# Tag Info

0

An object can be at rest in space, infinitely away from any other massive body.

1

No the time taken does not depend of the velocity attained by the first ball(if they are ideally rigid) it rather depends on the elasticity or rigidity of the balls. So for ideally rigid bodies, the time taken to transfer approaches 0. Nothing would happen with an increase in distance between the two balls. See: Is the reaction force for a stone hitting a ...

0

It is a hypothetical condition as inertial will never let this condition happen. For the sake of argument I am using impulse. faster you stop an object more will be the force. Example using gloves to stop a fast ball in sports. $$F_{impact}*t=mv-mu$$ $$F_{impact}=\frac{mv-mu}{t}$$ $$F_{impact}=\lim_{t \to 0}\frac{mv-mu}{t}$$ According the equation the force ...

3

I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the $dt$ or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ...

0

To drive a nail into a piece of wood you need to overcome the force of static friction and the force required to push aside the wood (make a hole). When an object of mass $m$ and velocity $v$ hits a nail, either the nail moves, or the object decelerates very quickly. This sudden change in momentum is what drives the nail. We know that $$F\Delta t = m\Delta ... 0 The momentum operator is the generator of shifts. In 3 dimensions (\hbar =1 ) \begin{equation*} (\exp [i\mathbf{a\cdot p}]f)(\mathbf{x})=f(\mathbf{x+a}) \end{equation*} Expanding in \mathbf{a} \begin{equation*} i\mathbf{a\cdot p}f(\mathbf{x})=\mathbf{a\cdot }\partial _{\mathbf{x}}f(% \mathbf{x)} \end{equation*} or \begin{equation*} ... 0 I think I figured it out. Bernoulli's assumption is incompressible flow. The equation yielding from momentum conservation always holds. When velocity is low (incompressibility holds), the two equations yields similar results. 1 For the forces between elementary particles we have Feynman diagrams, where there exists a mediating particle for the interaction. In the simplest diagrams: for the strong it is the gluon, for the weak it is Zs and Ws and for the electromagnetic it is the photon. Here is Bhabha scattering, where the electron and the positron ( attractive force) are first ... 0 The factor \frac 12 comes from the relation \vec v \cdot\nabla \vec v = \nabla \frac{\vec v^2}{2} + (\nabla\times\vec v)\times \vec v in the momentum conservation equation$$\rho \left(\frac{\partial \vec v}{\partial t}+\vec v \cdot\nabla \vec v\right)=\vec g-\nabla p$$(Sorry to post this as an answer, but I can't comment your post yet because of ... 2 The momentum operator is not -i\partial_x, rather, that is the representation of the momentum operator on the position basis: namely$$ \langle x|\hat{p}|\psi\rangle = -i\frac{\partial}{\partial x}\psi(x). $$Otherwise, the momentum operator is just defined by action on its eigenstates as \hat{p}|p\rangle = p|p\rangle. I understand the complex ... 1 Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane? In a sense neither the real part nor the imaginary part have physical significance, as these quantities do not directly appear in observables. One way to see this is that any solution \left|\psi \right\rangle to ... 2 Actually A.Zee's book on "QFT in a nutshell" has a very nice explanation on this on chapter I.5, I will breifly sketch it (this is a very rough skeptch),$$Z=\int DA e^{iS(A)} =e^{iW(J)}$$where W(J) is given by,$$W(J)=-1/2 \int \int d^4xd^yJ(x)D(x-y)J(y)$$where D(x-y) is the photon propogator and J(x) and J(y) refer to two lumps of matter Plugging in ... 1 For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ... 