A bullet through a wall

  • #1
A bullet with an initial velocity of v0, temperature t0(melting temperature), goes through a wall, and its velocity reduces to v. You know Lb(latent specific heat). You are asked to derive the ratio between the final mass of the bulet and the initial one.Oh, and you know that the energy lost by the bullet it's transformed in a ratio f(<1), to heat.
Here's what i have done:
We know that F=dp/dt, implies F=dm/dt*v + dv/dt*m;
We also know that dL=-fdQ, dQ=-dm*Lb(because dm is negative), which leads to dL=f*dm*Lb, dL=F*dx=F*v*dt=(dm/dt*v + dv/dt*m)*v*dt;
dL=dm*v^2 + m*v*dv
i earlier stated dL=f*dm*Lb, combining the last two gives:
dm*v^2 + m*v*dv = f*dm*Lb
m*v*dv=dm(f*Lb-v^2)
(v*dv)/(f*Lb-v^2)=dm/m
u=f*Lb-v^2 => du=-2v*dv => v*dv=-du/2
-1/2 * (du/u)=dm/m
=> -1/2 * ln (u2/u1) = ln (mf/mi)
(u2/u1)^(-1/2)=mf/mi
sqrt(u1/u2)=mf/mi
sqrt((f*Lb-v0^2)/(f*Lb-v^2))=mf/mi
the problem here is like this: the "-" sign demands that v0 cannot be greater than sqrt(f*Lb) which is illogical. I don't know what i have missed.
 

Answers and Replies

  • #2
Did you not understand what i said?
 
  • #3
Andrew Mason
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A bullet with an initial velocity of v0, temperature t0(melting temperature), goes through a wall, and its velocity reduces to v. You know Lb(latent specific heat). You are asked to derive the ratio between the final mass of the bulet and the initial one.Oh, and you know that the energy lost by the bullet it's transformed in a ratio f(<1), to heat.
Perhaps you could give us the actual wording of the question. It is very confusing the way you have it here. How are we supposed to determine the final mass of the bullet? What is the initial mass? Why does it change its mass just because it slows down in the wall?

AM
 

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