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-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 ...


1

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

Yes, the derivation is correct. The intuition should be that there should be no Doppler shift, since the source and detector positions are fixed (light path length is not changing).


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 ...


1

When we think about balls bouncing off walls and light bouncing off mirrors, we assume that the there will be a momentum exchange, but only for components that are perpendicular to the plane. If the mirror has some velocity component in the perpendicular direction, it affects the interaction. It can add or subtract momentum from the reflected particles. ...


4

OP is pondering if the corresponding Hamiltonian formulation is affected if the Lagrangian density $$\tag{1} {\cal L}~\longrightarrow~\tilde{\cal L}~:=~ {\cal L}+\sum_{\mu=0}^3d_{\mu}F^{\mu}$$ is modified with a total divergence$^1$ term $d_{\mu}F^{\mu}$, so that the definition of canonical momentum$^2$ $$\tag{2} p_i~:=~ \frac{\delta L}{\delta ...


2

Theoretically, you can try to move your arms downwards (for a short time), that would tend to move the rest of your body upwards. The mass of the arms is approximately 10% of the total weight of the man's body (http://www.timesdaily.com/archives/weighing-in-on-individual-body-parts/article_4729f5a7-c039-5649-910e-ee18a03435e0.html ). To have the rest of your ...


5

There is the principle and the practice. Let's look at the principle. If you have two arms, each with mass $m$, and length $\ell$, we could restate your question as saying: "how, and how fast, do I have to move such arms to make my body lift off?". That's quite easy. Assume you are holding your arms out sideways. Their center of mass is at $\ell/2$, and ...


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 ...


0

Let's say $\Phi$ is a delta function, $\Phi(k)=\delta(k-k_0)$. Presumably, you want this to be an eigenstate of the momentum operator with momentum $\hbar k_0$. With the convention you've chosen, we can convert this to a real-space wavefunction (I'm ignoring normalization for convenience): $$ \Psi(r)= \int dk \delta(k-k_0)e^{ikr}=e^{ik_0 r} $$ We can then ...


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 ...


3

Two different Lagrangians give different canonical momentum. If two different Lagrangians differ by a surface term then they differ by a total divergence. And they thus yield the same actions hence have the same equations of motion. When you do integration by parts you produce a surface term (the difference between the two). Imagine subtraction the two ...


-1


0

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

You can actually think of the "little rain drops" that you are picking up as providing some resistance. When you travel at velocity $v$, and have area $A$, you are "picking up" all the material in a cylinder with volume $V=vA$ per unit time. That volume of material needs to be accelerated to velocity $v$, requiring a force $F\Delta t \propto m \Delta v = ...


0

It's clear, higher you go, the Kinetic Energy increases. Let's see how : $v =\sqrt{2ax}$ where $x$ here is the $h$ and $a$ here is $g$. so we can write it like that: $v =\sqrt{2gh}$ And we have the Kinetic energy formula $K=\frac{1}{2}mv^2$ so, as you see, increasing $h$ causes a bigger $v$ that a bigger $v$ effects on the Kinetic Energy you will ...


1

When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...


1

You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...


0

Recall that force is equivalent to, $$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$ Where $p$ is the momentum, and $t$ is time. Momentum is given by, $$p=m \cdot v$$ Where $m$ is mass, and $v$ is velocity. When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small. When you hit the ground ...


1

then the created pions will be at rest correct? Well, they will be at rest in the Center of Momentum frame. But that is not the frame of reference that your problem is stated in. Momentum is conserved, which tells you that you have written the pion four-vectors incorrectly.


1

A Community Wiki Answer to capture another User James Large's most excellent summary made in the comment: Two key words to take away from the answers below are stress and strain. Good words to search for if you want to learn more. And remember, "It's not the fall that kills you, it's the sudden internal strain at the end." This is an excellent way of ...


3

It will increase an electron's momentum in the electron's current direction One must also take the gradient of that dot product so your conclusion isn't valid. Assume, for simplicity, that $$\phi = 0$$ $$\vec A = A_x(x,y,z)\; \hat{\mathbf x}$$ The first equation is then $$M\vec a + \frac{q}{c}\left(\nabla A_x \cdot \vec v \right)\hat{\mathbf x} = ...


0

I have no idea what you're on about, here. The one-sentence laymen's phrase for $p = \hbar~k$ is "Every moving object behaves like a wave, with wavelength equal to Planck's constant divided by that object's momentum." Usually you then add a one-sentence caveat, "The Planck constant is so tiny relative to most macroscopic masses that this wavelength is ...


0

After some in depth discussion with the users, we found that there are a lot of assumptions in the question that does not hold water: "The commutation relation of $E$ and $t$ is not even a commutation relation. Time is not an operator in quantum mechanics." (thanks ACuriousMind for reminding again), it is a parameter thus there is no (straightforward) ...


