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A Body (density $\rho_1$, elasticity modulus $E_1$ and volume $V_1$) crashes with constant velocity $V$ into another resting Body (density $\rho_2$, elasticity modulus $E_2$ and volume $V_2$). Both bodies are described by the equations of Motion

$$\rho_{1,2} \frac{\partial^2 u(x,t)}{\partial t^2} = E_{1,2} \frac{\partial^2 u(x,t)}{\partial x^2}$$

where $t$ is time, $x$ is the coordinate (for simplicity I assume 1-dimensional model) and $u(x,t)$ is the field of displacements in the Body. It holds for the stress $\sigma_{1,2}(x,t)=E_{1,2} \frac{\partial u(x,t)}{\partial x}$. This description holds in the interior of $V_1$ or $V_2$. If These bodies collide, I have a contact surface, in which stress must be continuous. But how I can formulate proper Initial and boundary conditions?

How I determine the stress Distribution in These bodies for this case? I assume that everything is without external fields, friction, etc. But how I can determine stresses in a Body during collision???

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  • $\begingroup$ Suppose the two bodies are identical, so that the interface remains stationary after the collision. Can you describe in words what you think is happening after the collision? $\endgroup$ Jul 14 '17 at 12:28
  • $\begingroup$ If collision is elastic, then I think the first Body which is colliding, will disappear with velocity $-V$ from the second Body. I don't know how to solve this Problem. $\endgroup$
    – kryomaxim
    Jul 14 '17 at 12:32
  • $\begingroup$ If you were moving with velocity V/2, from your frame of reference, what would the velocities of the two bodies be equal to? $\endgroup$ Jul 14 '17 at 14:02
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Thanks for comments and answers. Now I get good ideas for solving this model.

Maybe I can solve the model as follows:

D'Alembert solution: $u(x,t) = A(x+ct)+B(x-ct)$

Initial conditions: At $t=0$ there are no displacements and velocity is uniform within a region $\Gamma$, thus:

$u(x,0)=0=A(x)+B(x)$

$(\partial_tu)(x,0)=V1_\Gamma(x)=(\partial_tA)(x,0)+(\partial_tB)(x,0) = (\partial_xA)(x,0)c-(\partial_xB)(x,0)c$.

From the second condition it follows:

$A(x)=B(x)+\frac{V}{c} \int 1_\Gamma(x) dx$.

And then from the first condition:

$B(x) = - \frac{V}{2c} \int 1_\Gamma(x) dx:= - \frac{V}{2c} X_\Gamma(x)$.

Finally, the solution reads:

$u(x,t) = \frac{V}{2c} X_\Gamma(x+ct) - \frac{V}{2c} X_\Gamma(x-ct)$.

Substituting into the equation for the stress yields:

$\sigma(x,t) = \frac{EV}{2c}(1_\Gamma(x+ct)-1_\Gamma(x-ct))$.

From this solution it follows that the maximum stress is $\sigma_{max} = \frac{EV}{2c} = \sqrt{\rho E}V/2$.

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This is the same thing as a head-on collision with both colliding bodies traveling with at speed V/2. But that is the same thing as a single body with speed V/2 colliding with a rigid wall. When the body first makes contact with the wall, the leading edge of the body stop abruptly (v=0), while, further back, the rear part of the body will not yet even know that anything has happened (v=V/2). A compression zone will begin forming at the front end of the body. Throughout the compression zone, the velocity will be zero. To the rear of the compression zone, the velocity will still be V/2. The backward growth of the compression zone will occur at the speed of sound, $c=\sqrt{E/\rho}$. This problem can be solved using the d'alembert form of the solution to the wave equation.

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  • $\begingroup$ When things fall from great hights, high stresses arise during impact. But I don't know why. Maybe I have some missing foundations of mechanics. $\endgroup$
    – kryomaxim
    Jul 15 '17 at 19:35
  • $\begingroup$ I already indicated that the front end of the colliding object is in compression, so that certainly corresponds to a high compressive stress. The compression zone grows as time progresses. $\endgroup$ Jul 15 '17 at 20:46
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    $\begingroup$ If a cylindrical body traveling with a velocity v collides head on with a rigid wall, the strain in the compression zone is given by $\epsilon=v/c$, where c is the speed of sound, equal to $\sqrt{E/\rho}$. This follows from a momentum balance on the system. Ahead of the compression zone, the strain is zero and the velocity is unchanged. The stress within the compression zone is equal to the strain times E. So, $$\sigma=\sqrt{E\rho}v$$. The contact force is $$F=\sigma A=\sqrt{E\rho}Av$$ $\endgroup$ Jul 16 '17 at 0:26
  • $\begingroup$ In my previous comment, I meant "behind the compression zone..." $\endgroup$ Jul 16 '17 at 11:27

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