Force impacts resitution, any help would be great

1. Aug 22, 2008

cheff3r

there are three parts to my problem (you can probably tell i haven't done much work on resitution before)

A bioengineer studying helmet design uses an experimental apparatus that launches a 2.4 kg helmet containing a 2 kg model of the human head against a rigid surface at 6 m/s. The head, suspended within the helmet, is not immediately affected by the impact of the helmet with the surface and continues to move to the right at 6 m/s, so the head then undergoes an impact with the helmet. If the coefficient of restitution of the helmet’s impact with the surface is 0.85 and the coefficient of restitution of the subsequent impact of the head with the helmet is 0.15

a) what is the velocity of the head after its initial impact with the helmet?

b) If the duration of the impact of the head with the helmet is 0.004 s, what is the magnitude of the average force exerted on the head by the impact?

c) Suppose that the simulated head alone strikes the rigid surface at 6 m/s, the coefficient of restitution is 0.5, and the duration of the impact is 0.0002 s. What is the magnitude of the average force exerted on the head by the impact?

So we know these formulas (possibly more that i didn't think of)
0.5*m1*v1^2+0.5*m*v2^2=0.5*m1*v3^2+0.5*m2*v4^2
F=ma
v3-v4=-e(v1-v2)

part a)
this is the part i'm struggling on
i was going to use v3-v4=-e(v1-v2) for the helmet and the wall to work out velocity of helmet after impact first but that is wrong since it has not taken into account of the head hitting the helmet

my solution part b)
F=ma which is equal to F=mv/t
F=2v/0.004 (v being answer to part a which i cant work out)
F=500v (is this the right method?)

my solution part c)
using equation v3-v4=-e(v1-v2) (coefficent of restitution formula)
v1=6m/s
v4=0m/s
v2=0m/s
e=0.5

v3=-0.5(6-0)+0
v3=-3
v3=3 m/s in oppostie direction (is this right?)

F=ma which is equal to F=mv/t

F=2*3/0.0002
F30,000N

2. Aug 23, 2008

dynamicsolo

Let's deal with part (a) first. You will not be able to use the conservation of kinetic energy equation you wrote because that only applies to elastic collisions (by definition), for which the coefficient is 1.

The collision in the simulation must be broken into two phases, the first involving the rebound of the helmet from the rigid surface, and the second being the collision of the simulated head with the rebounding helmet.

In the first part, the rigid surface is probably fixed in place (this is generally the case in collision simulators), so its velocity before and after the impact is zero. This makes the ratio for this portion of the collision

$$e_1 = 0.85 = \frac{v'_{helm} - 0}{0 - v_{helm} } = \frac{v'_{helm}}{-6 }$$

This can be solved directly for the single unknown.

Now, for the second part of the collision, the "head", continuing forward, meets up with the rebounding helmet, for which the ratio is

$$e_2 = 0.15 = \frac{v''_{helm} - v''_{head}}{6 - v'_{helm} }$$

with $$v'_{helm}$$ being your result from the previous equation (don't forget to include the correct sign).

Now this second equation can't be solved by itself because it has two unknowns, the final velocities of helmet and head. Fortunately, we do have a second equation from conservation of linear momentum:

$$m_{head} \cdot 6 + m_{helm} \cdot (-|v'_{helm}|) = m_{head} \cdot v''_{head} + m_{helm} \cdot v''_{helm}$$

(you knew you were going to need those masses somewhere in here...). The answer they're after is for $$v''_{head}$$

You could now do part (b), for which you need

$$F_{ave} = \frac{m \cdot \Delta v}{\Delta t} = \frac{2 kg. \cdot (v''_{head} - 6 \frac{m}{sec}) }{0.004 sec}$$

This should give you an idea of what to do for part (c): the rigid surface remains at zero velocity and the unprotected "head" has a different coefficient.

Last edited: Aug 23, 2008
3. Aug 23, 2008

cheff3r

you make it look so easy, thanks heaps and yeah i can do (c) now (correctly this time)
i drew a picture in my answer i think it helps