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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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though momentum as well as energy is conserved but definitely the sum of individual momentum of particles is not equal to sum of individual K.E. of the particles. also there may be different value of K.E. for same momentum. So can not make any result by manipulating the equations.

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Both energy and momentum are conserved as always. But to understand why this statement is true you have to look at the system as you described it a little more closely: In order for blocks A and B to stick together after the collision, the force between them should be zero when the velocity difference is zero -otherwise that force would continue to ...

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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 ...

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The thing to keep in mind is that for inelastic collisions, energy is not conserved within the system. Some energy is lost as thermal energy, and after all, it takes some energy to get the blocks to stick together. This doesn't mean that the Law of Conservation of Energy is false, though. It just means that energy has left the system that you're studying ...

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The astronauts do have a potentially deleterious affect on the vibrational environment on the International Space Station. They have to exercise multiple hours a day to keep bone and muscle loss down to a tolerable minimum. The unmitigated vibrations from all that exercising would be harmful to very sensitive micro-g experiments. Before I get to the ...

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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 $f(x - \Delta x) \approx f(x) - \Delta x ... 0 I think I saw a video about this problem... but I couldn't find a link. I believe it has to do with where the bullet hits the "free" object. Whether it hits on a line through the center of mass, or whether it hits near an edge (and thus has both momentum and angular momentum.) Maybe someone else can find the link. 3 In principle there is an effect, but firstly it's tiny and secondly it averages to zero. The mass of the ISS is about 420 tonnes, or about 5000 times the mass of an astronaut. That means if an astronaut pushes themselves off a wall at 1 m/sec the ISS moves in the other direction at about 0.0002 m/sec. But the ISS isn't very large so after only a couple of ... 0 If you have an elastic collision between objects A and B and where 'kinetic energy is conserved', does this mean object 1 will always have the same velocity it had before the collision? The speed and the direction of A after an elastic collision$v_f$depends on the mass of B: the ratio$v_f/v_i$varies from -1 to +1: it is equal to -1 ($v_f = ...

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Mass cannot just cease to exist. It can detach from the main body or annihilate with anti-matter to produce EM radiation. The effect on the velocity in such cases will depend on the details of the process (for example, whether the mass is lost in all directions like for evaporating water droplet or in one preferred direction like for rocket).

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The concept of an isolated system is an approximation. Like all approximations, it applies to some systems better than others. A billiard ball feels only a weak frictional force in the form of rolling resistance, whereas a wall typically has foundations buried in the ground which can provide a strong resistive force. Thus, the former system is closer to the ...

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It all depends on what you want to study. The billiard balls are generally viewed as an isolated system for the purposes of explaining elastic collisions, but you could as well introduce friction with the pool table, and the consider the system balls+table as the isolated one. This just means you have to consider the friction. In the case of the car hitting ...

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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 ...

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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 ...

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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 ...

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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 ...

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Let $v_1$ and $v_2$ be the particle's velocities in the center of mass coordinate system after the collision. by conservation of momentum and energy we have \begin{gather}\tag{1}m_1v_1+m_2v_2=0 \\ \tag{2}m_1v_1^2+m_2v_2^2=m_1(v-v_c)^2+m_2v_c^2 \end{gather} Isolating $v_1$ in $(1)$ and substituting in $(2)$: ...

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$m_2$ 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 ...

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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. ...

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Kinetic energy of the billiard ball will be maximum when the ball will hit the billiard ball in straight line with a given velocity.$$mv=mu+MU...........1$$And,$$mv^2=mu^2+MU^2............2$$Now the kinetic energy $\frac{1}{2}MU^2$ is a fraction of the kinetic energy $\frac{1}{2}mv^2$$\frac{1}{2}MU^2=k\frac{1}{2}mv^2............3$$From 2 and 3 find$k$in ... 0 The way I would approach this problem is through the impact parameter,$b$. You can find the definition on Wikipedia. The first thing to note is that for$b > r + R$, there is no collision. Here$r$and$R$are the radii of the two balls. For$b$less than this limit, you can use geometry to determine the direction of$\mathbf{U}$. This is because the ... 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 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 ... 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 ... 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:$p=|\mathbf{p}|=\sqrt{p_x^2+p_y^2+p_z^2}$. 0 In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well as their rotations. If we assume that no exchange between these two forms of kinetic energy occurs, i.e. that both forms are separately conserved, we have $$... 1 I interpret your question as What type of equation is the equation$$\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 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 ... 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 ... 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 ... 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 ...

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