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30

As a very rude guess, fresh snow (see page vi) can have a density of $0.3 g/cm^3$ and be compressed all the way to about the density of ice, $0.9 g/cm^3$. Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters. Going from $30 m/s$ to $0m/s$ (as @Sean suggested in comments), you'd have ...


18

The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$ x=-\frac{gt^2}{2}+v_0t $$ $$ ...


17

@Señor O gives a very good answer, but he assumes an ideal deceleration. Based on a viewing of the scene, Anna sinks a little under a meter, while Kristoff doesn't sink more than half a meter. Since they fell about 200 feet (about 60 m), my initial estimate for their impact velocity is (assuming no air resistance): $v = \sqrt{2gh} = \sqrt{2*60*9.8} ...


17

This is another chance to use one of my favorite approximations ever! I first offered it as an answer to a question about how deep a platform diver will go into the water. Now is the chance to use it again! Issac Newton developed an expression for the ballistic impact depth of a body into a material. The original idea was expressed for materials of ...


11

Nice theoretical answers (I can certainly appreciate them, I'm a mathematician). But why delve into theory when experiment is available? In this video you can see a skier jump from more than 200 feet and get head first into the snow, without a helmet. The video starts with the aftermath, if you want to see the jump right away fast forward to about 1 ...


10

John correctly stated that this is possible because re-configuring our bodies allows us to change our moment of inertia, but not our mass. As the question was about an intuitive explanation, consider adding a series of floating weights to get an analogous situation for translational motion: The astronaut stretches their arms above the head, grabs a weight, ...


6

No, it doesn't imply that $a = 0$. If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there: $$x(t_0 + dt) = x(t_0)$$ which means that at $t = t_0$ $$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$ thus $\frac{dv}{dx}$ is undefined at $t = ...


6

The rotational energy of a body is given by: $$ E = \tfrac{1}{2}I\omega^2 $$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$ I = \frac{2}{5}mr^2 $$ You already have the mass, and you can Google for the ...


5

You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When $v=0$,make sure ...


4

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ...


3

If the cable of the "elevator" is not connected to a point on earth, then the satellite must be in a geostationary orbit (or it will float away); this implies that if you now attach something to the platform (increasing the pull on the cable) you will pull the satellite down to earth. And as @lionelbrits pointed out, the pulling part of a space elevator ...


3

"Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ...


3

It seems helpful to consider an extremely simple scenario. Suppose an astronaut is floating near two balls of lead; in this case the closed system consists of the astronaut together with the balls. She can pull the balls together without changing the momentum or angular momentum of the system. She can then rotate them in the center with almost no change, and ...


3

Why am I not accelerated by the reaction force applied by earth on me? Because the net force on your centre of mass is zero. The upward force on your feet is of the same magnitude as the downward force of gravity. Your major leg bones and spine are in compression because of the opposing forces. I know that these forces will not cancel each other ...


3

Even classically, forces arise from field being propagated at the speed of light. A physically relevant object is the energy-stress tensor, whose components represent energy density and momentum current density, so indeed momentum can be interpreted as a current that is conserved over time (as a consequence of symmetries). This point of view is also ...


2

Let's say car and bike be at rest at $1pm$ so, $v_c=0$ and $v_b=0$. Calculations for motion of car: Since car is moving with constant acceleration, At 1:00:00pm, $v_c=0m/s$, $S_c=0m$ At 1:00:01pm, $v_c=4m/s$, $S_c=4m$ At 1:00:02pm, $v_c=8m/s$, $S_c=12m$ At 1:00:03pm, $v_c=12m/s$, $S_c=24m$ At 1:00:04pm, $v_c=16m/s$, $S_c=40m$ Calculations for motion ...


2

In the framework of General Relativity, where the inertial frames are the ones in free fall, you can think that the Earth is accelerating upward, so it is not you who is pushing on Earth but it is Earth that is "running you over" because of its accelerated motion. Luckily enough, if we are standing on ground, we can avoid impulsive forces and our bodies are ...


2

When you push something and it remains at rest your muscles transfer energy through isostatic muscle contraction/respiration. This means that even though the muscles don't move they convert the glucose into respiratory energy for muscle contraction that will be dissipated eventually by heating the surroundings. The only work done is that in contracting the ...


2

Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass: $$ m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1 $$ $$ \frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2} $$ and ...


2

Variations of this problem show up all the time. If you start with the spring "locked" and the spheres charged, then release the spring, it will expand to the new length and when it gets there the spheres will have a velocity - essentially you have a simple harmonic oscillator and the point of (new) equilibrium is the point where the oscillator moved ...


2

$F = \frac{dp}{dt}$ means that force is the rate of momentum transfer per unit time. Lets say we have mass $m_1$ moving to the right, and mass $m_2$ is on the left side of $m_1$ with zero velocity. If $m_1$ put a force to pull $m_2$, that force will create the acceleration on $m_2$ and increase its velocity, this also means the change in momentum. At the ...


1

Since the force is radial, you are not changing the angular momentum, but you are adding potential energy: this tells you what must happen to the tangential velocity (decreases) and radial velocity (increases). I will leave it up to you to figure out by how much.


1

But if I take the absolute value of both sides, and drop summation It might be dangerous just to "drop" summation. You have to include all particles moving before and all moving after collision. In your case of only a ball and a wall it would reduce to: $$\sum m v_{1x}=\sum m v_{2x} \Rightarrow\\ m_{ball} v_{1x,ball}+m_{wall} v_{1x,wall}=m_{ball} ...


1

The steps in the OP's suggested solution are deeply flawed. First: you cannot take the absolute values of the two sides of a vector equation. Momentum has both size and direction, and both must be taken into account when doing an addition. Would you examine your bank statement for the month, while treating the deposits and withdrawals differently. Is a ...


1

After the mass exits the pipe, the tube will start to rotate from the recoil. At $t=0$, there is zero angular momentum, $L=0$. Let's take it that the pipe rotates about its centre-of-mass, and use that point as the origin from which to calculate the angular momentum. At time $t+dt$, the puff of gas has angular momentum $L_{gas} = dm \times l/2 \times v0$. ...


1

If we assume that mechanical energy (K+U) is conserved in both the earth frame and the initially co-moving, constant velocity frame, then it's not the differences in velocities which are the same; it's the differences in the squares of the velocities which are the same. $$\frac{1}{2}v_{1e}^2=\frac{1}{2}v_{0e}^2+2gh$$ and ...


1

Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar.


1

First off, when two values are directly proportional, it not only means that they are related, but also that they are related by a constant of proportionality (i.e. as one changes, the other changes proportionally). For example, in your second equation, the centripetal force is directly proportional to the radial distance to the mass and proportional to the ...


1

The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed. ...


1

Even I didn't get you but I may help you how much I can by describing your case. Your case have two bodies which are being rubbed against each other in opposite direction with constant acceleration. The definition of friction is, "The resistance which either one of the bodies offers to this motion is called the force of friction and is said to be due to ...



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