before you changed your velocity. You can call it indifference, nothing cares how you move unless you interact with it. Or you can call it relativity. What this does is reduce the case 1) of things at constant motion ...

1

In addition to the good answers above, I'd like to add a point about the nuclear reactor losing mass. It's true, it does, albeit at a really tiny rate. The reason that this doesn't contribute to propulsion is that the mass is loss in all directions so the net effect is zero.

2

No, This device would just oscillate (vibrate) When the first two electromagnets switched on the metal ball (which is presumably made out of a ferromagnet) would accelerate towards them but the craft would also accelerate towards the ball (At a lower acceleration as it has more mass). It would then switch to the next electromagnet pair and do the same ...

0

Before explosion the bomb is at rest. Its total momentum is zero. As it explodes, it breaks into many parts of masses $m_1,m_2,m_3$ etc which fly of in different directions with velocities $v_1,v_2, v_3$ etc. these diff parts have different momenta $m_1v_1,m_2v_2, m_3v_3$,etc. For eg,- If the bomb explodes in two parts then both of them fly into opposite ...

1

A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a ...

0

It is difficult to understand conservation from symmetry. But the opposite is much simpler. conservation means the invariance of equation of motion in its form under certain transformation. and the invariance of equation of motions arises as an implication of the underlying symmetry. for example i am taking one equation X2 =1 the solutions are +or-1 which ...

1

The force of humans on the earth is already there before they jump, because people are standing on the earth. With their weight distributed over the entire earth, their jump (which might briefly increase the force by 2-3x) will have no effect. Since you said they would be jumping in the sea, there will be a very small increase in the sea level. Mass of all ...

-1

Ridiculous idea! How could that possibly work?

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You are in a boat. You throw a ball outside boat with speed $v$. It will push the boat foward by conservation of momentum. Thus, assuming initially boat and ball are at rest, the boat will have opposite momentum of the ball, and then boat will move foward. $$\mathbf p_{boat} = -\mathbf p_{ball}$$ This is mathematically OK. There is no doubt that boat ...

0

Of course the the boat will move, you are converting the electrical energy from the battery into the rotational energy of fan, which produces almost a rectilinear motion of air particles, thus the momentum of air particles is changed by the the motion of blades of fan, which means that the blades of fan would be applying a force on the air particles, and the ...

0

The fan moves air (pushing it from the front of the boat to the back). Conservation of momentum say that "if something moves one way, something else must move in the opposite direction". In this case - air moves back, boat moves forward. If you were holding the fan in your hands you would feel a force (as it is pushing against the air). That force is ...

2

There is no doubt that such a system would move, as other people here say, moreover, such a system can be quite practical and is actually used in marshy/shallow water (https://en.wikipedia.org/wiki/Airboat )

0

The boat may obviously move forward. That's how airplanes move, too. The turbines etc. make the fluid move backwards so the boat or aircraft has to move forward. A more interesting fact is that it may also move forward by creating "wind in the forward direction" by its own fan if this wind is reflected from the sail. Mythbusters have checked this question ...

3

If your fan-boat is in vacuum, they won't move. In the air, they will. Your assumption conflicts with your intuition is because you isolated the system from the air, which should not.

1

Let the balls meet at the height $d$ from the ground then at the time of collision, the velocity of the ball (mass $m$) is given by third equation of motion as follows $$v^2=(0)^2+2g(h-d)$$ $$v^2=2g(h-d)\tag 1$$ & the time taken by ball (mass $m$) to reach at the point of collision is given by second equation of motion as follows ...

1

Let $t$ be the time from the dropping the fist ball until the collision of the balls. Then, $v_1=gt$ and $v_2=u-gt$. Moreover, $d=ut-\frac{1}{2}gt^2$ and $h-d=\frac{1}{2}gt^2$, so that $ut=d+(h-d)$, which gives $t=\frac{h}{u}$. Because $mv_1=m_1v_2$, we must have $m\left(\frac{gh}{u}\right)=m_1\left(u-\frac{gh}{u}\right)$, or ...

1

There exist particles which are their own antiparticle, even at the elementary particle level. Photons, and gluons are their own antiparticle and thus no need to come in pairs.

1

If you count a $\pi^0$ as a particle, you can create and destroy them individually. This is similar for any particle that is its own antiparticle (like the photon, the $Z^0$ and the Higgs.) You can have an arbitrary number of these.

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The creation of particles has to respect a number of conservation laws. For example charge has to be conserved, so it you create an electron with a charge of -1 you have to create a particle with a charge of +1 to balance it out. Likewise lepton number is conserved (in the Standard Model at least). An electron has a lepton number of +1, so if you create an ...

0

Jean Bouridan, rector of the University of Paris around 1350, was the first philosopher, to my knowledge, who specifically stated the current concept of momentum. He said that impetus was proportional to the product of weight and speed. Momentum is considered to be the product of mass and velocity (velocity has direction as well as magnitude). Momentum is ...

4

You've got it a little backwards - physicists first defined the quantity $m \cdot v$ because it quantified the amount of "motion" an object possessed. They named it "momentum". Modern physics is primarily concerned with the quantity $m \cdot v$ (and the updated versions of that quantity in more recent frameworks of physics) because it is conserved. This ...

1

What is missing: this must be recognized as a relativistic process and be treated accordingly. Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using ...

2

General Question: Why should I use just the friction force rather than the net force to integrate over distance when conserving energy? Answer: In energy conservation problems each way of storing energy generally gets it's own term. In the example problem there is a gravitational potential energy term (GPE), a kinetic energy (KE) term, and a friction ...

5

The Earth's tilt doesn't change so much as it's position around the sun changes. Notice the North Pole sees more sun in summer than in winter cause it's tilted towards the sun in summer but not in winter. The north pole always points in the same direction into space pretty much. It wobbles slightly, and quite slowly, completing a full wobble every ...

4

The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...

3

Well, since unburned fuel is not distinguishable from dry mass, we could replace the above equation with $$\Delta v = v_{ex}\ln(\frac{m_i}{m_i - m_b}) - gt$$ where $m_b$ is the mass of the burned fuel. Assuming constant burn rate $r_b$, it would just be $$\Delta v = v_{ex}\ln(\frac{m_i}{m_i - r_bt}) - gt$$

1

It is not enough for it to be moving - it needs to accelerate (or decelerate). An accelerating charged particle will emit radiation and it will lose energy as a result. An excellent example would be the loss of energy of charged particle in synchrotron accelerators. They emit... synchrotron radiation. This is either a boon (e.g. the Diamond light facility ...

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