# Re: Physics Question

From: Doug Jones (djones@xcor.com)
Date: Sat Feb 23 2002 - 14:38:58 MST

John Clark wrote:
>
> <hal@finney.org> Wrote:
>
> >When the 4000 pound car travelling at 35 mph hits the 3000 pound
> >minivan, if the collision were elastic and no energy was lost in the
> >crushing metal, the result would be the minivan going at 40 mph.
>
> The only way for the original car to transfer all its momentum to the target
> car is if the two automobiles have exactly the same mass, if the collision
> was perfectly elastic (and in the real would it would not be) it would transfer
> all its energy too. Under these ideal conditions when they hit the original car
> would come to a complete stop and the target car would move at the original
> car's former speed, but regardless of the weight of either car there is no way
> the second car can end up moving faster than the first after a collision.
>
> John K Clark jonkc@att.net

Sure the second object can go faster than the first- but the first
object is still moving downrange also. Consider an elastic collision
between a 10 kg "cueball" and a 1 kg target. The target is at rest in
our arbitrary coordinate system, and the cueball is moving at v=10 m/s.
Before the collision, the target's speed _relative_to_the_cueball_ is
-10, afterward it is +10. With both energy and momentum conserved:

M1*V1 + M2*V2 = M1*V3 + M2*V4

M1*V1^2 + M2*V2^2 = M1*V3^2 + M2*V4^2 (both sides *2 for simplicity)

V4 = V3 + (V1-V2) (definition of elastic collision)

M1*V1 + M2*V2 = M1*V3 + M2*(V3+V1-V2)

((M1-M2)*V1+2*M2*V2)/(M1+M2)=V3

With the initial conditions:

M1 = 10
M2 = 1
V1 = 10
V2 = 0

then

V3 = ((10-1)*10+0)/11 = 8.18
V4 = 8.18 + 10 = 18.18 (note that the velocity diff is of same
magnitude, opposite sign)

10*10^2 + 1*0^2 = 10*8.18^2 + 1*18.18^2
1000 + 0 = 669 + 331

This is how a golf club works. Large flimsy objects such as cars and
humans tend to collide rather inelastically, though, and my earlier
analysis in a previous post stands.

```--
Doug Jones, Rocket Plumber
```

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