In order for one grain of sand (measuring 0.0165' in diameter) to be subjected to a pressure of 6000 psi, at what rate of speed must it impact a metallic surface (steel)provided that surface were angled at 45 degrees?

To answer this you need to make an assumption about how much compressoin the grain of sand undergoes when it hits the surface. The less the compression, the greater the force, and the smaller the area that force works over. So the harder you assume the sand is, the greater the pressure that the leading edge of the grain is subjected to wghen it hits the wall. But we can scope this if we ake a few simplifying assumptoins.

1. Assume the sand gain is a sphere.
2. Assume that the sand grain deforms like a tennis ball when it hits the wall, and that at max deformation if is compressed to 1/2 its original width.
3, Assume the force of imnpact is spread across the full width of the sphere.

Start with the concept of impulse and momentum:


F \Delta t = m \Delta v


combined with the equation of motion

\Delta v^2 = 2 a d


where d is the deflection of the sphere. From these two equations you get:


F = \frac {m \Delta v^2} {2 d}


If you assume this force is spread over the cross-sectional area of the sphere A, then the pressure is:


P = \frac F A = \frac {m \Delta v^2}{2Ad}


A couple of other equations that come in handy:


A = \pi \frac {D^2} 4 \\
m = \rho \frac {D^3} 6


where \rho is the density of quartz. Now make an assumption for home much deformation d there is in the sand grain - for example you might assume that d = D/2. You will need to find the density of quartz (or the weight of the grain of sand), and then you have everything you need to solve for v. Post back and tell us what you get.