Kinetic energy transfer from shockwave to secondary body

[KataruZ98]

Homework Statement

I have an object A possessing a known mass of 10kg and density of 1,000kg/m^3 exposed to a shockwave in a way the latter impacts A over an area of one square meter at a right angle. The pressure of the shockwave at the point of contact is 10PSI.

Relevant Equations

Kinetic energy transferred by the shockwave to body A

I would guess that by multiplying the pressure exerted by the shockwave on the body, and then the resulting force - here ~69 Newtons - per the distance the shockwave passed through when traversing body A, I could get the work done but I’m not sure if it’s that easy and whether or not I should consider the shockwave accelerating when passing from a less dense to denser medium.

 

[haruspex]

Thinking of it per unit area (doubling the area should yield the same velocity gain) we have a pressure, a density, a distance and a velocity. What does dimensional analysis say?

 

[KataruZ98]

Hm, I’m kinda lost honestly. Unfortunately I’m not well versed.

 

[haruspex]

Hm, I’m kinda lost honestly. Unfortunately I’m not well versed.

Are you unfamiliar with dimensional analysis? Look it up.
It uses M for mass, L for length, T for time,…
Pressure is ML^-1T^-2
Density ML^-3
Distance L
Velocity LT^-1
How can you combine the first three, raising each to some power and multiplying the terms together, to make the last?

 

[KataruZ98]

Are you unfamiliar with dimensional analysis? Look it up.
It uses M for mass, L for length, T for time,…
Pressure is ML^-1T^-2
Density ML^-3
Distance L
Velocity LT^-1
How can you combine the first three, raising each to some power and multiplying the terms together, to make the last?

Well I would say I should divide density by the product of pressure and distance - though this leaves a T^-2 as denominator.

 

[haruspex]

Well I would say I should divide density by the product of pressure and distance - though this leaves a T^-2 as denominator.

Then that cannot be the answer.
The method is to let the answer be of the form velocity=pressure ^adensity^bdistance^c. In MLT notation that becomes $LT^{-1}=(ML^{-1}T^{-2})^a(ML^{-3})^bL^c$.
Three equations, three unknowns.