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Use conservation of momentum, which tells you that the total momentum (the sum of the momenta of the two particles) before and after collision must be the same. Also note that the momentum is a function of the vector velocity, which means that you can make two independent analyses, one on the $x$-axis, and one on the $y$-axis. Both should respect ...


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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 $L_{gas} = dm \times l/2 \times v0$. ...


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As far as I know, when a wheel rolls w/o slipping, it always has friction in the opposite direction of its movement. Here you go wrong . In pure rolling (that is no slipping ) the bottom point is at rest wrt ground . So there is no kinetic friction acting on it as there is no relative motion .But static friction acts . In your question gravity ...


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Friction acts on objects at rest too . The definition meant that if there is relative motion between two objects then friction will act as a resistance between them . If you find two objects at rest even when an external force is applied on it then it means friction is acting on them . Had there been no friction there would have been relative motion and the ...


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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 ...


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In lower gravity, you could expect to swim faster I am not answering the other questions as I do not have much more to say which is not already said in other answers. But I do disagree with their conclusion that swimming would be the same. Regarding swimming, one would need a better understanding of swimming motion to decide how much effect can be expected ...


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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} ...


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This was a tough question to crack and I wanted the Stack Exchange to carry the answer, as the community is a great resource. If you dont realize that her speed is relative to the plank, you'll be wondering what kind of problem this is. Answer: There is no external force on the system, so the sum of momentums stays = 0. And velocity of the girl relative ...


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Considering both the discs as the system , we can conserve angular momentum about their collinear axis of rotation . The torque due to friction will decrease the angular velocity of the disc having more angular momentum (before the collision ) while the torque will increase angular velocity of the one which had lesser initial angular momentum . I am assuming ...


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I've read a little bit on the science of a space elevator and it's a surprisingly difficult problem. To have a working space elevator, it would need to be at least to the Geosynchronous orbit, 22,000 miles up, probobly a bit beyond that for buoyancy. The highest balloon is some 25 miles - so that's less than 1/10th of 1% of the distance. The strongest ...


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"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 ...


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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 ...


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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 ...


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The simplest way to look at this is to consider separately the horizontal and vertical velocity/position. For a projectile launched at angle $\theta$ and velocity $v$, the components are: Horizontal velocity $$v_h = v\cos\theta$$ Vertical velocity $$v_v = v\sin\theta$$ The position at time $t$ is then given by $$(x, y) = (v_h\cdot t, v_v \cdot t - ...


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logically in the frame of centre of mass the accn of body is zero so momentum is conserved and as mass has not changed initial velocity is equal to final velocity in frame of centre of mass


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In addition to Pentane's solution, this can be done as a projectile problem as well. The following is a basic kinematics equation $$v_f^2 = v_i^2 + 2 a \Delta x$$ At the top of your jump $v_f=0$, so you can solve for $v_i$ and get the answer that way.


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Swimming would be nearly identical to a 1g planet, other than the splash being bigger. The forces involved in swimming are largely horizontal, so as long as there is some gravity to keep the water where it belongs you are acting against the viscosity of the water rather than the weight of the water. Might be a problem at very low g as you would splash away ...


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Start with the two pertinent conservation laws for elastic collisions: kinetic energy and momentum. Remember that momentum is a vector. In the center of mass frame, the total momentum is zero. That will get you started. Do the work for two particles first. As an aside you should try to show the total momentum is zero in the CoM frame by example by taking ...


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You can use conservation of energy: the kinetic energy when they land has to be the same as the kinetic energy when they left the ground. The kinetic energy when they land is also equal to the potential energy when the person is 60 cm off of the ground. So you can use: $$mgh=\frac{1}{2}mv^2$$ and the mass cancels out leaving: $$gh=\frac{1}{2}v^2$$


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The resistance to motion can also occur when the object is at rest. I push on a book, and it doesn't yet move. The object is not in motion, so there must be force that balances my push. That force, which is resisting motion, is friction. When the two objects are in relative motion we call it kinetic friction. When they are at rest relative to one ...


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Philosopher who said to you that the pressing force taken for friction's calculation is Normal force or reaction force exerted by earth on body? Actually, the pressing force is called the normal force (not reaction force) Since it is the force perpendicular or "normal" to the surfaces which affects the frictional resistance, this force is typically called ...


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Interesting question. 1 and 2 I agree with you. 3 - I think swimming would be similar to on earth - from the point of view of floating on the surface what counts is the density and our bodies and water have similar density - we are a bit less dense and float - swimming we force our bodies to go through water against the resistance of the water, which ...


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1) You'd be able to jump pretty high correct? Two or three times whatever you could on earth? Since you would have the same strength as on earth, the initial kinetic energy of your jump would be the same as on earth. Since your mass is the same as on earth, your initial jump velocity is the same as on earth. Thus the height to which you jump would be ...


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About 50 years ago in Reader's Digest there was an article about a Soviet airplane pilot who bailed out at high altitude. He fell into a snow-filled ravine and survived. If the angle of the snow is high enough it is no big deal. At Squaw Valley I have seen skiers do drops that might have been 100 feet. If the landing is steep enough it is OK. It is ...


