# Tag Info

## New answers tagged momentum

-1

I read the person's question about the distribution of velocities of large objects as meaning the following: Einstein sp. relativity leads us to believe that all "speeds" are essentially meaningless. For instance, the classic moving-train cartoon in all the books on his theories show that when someone on the ground (on earth) measures the speed of a bird ...

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There is no reason squaring equation (2) should give equation (1), because they are independent equations. You can use this fact to solve for $v_1$ and $v_2$; if this weren't so then using both conservation of energy and momentum would be rather useless.

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While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile. The mass and velocity of the projectile are $m$ and $v$ respectively, and the mass of the target is $M$. Since the target is stationary the initial momentum is just $mv$. Assuming the ...

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I haven't found a really good shortcut, but the following can make the integration much simpler in some cases. The time independent Schrodinger Equation: $$\frac{\hat{p}^2}{2m}\Psi+V\Psi=E\Psi$$ $$\frac{\hat{p}^2}{2m}\Psi=(E-V)\Psi$$ $$\hat{p}^2\Psi=2m(E-V)\Psi$$ So.... $$\langle p^2\rangle = \int\Psi^*\hat{p^2}\Psi dx = \int\Psi^*[2m(E-V)\Psi]dx$$ ...

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The forces act on BOTH the bodies involved, not on the same one! That's why the statement of Newton's third law is: The third law states that all forces exist in pairs: if one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal and opposite: FA = −FB Source:Wikipedia. So it isn't ...

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First of all, mathematical definitions of force and momentum aren't really very intuitive or common-sensical. Just ask Aristotle for his common sense laws of forces! The fact that momentum is conserved in closed systems is a highly non-trivial fact, as is the Third Law. The reason that these laws exist at all is because you can't really 'see' or' feel' ...

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When evaluating these matrix elements, ask yourself if you know how the operator in the middle acts on the ket. If you do know, then go ahead and evaluate it. If you don't know, you can re-write the operator in a different form, or re-write the ket (state) in a different basis. Let's take number 4 as an example: $$\langle 2p, m=1 | \hat{l_x} | 2p, ... 1 What you need to do is use the conservation of momentum to get the velocity of the combined system:$$ m_1v_{1,i}+m_2v_{2,i}=\left(m_1+m_2\right)v_f $$This conservation law shows that the final velocity of the two blocks will still be proportional to the initial velocity of the one block (i.e, v_f\propto v_i). Getting this into the fractional change ... 2 I assume you're thinking about Minkowski space, i.e. the metric \eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1). You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have$$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$Thus, there is no ... 1 Alternatively, and qualitatively, think about the components of velocity (in the x y directions) have changed. Along the x axis, velocity has reduced, so the re has been a force in the -x direction. In the y axis, velocity has changed sign, so there must have been a force in the -y direction. Hence the total force is down and to the left, ie quadrant ... 1 The total impulse is the change in momentum (note that this is a vector equation):$$ \vec{I} = \vec{p}_{final} - \vec{p}_{initial} $$You know the momentum before and after the collision so you can calculate the total impulse, both magnitude and direction. Impulse if force times time, so the direction of the force will be the same as the direction of the ... 1 This post has some information about impulse that you might find useful. Homework Question involving Momentum You will not find conservation of momentum useful here. True, the total momentum of object + wall is unchanged by the collision. But the momentum of the object does change. Since \Delta P = J = F_{avg} \Delta t, the direction of F_{avg} and ... 1 For (a), force equals the rate with which momentum changes$$\frac{d\vec p}{dt} = \vec F$$Since the force on each cart is equal, constant, and applied for the same amount of time, the change in momentum for each cart is...? For (b), keep in mind that the less massive cart will have greater acceleration during the time the force is applied. 2 You've probably learned about a quantity called "impulse" - try using that to solve the problem. Let J be impulse, defined for constant forces as J = F t where F is the force applied and t is the time for which the force is applied. Since F=ma, we can substitute this to get:$$\begin{aligned} J &= Ft \\ J &= mat \\ J &= m(at) \\ J ...

