# Time taken to reach common velocity - collision

Say for instance I have two objects with different masses that collide in-elastically the resulting velocity will be $$v =\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$.

What I don't fully grasp is how to calculate the time taken to complete the collision, i.e the time taken to reach the common velocity. I'm guessing it depends on the geometry and materials of both objects, probably rigid objects take almost no time whereas more elastic objects take longer.

The maths likely has something to do with the elasticity constant (Hooke's law) of both materials.

Thanks

EDIT:

For the sake of simplicity lets just assume we're dealing with a head on collision on a one dimensional view.

• You won't be able to get such data from conservation of momentum; it requires detailed information about the forces which collision calculations usually integrate over. – probably_someone Jun 21 at 17:30

## 1 Answer

This problem has the exact solution. For two elastic balls of radius $$R_1$$ and $$R_2$$, the contact time at impact is $$\tau=\frac {4\sqrt {\pi}\Gamma (2/5)}{5\Gamma (9/10)}(\frac {\mu ^2}{k^2v})^{1/5}, k=\frac {4}{5D}\sqrt {\frac {R_1R_2}{R_1+R_2}}, \mu =\frac {m_1m_2}{m_1+m_2}, D=\frac {3}{4} (\frac {1-\sigma _1^2}{E_1}+\frac {1-\sigma _2^2}{E_2})$$ $$v$$ is a relative speed of colliding balls,$$E_i$$ is Young's modulus, $$\sigma _i$$ is Poisson’s ratio, $$i=1,2$$ . See L.D. Landau, E.M. Lifshitz, Theory of Elasticity