0 A constant force pushes a mass m along a distance x. What is the final velocity? The acceleration is a=\frac{F}{m} and the kinematic relationship between speed and distance is$$ \left. x = \frac{v^2}{2 a} \right\} v = \sqrt{2 a x} = \sqrt{ 2 \frac{F}{m} x}$$The final momentum is$$ p = m v = \sqrt{2 F x m} $$So momentum is proportional to ... -1 From definition, force is a momentum change rate in time \vec{F} = \frac{\text{d}\vec{p}}{\text{d}t}. So if the force working on both is equal then their momentum change would be the same so answer C, but... in my opinion the correct answer is D. There is no information given about friction (if we assume that blocks are moved on surface), whether is or is ... 1 Imaging the balls on a string. You are launching N balls per second, at a velocity u. This means the distance between the balls is u/N. And N balls per second will pass a certain point in space. Now if the car is moving at a velocity v (same direction as u), fewer balls per second can hit it - because subsequent balls on the string have further to ... -1 It is just an integration by parts considering that boundary terms vanish. 0 The ball only feels an impulse along the normal direction and not the tangential direction. Hence there is only a change in momentum in the normal direction and not the tangential. It is probably worth noting that although the overall momentum is conserved when a ball strikes a very large wall the momentum of the ball does change (and so will that of the ... 0 Thinking out of the box i have brought an answer to this that works on paper. By using the forces listed in the original question (Balloon lift @ 6lbs and payload @ 8oz) i have included the following forces as being necessary to engineer a solution. First the weight+gravity for the pulley (I list at 4oz), secondary pulley system (.5oz friction force i will ... 0 simply no useful notion of 'force' in quantum systems in general. Yes you can do some calculations and sometimes handwave what a force is in a certain situation, but in most matters its of little particular use. Take the collisions at the LHC - there is no useful notion of 'force' you can ascribe to what is going on when all those particles collide. Yes, if ... 0 Remember subscripts! To avoid confusion. For object S_2 you find the expression T=mg. But remember to write it as T_2=m_2g since there are more T's and m's in this system. Now find T_1 in the same way as you found T_2. You have already explained the forces acting on S_1. (As @Vishwaas points out in a comment, the ramp's angle is needed here ... 2 1)The conservation law of linear momentum does involve the velocity, not just the speed. That said, however, it is ok to use speed in the problem statement. There are no external forces acting on the system bullet-gun. Therefore, if the bullet moves along the positive x-axes the gun recoils in opposite sense along the same direction. This is implicitly ... 0 No, you're not right. Anyway, to solve this problem I need velocity, and not speed. Why exactly do you need velocity, considering that the problem asks for the speed of the gun? In any case, unless there is a predetermined coordinate system you're supposed to use, it's fine to specify a velocity as something like "1\ \mathrm{m/s} opposite the ... 1 Keep in mind that the equation$$ E^2 = p^2c^2 + m^2c^4 $$is derived from the relations$$ \begin{align} E = \gamma mc^2,\qquad p = \gamma m v. \tag{1} \end{align} $$Therefore$$ p = E\frac{v}{c^2}.\tag{2} $$Although (1) is only defined for massive particles, it turns out that (2) remains valid when v=c, i.e. for massless particles. Indeed, we get$$ E= ...

