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## Hot answers tagged momentum

11

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...

6

When the mass reaches its lowest point, the steel wire will have increased in length from $L$ to $L+x$. So equating the strain energy of the wire with the initial gravitational potential energy of the ball: $$\frac{1}{2}kx^2 = mg(L+x) \approx mgL$$ which rearranges to $$x = \sqrt{\frac{2mgL}{k}}$$ Note that $$k = \frac{EA}{L}$$ where $E$ is ...

4

According to the MTU webpage Speed of Sound in Air, some things to consider: if the ideal gas model is a good model for a real gas, then you can expect, for any specific gas, that there will be no pressure dependence for the speed of sound. This is because as you change the pressure of the gas, you will also change its density by the same factor. ...

3

Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ...

3

Thus, work is done on the car, right? No, the car does (positive) work on whatever is stopping it. Alternatively, you could say that negative work is done on the car, but still, the meaning is the same: the car loses energy and something else gains that energy. What that something else is, and what type of energy it gains, depends entirely on how the ...

3

In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ...

3

He's doing a linear approximation. Suppose $\Delta x$ is very small. Then $\langle x - \Delta x | \alpha \rangle$ is almost equal to $\langle x | \alpha \rangle$, but not quite, because $\Delta x$ isn't zero. So we do a first order approximation: Let's write $\langle x | \alpha \rangle$ as $f(x)$. Then $f(x - \Delta x) \approx f(x) - \Delta x ... 2 Let's make a concrete example with numbers: Suppose that$v_a = 6m/s$and$v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$. According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ... 2 Your last expression (4) is equal to (2), you just have to realize what does it say.$\lambda$isn't$\tau$and $$u^c = \frac{d\xi^c}{d \tau} = \frac{d\xi^c}{d \lambda} \frac{d\lambda}{d \tau}$$ If you look back to your Lagrangian and how it was derived, you should be able to say what is$d\lambda/d\tau$. To be very explicit, the action of a free ... 2 Momentum is a vector. For example, in 3D$\mathbf{p}=(p_x,p_y,p_z)$. The magnitude of the momentum vector is a scalar:$p=|\mathbf{p}|=\sqrt{p_x^2+p_y^2+p_z^2}$. 2 To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ... 2 If there was no friction, your breaks couldn't clamp down on your rotors to slow the car, and the car's tires couldn't "stick" to the pavement. Your engine is generating energy, and it does cause pressure in the braking system. That braking system, triggered by you and your foot (and assisted by the car) converts that motion to heat through friction between ... 2 If I eliminate the associated Kinetic Energy of the car, where does this energy go? While conservation of energy dictates that the energy due to the car's motion must be conserved, it does not say how. An old-style braking system converts that kinetic energy into heat. A more modern regenerative braking system converts that kinetic energy into a ... 2 In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$\sum F = \frac{\Delta mv}{\Delta t}$$ Think of$\int F \mathrm{d}t$being equal to the change in momentum, that is: $$\Delta mv = \int ... 2 (Summary: In this post I argue that you need at least an energy of m v_1^2(1-\cos(\theta)) in the idealized kinematics situation to deflect an asteroid of mass m and velocity v_1 by an angle \theta using rockets, without changing the magnitude of v_1.) Energy isn't the be-all and end-all of motion. The problem is that momentum also has to be ... 1 m_2 will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, m_1 and m_2 have the same speeds leaving the collision as entering it. In other words, the speed of m_1 is v-v_c and the speed of m_2 is v_c after the collision. In order to simplify things and to ... 1 The force can be surprisingly large, but \Delta t is not zero, and the force is not infinite. Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration ... 1 what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small? Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest: Ball A will exert on b the Impulse of a Force J and its velocity, ... 1 The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for v_2. Rewriting equation (1) gives v_1=\left(m+M\right)v_2/m which inserted into (2) yields$$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$\$ This ...

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