inelastic collision

Question

This is all in one dimension. A bullet mass $m$ collides at time $t = 0$ with a cube of wood mass $M$ and edge $a$, free to slide. Penetrating the wood the bullet is retarded by a force of magnitude $kw$ where $w$ is the velocity of the bullet relative to the wood. Problem is to prove the minimum initial velocity $U$ (normal to the surface) to make the bullet pass right through the block, and derive the time and speed at which it emerges.

We are told the answer to the first part which is $U$ must be $>\frac{(M+m)ka}{Mm}$ so have to prove this.

Answer 1

Let's say that the cube is moving with velocity $v$ with respect to the ground, and the bullet is moving with velocity $v+w$ in the same reference frame. Newton's second law says the mass time acceleration is equal to the sum of the forces. For the cube:$$M\frac{dv}{dt}=kw$$ For the bullet: $$m\frac{d(v+w)}{dt}=m\frac{dv}{dt}+m\frac{dw}{dt}=-kw$$ Now take $\frac{dv}{dt}$ from the first equation, and plug it into the second. You will obtain a simple first order ODE for $w$. The solution is an exponential, which looks like $w=Ue^{-\alpha t}$. I will let you do the calculation for that, to figure out $\alpha$.

Now you can calculate what's the displacement inside the cube for the bullet: $$d=\int_0^t w(t) dt$$ In the case of lowest possible initial velocity for the bullet, it will take an infinite time to reach a distance $a$, so $$a=\int_0^\infty U_{\rm{min}}e^{-\alpha t}dt$$

For the second part, just say $$a=\int_0^{T(U)}Ue^{-\alpha t}dt$$ You calculate $T(U)$, that's the time required to pass through the cube. The relative speed at the exit will be $w(T(U))$. You can just plug into the equation fore $w$. If you want the absolute speed, with respect to ground, you will also need to calculate $v(t)$. Just plug in $w$ in the first equation, and integrate.

Answer 2

Let's consider the momentum equation $$mU = (m+M)v+mw,$$ and the action and reaction principle $$ M\frac{dv}{dt} = kw.$$ Then $$ \begin{split} -kw &= m\left(\frac{dw}{dt}+\frac{dv}{dt}\right),\\ w &= U-\frac{m+M}{m}v\\ v &= (U-w)\frac{m}{\left(m+M\right)}\\ \frac{dv}{dt} &= -\frac{m}{m+M}\frac{dw}{dt}\\ kw &= -\frac{mM}{m+M}\frac{dw}{dt}\\ \frac{dw}{dt} &= -\frac{k\left(m+M\right)}{mM}w\\ w &= Ue^{-\frac{ k\left(m+M\right)}{mM}t}=\frac{dx}{dt}, \end{split}$$ and thus $$ \int_{t=0}^t\frac{dx}{dt}dt = x = \frac{UmM}{ k\left(m+M\right)}\left[1- e^{-\frac{ k\left(m+M\right)}{mM}t}\right].$$ In the limiting case, $t = \infty$ and $x = a = \frac{UmM}{ k\left(m+M\right)}$. Normally the right hand expression is $\gt x$ so for $x$ to be $\geq a$ it is needed that $$ U\gt\frac{ka\left(m+M\right)}{mM}\qquad\mathrm{QED}. $$ ……………………………………………………………………………………………………………….

The bullet emerges at speed $v + w$ and from the above expression for $a$ we have $$ a = \frac{UmM}{k\left(m+M\right)}\left[1-\frac{w}{U}\right]. $$ So $$ w = U - \frac{k\left(m+M\right)a}{mM} $$ and also $$ v = \frac{m\left(U-w\right)}{m+M}. $$ Adding these equations and cancelling some terms out gives $V+w = U-\frac{ka}{m}$. QED.