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The restoring force due to a spring is defined to be the force for a unit displacement.The force acts in a direction that tries to restore the mass back to its equilibrium position. (Note since the spring is mass-less you have to have a mass in place since you don't apply forces on mass-less objects)


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One of the most important of all dynamical problems is that of a mass attracted toward a given point by a force proportional to its distance from that point. Let a spring be in its relaxed state- neither compressed nor extended. One end is fixed & on the other end, a point-like object - a block,say - is attached. If we stretch the spring by pulling the ...


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The word 'restoring' is synonymous with 'opposing' in that it matches the applied force, but in the opposite direction. But more so 'restoring' implies that energy is being stored - potential energy - which can subsequently be retrieved. The potential energy is the integral of force over the path of deflection: $$E_p=(1/2)kx^2$$ The energy imparted by the ...


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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 ...


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Energy conservation of the system in your example restricts solutions which will make energy dependent on time. Conservation laws are conclusions of symmetries of equations of motion, and for system of springs it seems that energy conservation is a valid law. It is not possible to write down trajectory that would have any sort of damping and solve equations ...


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$F = \frac{dp}{dt}$ 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 ...


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The $d$ in front of momentum and in front of time means infinitesimial change of time $$dt = t_{final} - t_{initial}$$. Therefore the change in momentum over the change in time equals the force! Also momentum is equal to $m\cdot u$ ($u = \text{velocity}$) . So the change in momentum is equal to$$ dp = m\cdot u_{final} - m\cdot u_{initial}$$ . We also know ...


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To you, lets apply the second law. As it states, actually $\sum F=m a$ NOTE THE SUMMATION, which means the resultant force applied. If you're just sitting or not moving, both normal force and gravity exerted in opposite direction, for its magnitude, they equals. You're still having problem with how will the Earth move ha? The third law tells, You give the ...


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I know that these forces will not cancel each other out since they are applied on different bodies and not a single ... That's correct. ... but I ask if net force is not zero then according to the equation, a=Fm, I must have some acceleration produced in my body. First off, it's $F=ma$ (or equivalently, $a=F/m$), not $a=Fm$. One thing you are ...


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So yeah, there are these things called vectors which represent arrows; they are made up of components. For example, we would not say "velocity (horizontal)" and "velocity (vertical)" but we would add labels $v_x, v_y$ where $x$ is usually a horizontal-label and $y$ is usually a vertical-label; then we would write the arrow as $\vec v = [v_x, v_y]$, packaging ...


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The motorcycle is moving at $10\frac{m}{s}$, so his distance from a starting point of 0 each second can be modeled as: $$y = 10x$$ The car is moving at a constant acceleration, so that distance from an origin of 0 can be modeled as: $$y = 2(x^2 + x)$$ Therefore, when they are equal to each other, they are at equal distances from the starting point. ...


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You made a mistake in the direction of your drag force (vector). You calculated the magnitude of this force correctly, then applied that magnitude both along your horizontal and vertical basis vectors for a total of $\sqrt{2}$ of the desired drag force, and pointing in the wrong direction! The missing direction is given by $-\vec{v}^0 = -\vec{v} / \left| ...


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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 ...


0

Schrödinger's cat and the double-slit experiment might be famous things to play with. Of the latter it was said: a phenomenon which is impossible […] to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery [of quantum mechanics].


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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 ...


1

Consider the well-known mechanical harmonic oscillator, where you combine Newtons 2nd law $F = m\ddot{x}$ for the motion of an object with mass $m$, with Hooke's law $F = -kx$ for the force from a spring with stiffness $k$. The resulting differential equation is: $$ m\ddot{x} + kx = 0$$ This is a linear differential equation, because if we add the equation ...


0

I think you just need to treat two forces separately relative to each body as $\vec{F}_1=-\vec{F}_2$ and $\lvert|\vec{F}_1\rvert|=\lvert|\vec{F}_2\rvert|$. Having in mind that $\vec{F}_1(t),\vec{F}_2(t)$ both are functions of time because the motion goes with constant acceleration.


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I think it helps if you look at the problem from more of an intuitive perspective. When you're standing still on Earth, you're clearly not accelerating, right? (In this example we are ignoring the energy and force exerted by the molecules in our body, which are all moving to some extent regardless of where we are.) If you were accelerating, you'd see some ...


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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 ...


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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 ...


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You mean only the gravity is considered? If so, here's the solution. Let's assume the mass of earth is $6 \times 10^{24} kg$. If your speed is much lower than the first cosmic velocity, regard it as the normal projectile motion. For higher speed($\leq v_{1stcosmic.}$) x(t) is like followings. where the f(t) means the distance to center of earth in meter. t ...


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I hope I have found how to evaluate the relative acceleration. If we place the origin in the centre of Earth according to a geocentric model, the curve described by Venus is $$\mathbf{r}_{VE}(t)=R_E\begin{pmatrix}\cos(\omega_{E}t+\varphi_1)\\\sin(\omega_{E}t+\varphi_1)\end{pmatrix} + ...


1

Nothing is incompressible, but most liquids and solids have a very low compressibility i.e. a very high bulk modulus. The reason for this is that in liquids and solids the atoms/molecules are in contact with each other. To squeeze them closer together you need to deform the bonds in molecules and/or the electron distribution around atoms. Both processes ...


-1

Here is my answer. I think it's self explanatory enough: In the last sentence there must be a minor correction: If someone keeps omega1 constant, then for the same speed they have to cycle faster so they can maintain the same power. If someone keeps omega2 constant, then bigger alpha mean less forward speed but less power too.


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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 ...


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. ...



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