2

I have a gut feeling that my reasoning is indeed faulty, but I'm unable to figure out why. If a 500N (net) force acts on a 30kg object, the acceleration of the object is $$a = \frac{500}{30} \mathrm {\frac{m}{s^2}} \approx 1.7g$$ which gives a 0 to 100 km/h time of 1.67 seconds thus beating all of the quickest supercars. In other words, there's ...


0

In general, I might prefer your solution to the official one, the more so as their numbers are wrong, as Alex noted. There is, however, a problem with your solution as well: the athlete could not achieve the speed of 250/3 m/s (I guess it's 300 km/hour), even if (s)he did not have to push the luge. So I guess before s(he) jumped onto the luge her (his) speed ...


2

As the athlete pushes off the ground, she and the luge would both be accelerating relative to the ground. This force of 500N for 5 seconds would result in a final momentum of 2500Kgm/s. If the system has 100kg total mass then you simply divide the momentum of 2500kgm/s by the mass 100kg to find the velocity which I believe is 25m/s not 250m/s. Also note that ...


1

I agree with James' explanation of the difference between a "theory" and a "quantity" but just to be more focused on the two keywords in the original question, "linear momentum" and "thrust". They are not the same concepts because they do not have the same units. The thrust is a force (caused as a reaction to the opposite force). The SI unit of a force is ...


1

"momentum" and "thrust" are not theories: They are quantities that can be directly or indirectly measured. We make theories about measurable quantities like thrust and momentum. "Thrust" is another name for the force that is exerted on a vehicle by its own propulsion machinery. Momentum is the scalar product of the mass of a body and its velocity. It's ...


0

If I may explain by means of a few "detours". I am structural engineer (not a doctor), but I do have a good sense of forces and energy dissipation through a structure. In this case the structure being the human body. Your question does not only apply to skydivers who get fatally injured, but every injury relating to an impact. The only difference is the ...


6

It is not that simple. Injury arises from a variation of acceleration with position over parts of the body. In the case of a fall, when the first part of you hits the ground (say your feet) and stops suddenly, there is nothing decelerating the rest of your body aside from the force that the feet can transmit to it. So this situation gives rise to compressive ...


8

Assuming terminal velocity of 200 km/hour, the scenario seems equivalent to stepping out of a car that's travelling at 200 km/hour. In that case it's not the fall (hitting the road) that kills you, it's the friction (i.e. sliding or tumbling along the road). There might be a minimum of friction initially (when you're falling parallel to a vertical wall) but ...


14

I have slid down a much smaller version of this at Burning Man. Paha'oha'o was a 30 foot tall volcano art piece which you climbed and then "sacrificed" yourself by dropping into a pit featuring a slide just like you mention. The drop features a 10 foot free-fall, just enough to take your breath away, after which the careful curve of the slide gently catches ...


7

Probably the closest to what you are asking about is the story of Ivan Chisov's survival (see Ivan Chisov); but there have been several other similar cases (see for example 10 Amazing Free Fall Survivors).


35

The answer is Yes and your thinking is correct. You try to differ between impact and sliding on a curve. In fact the impact is just a sudden large force, while a curved (e.i. circular) motion similarly applies a force, just much smaller but also over a longer period of time. The key in surviving any fall is to reduce the force on your body at "impact". A ...


26

Let's make life easy for ourselves by assuming that the slide is an arc of a circle: We also assume the slide is made out of something with a very low friction, so the skydiver maintains a constant speed $v$ all the way round. The reason that using an arc of a circle makes life easy is that the acceleration felt by the skydiver is simply: $$ a = ...


2

Yes. In fact it would be better to imagine that you skydive towards a "track" that you can strap a "chair" onto, and then the chair is stuck on the track. the acceleration to keep you in a circular orbit of radius $R$ is only $v^2 / R;$ with terminal velocity being about $v \approx 56 \text{ m/s}$ a $1~g$ acceleration will be accomplished by a radius of ...


2

The way you do calculations like this is to work in the centre of mass frame. The setup you describe looks like this (the convention here is that velocities to the left are positive, and velocities to the right are negative): But suppose we view the collision from a frame moving to the left at $v/2$ i.e. we have to subtract $v/2$ from all the velocities. ...


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 ...


0

Your inertia is a vector mass x velocity If you are turning 10 degrees right your inertia in the new direction is Inertia X cosine(10) = inertia X 0.9894 The rest of energy is friction of turning A wheel on a axle is diffent as straight is bacially the circle


0

When you turn, there is a friction force between your bicycle tires and the road. That friction force is opposite the direction that your turning tire would like to go (it would like to slide straight), and it is this friction force that causes your bicycle to change directions. It shouldn't be surprising that the friction force did some amount of work on ...


0

In a collision there are two key points to consider, which I think you're getting a bit confused over. The first is conservation of energy. In an elastic collision, we have that $\sum_{bodies}{\frac{1}{2}m_{body} |\vec{V_{body}}|^2}$ is constant. Now this doesn't impose any conditions on the direction of $\vec {V_{body}}$, so if initially the bike has ...



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