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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 ...


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In a perfectly elastic collision, the final momentum of the system should be equal to the initial momentum of the system. It seems to be set up correctly, so I would say that you would need additional information for this.


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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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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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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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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 ...


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

$F = \frac{dp}{dt}$ 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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If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds. There is no way out. ...


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Yes, it would of course. And so the resultant weight (your's + the weight of what you are carrying) starts acting through this new and horizontally shifted (maybe vertically shifted too) Center of Mass. Notice that all this while your weight was being balanced by the normal reaction from the surface on which you were standing. Also, as only 2 forces were ...


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Model the ground as massless critically-damped vertical spring that the particle contacts at zero height. When the particle reaches zero height, it has some KE which is dissipated by the damping mechanism. When in contact with the spring, there are three forces acting on the particle, gravity downward and the damper and spring force upward. The net force ...


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When an object falls and hits the ground - which forces are involved to change its momentum? Vectorial sum of all the forces acting on the object will cause the change in momentum of the object. When the object was in free-fall, its momentum was already changing due to gravity(assuming negligible amount of air resistance) and then it hit the ground. ...


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In Newtonian Mechanics, if a body of mass $\mathtt{m}$ is in free-fall, then gravitational force is responsible for acceleration & hence changing its momentum. Simple, right? The equation of motion is $$\mathtt{m}\cdot a = \mathbf{F_g} = \mathtt{m} \cdot g$$ where $a$ is the net acceleration of the body. Things become intricate when you consider a ...


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If the photon is massless, how can it make an electron change momentum? Because, relativistically, momentum isn't proportional to (invariant) mass? Thus, particles with zero invariant mass can have non-zero momentum.


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To push something over, you apply a torque. If the thing you are pushing can provide an equal and opposite counter torque without moving, then it won't move. In your example, torque can be due to one of two things: Gravitational: a heavy object with its center of mass displaced relative to its rear support point needs torque to push over because as it ...


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Using dimensional analysis, and the relations $p=mv$ and $E=mgh$, we can write: $$[P]=\dfrac{kg^2\cdot m^2\cdot s^{-2}}{kg\cdot m^2\cdot s^{-2}}=kg$$ And kilogrammes are obviously not a unit of pressure, therefore your relation doesn't have the correct dimensions.


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For the mass $m>0$, spin $j$ representation of the (universal cover of the) Poincaré group, the theory of induced representations give the transformation law $$(u(A,a)\phi)(p) = e^{ip\cdot a} D^{(j)}(A_pAA_{\Lambda(A^{-1})p})\phi(\Lambda(A^{-1})p),\qquad \forall\phi\in L^2(\Omega_m^+,\text d\Omega_m^+)\otimes\mathbb C^{2j+1}$$ where $(A,a)\in SL_2(\mathbb ...


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In this cases, momentum is not conserved because of the action of gravity as an external force. When you have a pivoted rod, as in your problem, you can use basically two conservation laws: a) conservation of energy, if the collision is assumed as perfectly elastic; b) conservation of angular momentum about the pivot. As regards b), indeed, if we choose ...


1

It might help to think of an example. A simple is example is dust, specifically a collection of particles of a fixed rest mass, all at rest with respect to each other, and we can consider uniform dust, so they are equally spaced. If that's the only thing in our universe, then there is no momentum or stress in the frame of the dust and the energy is just ...


2

($\hbar$ omitted in the following.) That is not weird, it is one of the crucial properties of the Fourier transform $F(\bar{})$ that $$ F(\partial_x f) = \mathrm{i}p F(f)$$ i.e. differentiation by one variable becomes multiplication with the Fourier conjugate variable and vice versa. Because of this, Fourier transformation is a powerful tool to solve ...


2

In a 1D elastic collision, it is well-known that the relative velocities of the two objects (before and after the collision) are reversed. When you say reversed do you mean that each object keeps their own velocity just with a change of sign? That would not always be the case for a 1D situation. If a resting block $v_{1,before}=0$ is hit by a moving ...


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Introductory physics problems often limit the momentum economy to the motion of large particles or fragments (collisions and explosions) for simplicity of calculations. In reality, the momentum transferred to any surrounding gas (air) should ideally be part of the conservation. These introductory problems are constructed so that compression waves and huge ...


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In the case of an explosion, before the explosion the momentum of the bomb is zero, so according to law of conservation of momentum, the momentum after explosion should also be zero. So, momentum of the bomb before collision = momentum of the bomb after collision. As for sound and light energy, I think that it is the chemical energy of the bomb that is ...


2

Comments to the question (v2): A field $\phi^{\alpha}:[t_i,t_f]\times \mathbb{R}^3\to \mathbb{R}$ is the field-theoretic version of a (generalized) position variable $q^i:[t_i,t_f]\to \mathbb{R}$ in point mechanics. Note that the physical position space $\mathbb{R}^3$ typically plays very different roles in field theory and in point mechanics.$^1$ ...


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I think that the law of conservation will hold good in this situation because there is no external force acting on the system. Because like you said the impulsive force by the hinge is internal and no other force is a acting on the system.


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If I've interpreted your question correctly - the ball will collide with the rod at the opposite end to the hinge. This will lose energy via usual mechanisms. The rod will then have an instantaneous velocity, hence momentum, and will swing round the hinge. The ball will career off in whatever direction with the remaining momentum resulting from the ...


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You asked: Why do both vehicles experience the same magnitude of force? The larger principle at work is conservation of momentum. (Noether's Theorem, symmetry, and all that jazz.) During the small time frame of the collision we generally assume that there is no transfer of momentum into or out of the system of the car and truck. Changes of momentum ...


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The thrust for a rocket (can be demonstrated with global conservation of momentum on control volume set around the rocket) is, in general: $$\mathbb{T}=\dot{m}u_e + A_e\left( p_e - p_a \right) $$ where $\dot{m}$ is the mass flow rate $\left(\frac{M_p}{t_b} \right)$ (being $M_p$ the mass of propellant and $t_b$ the burning time) and $u_e$ the gases exit ...



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