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4

Probably the best way to think about this is to say that $$p = mv\\ F=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$ (Using the usual product rule for differentiation - thanks @ja72 for the suggestion). If velocity is constant the first term vanishes and your result follows.


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You have the right starting point with energy basically, but I'm finding your homework hint more useful than where you go from energy of a differential unit. It says "The energy density should be easy to identify." The energy density is: $$ \frac{\text{energy}}{\text{volume}} = \frac{\text{mass}}{\text{volume}} \frac{\text{energy}}{\text{mass}} = \rho ...


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First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ...


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The two equations are the same to first order, which is all that is important. If I were writing down the equation for the total momentum P(t+dt) myself, I would probably jot down the first equation (that of Morin) since I would be thinking of the instantaneous velocity of the rocket at time t rather than at time t+dt. But, again, the distinction is not ...


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If you draw similar triangles, then you'll find that $r_A/r_B = v_y/v_x$, and so the product $r_A v_x$ is equal to $r_B v_y$. Try drawing a line from the tip of your lower $\vec{v}$ vector to the tip of your lower $v_y$ component to see this.


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This is indeed confusing. The confusion comes from this very peculiar hypothesis: What if the person doesn’t apply a tangential friction force at his feet? It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay ...


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We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


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I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. This depends entirely on where one draws the system boundaries. One possible boundary is the rocket plus all of the exhaust gases it has released. The center of mass of the rocket + exhaust gas cloud system moves per the ...


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For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$ defined at the center of mass. Each body has mass and mass moment of inertia. These are $m_1$, $m_2$ and $Iz_1$, $Iz_2$. The contact is at point A with coordinates $(x_A,y_A)$ and with ...


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Yes and no. depends on the direction considered. Momentum is a vector quantity. Since there is no force in the x direction, momentum is conserved in that direction but not in the vertical direction because gravity (an external force for the system of Wedge and block) is acting. Yes, for your 2nd comment above. You can use COLM (conservation of linear ...


1

Defining precisely what are all the quantum numbers is a difficult question because it depends highly on the model under consideration, even for the standard model. In particular any U(1) symmetry leads to a quantum number, and similarly some U(1) subgroup of non-abelian groups that commute with all other interactions can also be associated to quantum ...


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For linear momentum to be a constant, its value should not change. Being linear momentum a vector quantity, it's value is completely described by specifying both magnitude and direction. In a circular motion, even though the speed remains a constant, the direction of velocity (which is tangential to the point on a circle) is changing throughout the motion as ...


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According to the rules of qft there are virtual photons in the vacuüm. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...


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The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...



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