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## Homework Statement

A cannon shoots bullets of mass m

_{0}and velocity v

_{0}at a massive log of wood of mass 100m

_{0}. the interval between the bullets is t

_{0}. the surface is smooth. the bullets are stuck in the wood. the penetration time of the bullets is very short, relative to t

_{0}.

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 0.5v

_{0}.

## Homework Equations

Conservation of momentum: ##\sum m_i v_i=\sum m_i v'_i##

Kinetic energy: ##\frac{1}{2}mv^2##

## 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 v

_{N}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 0.5v

_{0}will be reached after:

$$\frac{Nv_0}{100+N}\leqslant \frac{v_0}{2}\quad\rightarrow\; N\leqslant 100$$