# Tag Info

## Hot answers tagged collision

3

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 ...

2

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 ...

1

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 ...

1

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 ...

1

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 If the collision is inelastic, try this: $$mvh = I_m\omega_m + I_M\omega_M$$$I_M$, the moment of inertia of the solid sphere, is taken as that about its tangent at the point of contact with the ground. $$I_M = \frac {7MR^2}{5}$$ and, since the particle of mass$m\$ revolves about the centre of the sphere, $$I_m = \frac {mR^2} {2}$$ Here, the angular ...

1

Newton developed a formula for penetration depth of projectiles traveling at high speed. $$D\approx l_\text{bullet}\frac{\rho_\text{bullet}}{\rho_\text{wall}}$$ To a good approximation, the depth of penetration is constant.

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