# Bullets stuck in a log and speed it up

1. May 7, 2016

### Karol

1. The problem statement, all variables and given/known data
A cannon shoots bullets of mass m0 and velocity v0 at a massive log of wood of mass 100m0. the interval between the bullets is t0. the surface is smooth. the bullets are stuck in the wood. the penetration time of the bullets is very short, relative to t0.

1) What's the log's velocity after the first bullet.
2)What's the log's velocity after N bullets.
3) Express the number of the bullets in the log as a function of it's velocity.
4) How much heat energy was created in the log after N bullets.
5) How much time elapsed between the N-1 bullet hit and the Nth bullet hit.
6) How much time elapsed between the first bullet's hit and the Nth bullet's hit.
7) When does the log's velocity equal 0.5v0.

2. Relevant equations
Conservation of momentum: $\sum m_i v_i=\sum m_i v'_i$
Kinetic energy: $\frac{1}{2}mv^2$

3. The attempt at a solution
After the first bullet:
$$m_0v_0=(100+1)m_0v\;\rightarrow\; v=\frac{1}{101}v_0$$
The total momentum after N bullets is $P=Nm_0v_0$ . the velocity must cancel with a part of this momentum to give back the same P, so:
$$Nm_0v_0=(100+N)m_0\frac{Nv_0}{100+N}m_0\;\rightarrow\; v_N=\frac{Nv_0}{100+N}$$
The number of bullets as a function of the velocity:
$$\frac{v_0}{v_N}=\frac{100}{N}+1\quad\rightarrow\;N=\frac{100v_N}{v_0-v_N}$$
The velocity of the log after N-1 bullets is $v_{N-1}=\frac{(N-1)v_0}{99+N}$ (derived from the vN formula). so the energy loss between the (N-1)th and the Nth bullets is:
$$\Delta E=\frac{1}{2}\left[ m_0v_0^2+(99+N)m_0\frac{(N-1)^2v^2_0}{(99+N)^2}-(100+N)m_0\frac{N^2v^2_0}{(100+N)^2} \right]=\frac{m_0v^2_0}{2}\left[ 1+\frac{(N-1)^2}{99+N}-\frac{N^2}{100+N} \right]$$
The total energy loss from the first bullet to the Nth is the sum of these ΔE's, which is complicated. i don't think this is the best method.
The distance the log travells between the (N-1) and the Nth bullets is:
$$\Delta s=v_{N-1}\cdot t_0=\frac{(N-1)v_0t_0}{99+N}$$
The time between the (N-1)th and the Nth bullets is, therefore:
$$\Delta s=v_0\cdot t\quad\rightarrow\; t_{(N-1)\rightarrow N}=\frac{\Delta s}{v_0}=\left( 1+\frac{N}{99+N} \right)t_0$$
The time elapsed between the first and the Nth bullets is the sum of the above time, which is complex and isn't expected to be found in this way.
The velocity of 0.5v0 will be reached after:
$$\frac{Nv_0}{100+N}\leqslant \frac{v_0}{2}\quad\rightarrow\; N\leqslant 100$$

2. May 7, 2016

### ehild

Why do you do it so complicated? What is the input of KE during N shots? What is the resulting KE?
If you want to add the loss of KE in every step, it will be (KE(bullet) - KE(1)) +( KE(bullett+ KE(1)-KE(2)) +...

3. May 7, 2016

### Karol

$$\Delta EK=\frac{1}{2}Nm_0v_0^2-\frac{1}{2}(100+N)\frac{N^2v_0^2}{(100+N)^2}=\frac{50Nm_0v_0^2}{100+N}$$
But how to find the time from the first bullet hit to the Nth hit? i don't see a method other than adding the individual times, which depend each on a different velocity

4. May 8, 2016

5. May 8, 2016

### SammyS

Staff Emeritus
What is the initial velocity of the log ?

6. May 8, 2016

### ehild

The time between two subsequent bullet hits is not to. to is the time between the shots. The bullet hits the log between the shots, and before the (N-1)-th hit the log travels with VN-2 velocity, after the hit it travels with VN-1. To get the distance the log travels between the hits is quite complicated.
Edit: I think I have found the solution. The results are quite simple.

Last edited: May 9, 2016
7. May 8, 2016

### ehild

It is assumed zero.

8. May 10, 2016

### ehild

Assume the N-1-th shot happens at tsN-1=0. At that instant, the log is at xs(N-1)=D and its speed is V(N-2). What time is the impact?
What is the position xi(N-1) of the log at the impact?
Where is log at instant of the N-th shot? (What is xs(N) at ts(N)=t0?

9. May 10, 2016

### Karol

$$\Delta s=v_0\cdot t=v_{N-1}\cdot t_0+v_{N-1}\cdot t~~~\rightarrow~~~t_{(N-1)\rightarrow N}=\frac{N-1}{100}t_0$$
The total time for N hits:
$$\sum^N_1 t_0=\frac{(N-1)}{100}=\frac{1}{100}\left[ \sum^N_1-N \right]=\frac{1}{100}\left[ \frac{N(1+N)}{2} \right]=\frac{N(1+N)}{50}t_0$$

10. May 10, 2016

### ehild

Would you explain your notations?
And what do you mean on total time of hits?
What is your answer to question "How much time elapsed between the N-1 bullet hit and the Nth bullet hit"?
Your last line is totally wrong.