1

Once we have the position 4-vector $$x^\mu= \left( \begin{array}{c} ct\\ \vec{x}\\ \end{array} \right)$$ It is natural to define the momentum and energy in a fashion which is analogous to the Newtonian case (and reduces to it in the frame of the particle itself, when $\vec{v}=0$: $$p^\mu \equiv m\frac{d}{d\tau} \left( \begin{array}{c} ct\\ \vec{x}\\ ... 1 Momentum is vector and it has a direction. Since the motion is at 45° from the collision normal axis, the direction vector is$$\left[ \sin 45°, \cos 45° \right] = \left[ \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right]$$1 If you know the total force as a function of time, then you know the acceleration as a function of time also$$ a(t) = \frac{\sum F(t)}{m} $$Now you find the velocity and acceleration using direct integration$$ v(t) = v_0 + \int a(t)\,{\rm d} t \\ x(t) = x_0 + \int v(t)\,{\rm d} t If the forces are constant then you can convert the integral into ... 0 You should consider Newton's 2nd Law: \vec{F} = m\vec{a}  and relate the forces acting on the object to accelerations. Once you know the accelerations for different periods of time you have broken down the problem into a simple kinematics one of an object moving with constant acceleration. 1 The result is straight forward. As Landau and Lifshitz explain in p.41, when a body disintegrates into two pieces of masses m_1 and m_2 respectively, their momenta must be equal in magnitude and oppositely directed by the law of conservation of momentum. So, let each body have momentum p_0. Then, (16.1) and (16.2) say that the difference in the ... 2 I assume your car is front wheel drive. The phenomenon is simply Newton's third law in disguise. The car exerts a torque on its forward axle and the wheels exert the same magnitude, opposite sense torque on the car. Normally, the torque is not so big, because as soon as it is exerted on the wheels by the car, the wheels push backward on the road and the ... 10 General remarks. The momentum you define in the first equation, namely \begin{align} p = \frac{\partial L}{\partial \dot q} \end{align} is not necessarily the same momentum that appears in Newton's Second Law. This momentum is called the canonical momentum conjugate to q, and it can be quite different from the momentum you're used to (the one ... 0 Force really is the derivative of momentum. Except that it isn't. Have you taken a Statics course yet? Don't forget, in the equation\vec F = \frac{d\vec p}{dt} $$the left hand side is understood to be the (vector) sum of all forces acting on the particle, i.e., it is the net force$$\Sigma \vec F = \frac{d\vec p}{dt}$$Also, there are ... 2 So here, really, lies my question: Is there even a point to arguing about this? Perhaps there is a point in discussing this. In the Newtonian point of view, impulse and change of momentum are different concepts. Why? Force F(t) is a basic quantity describing instantaneous influence of one body on another, in general having a magnitude and ... 1 This has a simple closed-form solution. Denoting m_0,m_1 as the initial and final person's mass, v_r as the rice speed and \delta=m_0/m_1, if the bag is thrown in one single parcel, we have$$\Delta v_1=(\delta-1)v_r$$By the rocket equation, if the rice is thrown continuously, we have$$\Delta v_2=v_r\text{Log}(\delta).$$But$$\text{Log}(\delta)\leq ...

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You need a model for how you throw the rice. The obvious one is that you can expel any mass at the same velocity $v$ relative to you. Letting $M$ be your mass (without the rice), $V$ your velocity in the CM frame, if you throw it as one lump we have momentum conservation. You start with no momentum in the CM frame, so $10v=MV, V=\frac {10v}M$. If you ...

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Given a bag of rice of mass $m_b$ that you can throw with a maximum acceleration $\vec{a}_b$, by Newton's second law, the most force $\vec{F}_b$ you could exert on the rice is given by $$\vec{F}_b = m_b \vec{a}_b$$ By Newton's third law, the reaction force (acting on you) $\vec{F}_{you}$ is given by $$\vec{F}_{you} = - \vec{F}_b = - m_b \vec{a}_b$$ Again ...

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To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation: $$\vec F=m\vec a$$ So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, $m\vec v$, we can use the following equation ...

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Well I say charge indeed has momentum... Given $F = qE$ and realising that $F = \frac{dP}{dt}$ It follows $\Delta P = qE \Delta t$. We define $\Delta P$ to be the charge momentum and $E\Delta t = (iota)$$From this it is clear that the charge momentum (which equal mass momentum) are two different entities, in fact it can be showed that the two exist in ... 1 As Jan noted, the Hamiltonian should have a minus sign: H=\frac{(p-qA)^2}{2m} where p is the canonical momentum, and the expression p-qA is the kinetic momentum P. A homogenous magnetic field is an interesting case, because the vector potential in a given gauge does not exhibit translation invariance, but the physical system clearly does. The ... 0 Maybe an example helps. Let B be a constant magnetic field. Then we can take A=\frac12B×x. Now$$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}(p⋅A+A⋅p)+\frac{q^2A^2}{2m},$$and$$p⋅A+A⋅p=l⋅B$$where l=x×p. Thus$$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}l⋅B+\frac{q^2A^2}{2m}.$$Here we recognise the l⋅B-term as the Zeeman term. If we now ... 0 Hamilton's equations state \dot{P_i} = -\frac{\partial H}{\partial q^i}.In this case, this is \dot{P_i} = -\frac{\partial H}{\partial q^i} = -\frac{\vec{P}}{m} \cdot \frac{\partial \vec{A}}{\partial q^i}. So the canonical momentum is not conserved. 0 Hint: P is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero. So you just neet to check that:$$ \{P,H \}=0$$0 Since$P=p+qA$commutes with$H={P^2}/2m$, it is a constant of the motion. But this is not true for$p$, so I have the impression that there is some confusion about$P$and$p\$.

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