How does the movement of a block affect the penetration depth of a bullet?

[MCPO John-117]

Homework Statement

Problem from the book:
"A 7 gram bullet, when fired from a gun into a 1 kg block of wood held by a vise, penetrates the block to a depth of 8 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7 gram bullet is fired from the gun into the block. to what depth will the bullet penetrate the block in this case?"

even problem so the answer is unknown. the book the 4th edition of "principles of physics", problem 20 of chapter 8.

Homework Equations

P=mv
Ke=0.5mv²

The Attempt at a Solution

Attempt
so i tried a bunch of things. at first i tried to find an actual answer but after an hour of that i figured the answer must contain variables so i tried that instead. i just couldn't find a way to relate momentum and kinetic energy to penetration. i had equations for acceleration to force to energy but couldn't relate it to momentum and what happens when the block is free.

 

[Delphi51]

Welcome to PF, John. Wow, my compliments to your prof for coming up with a very interesting problem! You mention momentum and energy - surely a clue. In the frictionless case, the block is moving after the collision so some of the energy is KE of the block (and bullet) after the collision. That leaves a little less to be dissipated as friction in the block, so slightly less penetration distance.

In the stationary block part, I used
initial KE = F*d
to find the friction force F on the bullet. This assumes constant force and deceleration, but we likely couldn't find an answer otherwise. Anyway, that gave me a value for force (with a v² in it), which I used in the second part. Luckily the v's canceled out.

Good luck with it! I would be interested in knowing how it works out and if you need more help.

 

[venkatg]

First case:
========

When the bullet hits the block in the vice, all energy is expended by penetration. Assuming a constant deceleration (which is not the case in real life. A time varying force is exerted in a short time - impulse, that invokes transient vibrations), the work done by the bullet in overcoming the tearing forces inside the block.

Work done W = F * d
KE of bullet is 0.5*m*u^2
So, F * D = 0.5 * m *u^2 -------------------- (1)

Second case:
===========

When the same bullet from the same gun hits a block on a frictionless surface, the momentum and kinetic energy conservation results in the following 2 equations

m*u = (M+m)V --------------------- (2)

0.5 * m *v^2 = F * D + 0.5 * (M + m) *v^2 ------------------ (3)

Use the above equations (1), (2) & (3) to find "D" which is the penetration distance in the second case

 

[Delphi51]

Looks good! I would put the numbers in (1) to get F = .04*u²
and do the same with (2) to get V = .007u before substituting into (3). I only did 1 digit accuracy; you'll need 3 or 4.

 

[Delphi51]

Looks good! I would put the numbers in (1) to get F = .04*u²
and do the same with (2) to get V = .007u before substituting into (3). I only did 1 digit accuracy; you'll need 3 or 4.

 

[MCPO John-117]

wow thanks guys! looks good venkatg, similar to what i came up with.
i went a step further so please check my algebra :)

momentum conservation: mv_i=(M+m)v_f ----1----
so... v_f=mv_i/(M+m) ----2----

conservation of energy: 0.5(m)v_i²=Fd+0.5(M+m)v_f² ----3----
so... 0.5(m)v_i²-0.5(M+m)v_f²=Fd ----4----

the initial velocity and final velocities are the same in both momentum and energy conservation equations venkatg's 2 and 3 and so are the masses so i can plug -2- into -4-

0.5(m)v_i²-0.5(M+m)(mv_i/(M+m))²=Fd

plugging in for M and m which are given as 1kg and 0.007kg respectively i got

0.0035(v_i²)-(0.5035)(0.007(v_i)/(1.007))²=Fd ----5----

((0.003476)v_i²)/F=d ----6----
this is a reasonable place to stop but F can be broken into ma
((0.003476)v_i²)/ma=d -----7-----

((0.49652)v_i²)/a=d ----8----

i'll probably submit something like this as my answer if my algebra and theory are correct :)

 

[Delphi51]

I think you've got it right so far. From Venkatg's equation (1) you can get an expression for F to use in your step 6 that will result in an actual numeric answer for d. The v² cancels out.

 

[MCPO John-117]

I think you've got it right so far. From Venkatg's equation (1) you can get an expression for F to use in your step 6 that will result in an actual numeric answer for d. The v² cancels out.

yeah i thought about it and concluded that the Forces in each case are not the same. in the first case where the block is stationary the deceleration would be much higher than in the second case where the deceleration would be less due to the moving block. the moving block is kind of softening the force. this is what i think it may be wrong. i just did the calculation if the force is the same anyway.

the original penetration of the stationary block is 8 cm and the free moving block penetration would be 7.95 cm.

not a big difference but it is less which is what you'd expect.

 

[Delphi51]

Looks good!
I think the assumption that the force remains the same may be okay because the block does not move very fast or very far during the collision. Assuming constant deceleration, d = ½vt so for v = 300 m/s, d=.08, we have t = .0005. The block's average speed will be about 1 m/s and will go .05 of a cm in that time. Not perfectly accurate, but a reasonable assumption.

 

[venkatg]

yeah i thought about it and concluded that the Forces in each case are not the same. in the first case where the block is stationary the deceleration would be much higher than in the second case where the deceleration would be less due to the moving block. the moving block is kind of softening the force. this is what i think it may be wrong. i just did the calculation if the force is the same anyway.

the original penetration of the stationary block is 8 cm and the free moving block penetration would be 7.95 cm.

not a big difference but it is less which is what you'd expect.

Since the block is 1000/7 times heavier than the bullet, the forces are approximately same in both cases. The velocity gained by block in the second case is negligible compared to velocity of the bullet. The actual forces are fluctuating during the penetration. We have assumed an average here. So the actual value in the second case will be slightly less than 7.95 cms.