Last edited: May 10, 2016
11. May 10, 2016

### Karol

Sorry, missed some things on the last line.
The situation is like of a police car starting fast to catch a thief's car, which starts further in front but drives slower. the cars will meet at distance Δs.
The starting point of the police car is where bullet (N-1) hit the log. if the log didn't move, the next bullet, the N-th, would hit at the same place, which is the start for the police car.
$t_{(N-1)\rightarrow N}$ is the time elapsed between the (N-1) hit and the N-th:
$$\Delta s=v_0\cdot t_{(N-1)\rightarrow N}=v_{N-1}\cdot t_0+v_{N-1}\cdot t_{(N-1)\rightarrow N}~~~\rightarrow~~~t_{(N-1)\rightarrow N}=\frac{N-1}{100}t_0$$
The last line displays the time elapsed from the first to the N-th hit:
$$\sum^N_1 \frac{N-1}{100}t_0=\frac{t_0}{100}\left[ \sum^N_1 N-N \right]=\frac{t_0}{100}\left[ \frac{N(1+N)}{2}-N \right]=\frac{N(N-1)}{50}t_0$$

12. May 10, 2016

### ehild

The shots follow each other in t0 intervals. The elapsed time between two hits can not be shorter than t0. From your formula, the time between the first and second hit is 0.01 t0.
You determined the time between the N-th shot and impact. To get the time between the impacts, add t0.

Last edited: May 10, 2016
13. May 10, 2016

### Biker

I got this for the 5th question.

Where $T_2$ is the time elapsed between the N-1 bullet hit and the Nth bullet hit.
$T_2 = \frac{T_o (N+99)}{100}$

When I looked at your solution, It seems you have thought of the same idea as mine. (Which I think is correct, waiting for confirmation)
This should be the correct formula
$$\Delta s=v_0\cdot t_{(N-1)\rightarrow N}=v_{o}\cdot t_0+v_{N-1}\cdot t_{(N-1)\rightarrow N}$$

14. May 10, 2016

### ehild

Your result is correct.
Yes, that is the correct formula.

Last edited: May 10, 2016
15. May 11, 2016

### Karol

It's not, the log has moved a distance, not the bullet, so the term $v_0\cdot t_0$ on the right isn't correct.
My formula on post #11 is correct:
$$\Delta s=v_0\cdot t_{(N-1)\rightarrow N}=v_{N-1}\cdot t_0+v_{N-1}\cdot t_{(N-1)\rightarrow N}$$
But if i extract $t_{(N-1)\rightarrow N}$ from the "wrong" formula i get the right answer:
$$t_{(N-1)\rightarrow N}=\frac{99+N}{100}$$
So why, despite of it, is it correct?

16. May 11, 2016

### ehild

You do not get the correct result from your equation. In post #12 you got (N-1)/100 *to. It followed from your equation.

17. May 11, 2016

### Karol

Yes but if i add t0 it's correct.
But the main thing for me is that i don't understand the reasoning for the term $v_0t_0$.
I am beginning to understand but not yet. if i use:
$$\Delta s=v_0\cdot t_{(N-1)\rightarrow N}=v_{o}\cdot t_0+v_{N-1}\cdot t_{(N-1)\rightarrow N}$$
Then my drawing isn't adequate, it's adequate only for my formula:
$$\Delta s=v_0\cdot t_{(N-1)\rightarrow N}=v_{N-1}\cdot t_0+v_{N-1}\cdot t_{(N-1)\rightarrow N}$$

Last edited: May 11, 2016
18. May 11, 2016

### ehild

The time you got first was between the N-th shot and the N-th impact.
You said that you counted distance from the place of the N-1-th impact. That is a fixed place, and the next shot arrives there to time later. That is why you need to add to to your time.

19. May 11, 2016

### Karol

Exactly, that's what i understand, but what's the physical reasoning for the member $v_0t_0$ instead of $v_{(N-1)\rightarrow N}t_0$?
Mathematically it adds my missing t0, but i don't see the physics of it. is it just a mathematical trick? if so, how did you do this, how did you know in advance what to do (if it's merely a mathematical trick, with no physical meaning)?
Edit: i think i am beginning to understand, you translated time to distance, there is no complete physical meaning. you need to get t0 plus something, so for the first t0 you need a distance of $v_0t_0$, is that correct?

20. May 11, 2016

### ehild

The N-th bullet arrives to the place of the N-th impact t0 time later than the previous impact was. From the place of N-th impact, the N-th bullet moves with v0. The log is at distance VN-1to already. So the time the N-th bullet reaches the log is $\frac {t_0 V_{N-1}}{v_0-V_{N-1}}$. You have to add t0 to get the time between the impacts. It is $\frac {t_0 V_{0}}{v_0-V_{N-1}}$