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44

Is it even possible to hit 350Gs of force to a hard drive? Sure is. Drop it on the floor. You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level. Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in ...


16

Here's an application where an ability to withstand high shock is important. Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research). We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a ...


4

You are not the first person to ask this question. http://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't ...


3

In principle, you don't need to tune to 'exactly' this energy but having less than $m_h + m_Z$ suppresses this diagram. Having more should typically give you a higher cross section because there is 'more phase space' the final state particles can be in, i.e. the 'excess' energy will just be used as kinetic energy of the final state particles ($Z$ and $h$ ...


2

I believe b is the impact parameter. So b should be the perpendicular distance between the asymtote you drew and the fixed ion. The Wikipedia link has a picture that is pretty clear, although the picture in the link is illustrating Coulomb REPULSION, while your problem involves Coulomb attraction..


2

In a collision it's often the case that it's hard to measure exactly how long the collision lasts and exactly how the force between the objects changes during the collision. Squishy objects like nerf balls will collide relatively slowly while hard objects like billard balls will have a short collision time. However there is a well defined quantity called ...


1

An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision.


1

The law states this: Two objects masses and with velocities $m_1$, $m_2$, $\vec{v}_1$ and $\vec{v}_2$ collide with contact normal $\vec{n}$. The final velocities are $\vec{v}_1^\star$ and $\vec{v}_2^\star$ such that the coefficient of restitution $\epsilon$ is defined by $$\vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right) = -\epsilon \;\left( ...


1

The usual approach: write down conservation of angular momentum, linear momentum, and energy. Assume the impact is elastic and infinitely short duration. In that time the spring didn't move and the third particle didn't come into the equation. That means the problem can be reduced to two simpler problems: two particles that hit elastically (after collision ...



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