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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

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 ...


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 ...


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

The equation, $v_A-v_B = \frac{4}{5}(u-2u/3)$ is incorrect. The proper equation for the coefficient of restitution is given by, $v_A-v_B$ = $e(u_B-u_A)$, where $u$ and $v$ are velocities along the line of impact. I believe you came across the somewhat incomplete statement that the relative speed of separation after the collision is $e$ times the initial ...


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

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 ...


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 ...


1

Op has read the previous answers and there is confusion, the answers either just repeat the definition or say: The impulse takes into consideration both the effect of the force on the system, and the duration of time for which the force acts. -- or In the Newtonian point of view, impulse and change of momentum are different concepts... (and ...


1

I interpret your question as What type of equation is the equation $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) \cdot d\vec{x}(t') = \frac{1}{2} m ((\dot{\vec x}(t))^2 - (\dot{\vec x}_{0})^2) \; \; \;\; (1)$$ where $t$ is variable and $\dot{\vec x}_{0}$ fixed? The answer is, it is a first order integro-differential equation, not a mere ordinary differential ...


1

Here is a general figure of an hard spheres collision drawn in the center of mass of the mass $m_2$ before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy: $m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum: $m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$ Conservation of torque ...


1

Intro I'm the original asker of this question (9 months ago); thanks to the comments and answers I've gotten here, I think I've pieced together an answer that I'm happy with. Short forms used in this answer: CoLM = Conservation of Linear Momentum CoAM = Conservation of Angular Momentum KEB = the Kinetic Energy Balance CoE = Conservation of Energy ...



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