# Tag Info

## Hot answers tagged momentum

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: $$a = ... 14 I have slid down a much smaller version of this at Burning Man. Paha'oha'o was a 30 foot tall volcano art piece which you climbed and then "sacrificed" yourself by dropping into a pit featuring a slide just like you mention. The drop features a 10 foot free-fall, just enough to take your breath away, after which the careful curve of the slide gently catches ... 8 Assuming terminal velocity of 200 km/hour, the scenario seems equivalent to stepping out of a car that's travelling at 200 km/hour. In that case it's not the fall (hitting the road) that kills you, it's the friction (i.e. sliding or tumbling along the road). There might be a minimum of friction initially (when you're falling parallel to a vertical wall) but ... 7 Probably the closest to what you are asking about is the story of Ivan Chisov's survival (see Ivan Chisov); but there have been several other similar cases (see for example 10 Amazing Free Fall Survivors). 6 It is not that simple. Injury arises from a variation of acceleration with position over parts of the body. In the case of a fall, when the first part of you hits the ground (say your feet) and stops suddenly, there is nothing decelerating the rest of your body aside from the force that the feet can transmit to it. So this situation gives rise to compressive ... 5 There is the principle and the practice. Let's look at the principle. If you have two arms, each with mass m, and length \ell, we could restate your question as saying: "how, and how fast, do I have to move such arms to make my body lift off?". That's quite easy. Assume you are holding your arms out sideways. Their center of mass is at \ell/2, and ... 4 OP is pondering if the corresponding Hamiltonian formulation is affected if the Lagrangian density$$\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 ...

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 ...

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 ...

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 $$\Delta v = v_{ex}\ln(\frac{m_i}{m_i - r_bt}) - gt$$

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} = ... 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. 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 ... 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 ) 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 ... 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, ... 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 ... 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 ...

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. ...

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 ...

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 ...

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

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 ...

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

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." ...

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 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 = ... 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. ...

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. ...

Only top voted, non community-wiki answers of a minimum length are eligible