# Collision Time

1. May 12, 2005

Consider a collision between two spheres where there was an elastic force acting between the two bodies.
F=u*dv/dt=-h*x^3/2
where velocity v=dx/dt, h is a constant, u is the reduced mass (m1*m2/m1+m2), and x is the "change in distance" between the centers of the two bodies. Maximum compression distance is X_m, and initial velocity is v_i.

Show that the collision time, t, is given by
t=2(X_m)/(v_i)* [Integral from 0 to 1 of (1-k^5/2)^(-1/2) dk],
where k is a dimensionless variable replaced in the integral for x.

Just some idea to start this off would be much appreciated. Thanks!

Last edited by a moderator: May 13, 2005
2. May 13, 2005

### OlderDan

I think you just have to go with energy on this one. The force is elastic, hence conservative, hence you have a potential energy stored in the deformed spheres.

PE = (2/5)hx^(5/2)

That plus the kinetic energy has to be constant, which gives you velocity as a function of position, and vice versa. That relationship ought to make your DE separable into something of the form

f(v)dv = dt

that you can integrate to get the result you are looking for. I haven't done it, but that appears to be a path that will get you there.

3. May 13, 2005

I got the part about PE... i.e. after integrating the force function give, we get the work done/ potential energy...

btw, is work done on moving one of the mass/PE= (2/5)h*(m/u)*x^(5/2)?
and equating this to (1/2)*m*(v_i)^2, the x we find would be the maximum, i.e. X_m?

Also, question: How do you get a DE in the form f(v)dv=dt?? cos I tried but I cant seem to separate it out to get that stated integral..

And what does itmean by replacing the dimensionless variable k...?

Last edited by a moderator: May 13, 2005
4. May 13, 2005

### OlderDan

The algebra gets a bit involved, but I expect you can work that out. I don't see there being any mass terms in the potential energy. The work done to compress the spheres has nothing to do with their masses. This is analogous to a spring where the PE just depends on the spring constant and the displacement. Once you have the PE, you know the total energy is conserved so you can write

$$U = \frac{1}{2}\mu v^2 + \frac{2}{5}hx^{5/2} = \frac{1}{2}\mu v_i^2 = \frac{2}{5}hX_m^{5/2}$$

solve for x

$$x = \left( {\frac{{5\mu }}{{4h}}} \right)^{2/5} \left( {v_i^2 - v^2 } \right)^{2/5}$$

Then

$$x^{3/2} = \left( {\frac{{5\mu }}{{4h}}} \right)^{3/5} \left( {v_i^2 - v^2 } \right)^{3/5}$$

Substitute that into the DE and solve for -dt

$$\left( {\frac{4}{5}} \right)^{3/5} \left( {\frac{\mu }{h}} \right)^{2/5} \frac{{dv}}{{\left( {v_i^2 - v^2 } \right)^{3/5} }} = - dt$$

I have not taken it far enough to identify k, but it is just an integration variable you need to come up with to integrate the left side of this thing. The wording suggest that you would have an integral over x, but I suspect you might not need to change the integral over v to an integral over x. If you need to do that, you can get the differential relationship between v and x from the equation for x above and express v in terms of x by solving the x(v) equation for v(x). I wouldn't go there unless I had to, but it might be necessary to get it into the form they want.

You would want to take advantage of the symmetry of the velocity integral and find the time of compression for v to go from initial to zero, then double it, but apparently you don't have to do the integral, just find the appropriate variable of integration. It's not jumping out at me, but if you work from the total energy equation you can cume up with the ratio of maximum compression to the initial velocity and take it from there. Check all the algebra!!