0

The definition of momentum isn't $\gamma m \dot x$. The proper definition of momentum is that it is the generator of translations. Then you find that for massive representations of the Lorentz group (~timelike curves), $p = m \gamma \dot x$, while for massless representations (~lightlike curves), $p$ is arbitrary, as long as $E = pc$. Another way of ...

0

If one considers that the deBroglie relationship holds for photons we have $$p=\frac{h}{\lambda} = \frac{hf}{c} = \frac{E}{c}$$ which immediately gives us $$E=pc.$$ This is consistent with the Lorentz invariant energy four-vector magnitude which yields the mass of a particle: $$mc^2=\sqrt{E^2-(pc)^2}=0.$$

2

Yes. It is much easier to think of this in terms of conservation of momentum: Because light (and electromagnetic radiation in general) has momentum, you will have to gain momentum in the opposite direction to conserve total momentum --- just like if you were to throw the flashlight. It is difficult to think of this in terms of forces because we tend to ...

0

You want to use $$\hat x= i\hbar\frac{\partial}{\partial p}$$ in the momentum basis. This means that $$<p|\hat x|\psi>= i\hbar\frac{\partial}{\partial p} <p|\psi>$$ Thus, by hermiticity of $\hat x$, we evaluate $$<x|\hat x|p> = (<p|\hat x|x>)^*$$ $$=(i\hbar\frac{\partial}{\partial p} <p|x>)^*$$ $$... 0 Even in one dimension the operator p_r=-i\partial_r on the half line r>0 has deficiency indices (0,1). There is thus no way to define it it as a self-adjoint operator. In practical terms this abstract mathematical statement means that there is no set of boundary conditions thta we can impose on the wavefunction \psi(r) that lead to a ... 1 I think for 1-dimensional bound states, this is the proof : The expectation value of the momentum operator, \langle \hat{P} \rangle=\langle \psi|\hat{P} \psi\rangle=\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)dx=\frac{\hbar}{i}\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\partial}{\partial ... 0 First of all you cannot separate linear from angular momentum. They work together just like linear and angular velocities do (or forces and torques). I am going to answer your question from the perspective of geometry. The quantity of momentum is not so important as the geometrical construction that momentum implies. Let's see if you can follow: All ... 0 I suppose Newton may have devised the momentum equation to numerically express how objects of exact speeds (but different densities) would create different effects upon impact and perhaps how much energy would be needed to move such objects to a given speed. Consider the following: a wood ball (25g, 33.5 cc) hurled at a sheet metal target at an average ... 1 What you're looking for is an intuitive explanation or how you could visualize momentum. You can think of momentum as the quantity/amount of motion or "how much would I not want be in the path of this body." I'm going to try and provide some intuition through a few examples: A car of mass 1000 kg moving at 5 m/s would have the same "quantity/amount of ... 1 Newton (if I recall correctly) typically referred to the concept of inertia, which was an objects resistance to changes in velocity when subjected to external forces. You are right about him not thinking about it as just the speed of the object, because this is where the mass term comes in. Many people think of Newton's second law as being written as F = ... 3 Newton thought of momentum as "Quantity of motion" - as we can see in the translated version of 'Principia'. Particularly, he defined momentum in the following words: The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. So yeah, that is the definition of momentum. The question why we defined the ... 0 In an elastic collision, kinetic energy is conserved. That is, for a system of N particles,$$\sum_{j=1}^N\frac{1}{2}m_jv_{i_j}^2=\sum_{j=1}^N\frac{1}{2}m_jv_{o_j}^2$$where v_i denotes initial velocity and v_o denotes final velocity. This is incredibly useful when it comes to calculating the final velocities of the objects. In the case of an object ... 0 Your first equation is wrong, the relative speed of first stage to second stage in your equation is v+35, not 35. 0 You need to remember that in an elastic collision the equations due to the conservation of linear momentum and kinetic energy are: \begin{eqnarray} V_{1\:final}={m_1-m_2\over m_1+m_2}\,V_{1\:initial},\\ V_{2\:final}={m_1\over m_1+m_2}\,V_{1\:initial} \end{eqnarray}  a) First case: m_1=m_2. Here V_{2\:final}=(V_{1\:initial})/2. Then, the momentum ... 2 As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum p and the antiparticle comes in with the opposite momentum -p, and the two annihilate. Suppose the annihilation produced a single photon. The momentum of a photon is:$$ p ...

1

Suppose $a\overline{a}\rightarrow\gamma$ is possible for a particle $a$ with a definite nonzero mass, $p_a^2=m^2>0$ ("mostly-minus" metric, $c=1$). Conservation of momentum implies $p_\gamma=p_a+p_{\overline{a}}\implies p_\gamma^2=m_\gamma^2=0=p_a^2+p_{\overline{a}}^2+2p_a p_{\overline{a}}=2m^2+2 p_a \cdot p_\overline{a}$ However, the scalar product on ...

1

There is a mistake in equation (2). Its denominator should include the total mass of the system that you're considering, so the denominator should be '2m+m'. You correctly used this value for equation (1), but apparently incorrectly believed that since the position (and velocity?) of the lighter mass 'm' is zero that the value of 'm' shouldn't be included in ...

0

In the second case, you made a mistake in the denominator. You always put the overall mass. The lighter mass is part of the system that you are trying to find the center of mass of, even when it has a zero value of acceleration. So, the denominator should be 3m. In general, we put in the denominator the mass of every body that is in the system of bodies for ...

1

This question is actually more complicated than it might seem. To clarify the problem, let us consider a simplified model of the string: The string extends along the x-direction and is made up of masses connected by springs. For conceptual clarity, suppose these masses can only move up and down (along y; this could be enforced in a mechanical model by ...

0

If I understand the question you are partially asking what causes a skid. You only get so much friction between two surfaces. After that they slide. Even after they slide there is still friction. Static and kinetic friction The car is exerting a force. It has a negative acceleration. The car is loosing speed / momentum. The force that causes the ...

0

First you need to find the force of friction on the wheels of the car using the weight of the vehicle + the passenger. Then, calculate the acceleration. Then, calculate the force needed to accelerate the passenger by the same amount and you've found the force on your passenger.

1

Just because an object is in motion, that doesn't mean that there is a force acting on it. Newton's 1st law states that an object in motion will stay in motion unless acted on by an outside force. So the car's momentum is what keep's it going as it coasts. The forces acting on the car would be gravity, the normal force (the ground pushing up on it), and ...

Top 50 recent answers are included