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The behavior of the balls in Newton's cradle does provide a clue to understanding this phenomenon. Because of the spherical symmetry of the balls, and the fact that they make contact at a point, the propagation of compression pulses through the array is not like regular sound propagation through a solid medium. Sound in any medium is usually subject to ...


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This is essentially the same question as https://astronomy.stackexchange.com/questions/13302/ In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum is $\text{m1} \text{v1}$ and the initial kinetic energy is $\frac{\text{m1} \text{v1}^2}{2}$, since m2 is at rest. Let u1 and u2 be the velocities of the ...


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In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $ V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$


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Parts of the masses must come to rest if one or both of the masses change direction. The system of two masses has a centre of mass with linear momentum $m_A u - m_B u$ in the direction of $A$'s travel before the collision. Since there are no external forces acting that momentum cannot change and so there must always be parts of those masses which are moving. ...


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This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...


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Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...


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The acceleration at the point of reflection is actually quite complicated. It is caused by the elastic forces of the surface and the ball and has a complicated time dependence. However, the timespan in which the ball touches the ground is very short (especially if the ground and the ball are very rigid), therefore we can simplify the actual acceleration ...


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A rod of mass $M$ and length $\ell$ has mass moment of inertia $I = \frac{M}{12} \ell^2 $. The impact at a distance of $c = \frac{\ell}{2}$ from the center of mass imparts an impulse $J$, while an equal and opposite impulse $-J$ is applied to the projectile mass $m$. The projectile is going to bounce with velocity $v_B = v - \frac{J}{m}$. The center of mass ...


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It's not the fall that kills you. Your chance of survival is 100% :) In regards to the suspected impact after the fall, you will need to expand on the parameters to your question. Divers survive 50ft drops routinely. The elderly are killed from 4 foot drops routinely. I'm sure this wasn't the answer you were looking for, but I think the best answer ...


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Well let's say you weigh 220 lbs. Which translates over to 100 kg. The fall is 50 ft, so about 15.24 meters. The running thing that most people say is that it takes about 5000 Newtons of force to break a human bone, but we know that, this varies. We also know that it is not how hard you hit something that necessarily kills you, it's energy from the impact, ...



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