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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 ...

4

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 ...

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 ...

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 ...

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 ... 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}$$Thus, there is no ... 2 While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile. The mass and velocity of the projectile are m and v respectively, and the mass of the target is M. Since the target is stationary the initial momentum is just mv. Assuming the ... 1 First of all, mathematical definitions of force and momentum aren't really very intuitive or common-sensical. Just ask Aristotle for his common sense laws of forces! The fact that momentum is conserved in closed systems is a highly non-trivial fact, as is the Third Law. The reason that these laws exist at all is because you can't really 'see' or' feel' ... 1 What you need to do is use the conservation of momentum to get the velocity of the combined system:$$ m_1v_{1,i}+m_2v_{2,i}=\left(m_1+m_2\right)v_f $$This conservation law shows that the final velocity of the two blocks will still be proportional to the initial velocity of the one block (i.e, v_f\propto v_i). Getting this into the fractional change ... 1 As Jan noted, the Hamiltonian should have a minus sign: H=\frac{(p-qA)^2}{2m} where p is the canonical momentum, and the expression p-qA is the kinetic momentum P. A homogenous magnetic field is an interesting case, because the vector potential in a given gauge does not exhibit translation invariance, but the physical system clearly does. The ... 1 Alternatively, and qualitatively, think about the components of velocity (in the x y directions) have changed. Along the x axis, velocity has reduced, so the re has been a force in the -x direction. In the y axis, velocity has changed sign, so there must have been a force in the -y direction. Hence the total force is down and to the left, ie quadrant ... 1 The total impulse is the change in momentum (note that this is a vector equation):$$ \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$$Since the force on each cart is equal, constant, and applied for the same amount of time, the change in momentum for each cart is...? For (b), keep in mind that the less massive cart will have greater acceleration during the time the force is applied. 1 Once we have the position 4-vector$$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}\\ ...

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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 ...

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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 ...

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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 ...

1

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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