# Tag Info

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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 ...

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Background There is a good reference1 on the physics of sound/shock waves in solids (look at Chapter XI). I found the following (on page 688) very interesting and relevant to your question: In a solid or liquid, a shock wave with a strength of even a hundred thousand atmospheres is regarded as weak. Such a wave differs little from an acoustic wave: it ...

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When two objects collide, they transfer momentum because they exert an equal and opposit force on each other (Newton's third law), and $\Delta p = \int F dt$. In order to know how fast an object moves after a collision, we need to know the velocities and mass of the objects before the collision and how elastic the collision is (conservation of kinetic ...

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There are different forms of energy. Energy can be converted from one form to another but cannot be destroyed. In this case the kinetic energy of the hammer is driving the nail into the wood which is breaking the molecular bonds in the wood fiber. The energy is converted to heat energy as a result of the breaking of the bonds and the friction of the nail in ...

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For the first three cases involving the car one can think of the collisions in the frame of reference where the motorcycle is at rest. The car approaches the motorcycle at a speed of 110 miles per hour in the first case, 50 miles per hour in the second case, and 60 miles per hour in the third. Comparing the three cases is a little easier now. The force ...

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You've underestimated the effect, although your math is correct as far as it goes. At 22% of c, relativistic effects do rear their ugly heads, and the proper equation is $$KE = \frac{\frac{mv^2}{2}}{\sqrt{1-\frac{v^2}{c^2}}} = 7.6\times10^{19}\text{ J}$$ Divide by $63\times10^{12}$ and the ratio is 1,200,000 (1.2 million). And yes, this is an unreasonably ...

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The formula that WhatRoughBeast uses is not correct. It should be $$KE = \left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}} - 1\right) mc^2 = \left(\frac{1}{\sqrt{1-0.22^2}} - 1\right) (22,000)(3\cdot10^8)^2 = 5\cdot10^{19} J$$ Your calculation wasn't far off since relativistic effects aren't very large at 22% of the speed of light, so $KE = ... 1 Kinetics: In physics and engineering, kinetics is a term for the branch of classical mechanics that is concerned with the relationship between the motion of bodies and its causes, namely forces and torques. Kinematics: Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects), and systems of bodies ... 1 So I came up with a graphical solution to this kind of problem. It might help you understand the process of collisions, without giving you a direct answer. Consider an Cartesian coordinate system xy for measuring momentum. Draw the initial momentum vectors$\vec{A} = m_A \vec{v}_A$and$\vec{B} = m_B \vec{v}_B\$. Draw a circle with the two vectors as ...

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Rotate the digram so the line connecting the circles is horizontal at the moment that they touch - since you know dx and dy, you just take the arc tangent. Now you move the frame of reference so the point where the two balls meet is stationary. The actual speed of the center of mass is just the vector mean of the velocities of the two balls (if they have ...

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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 ...

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This is an interesting question. Basically it's very similar like any meteorite collision. The gas planet makes here no other difference, but that there will be no crater. Assuming a Jupiter-like planet and an Earth-like planet (Except, say... half the mass of Jupiter), what would happen when the two collide? For clarification: What would the actual ...

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The force upon both objects is equal, but since the mass of the baseball is less, the acceleration will be greater due to F=MA

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