tittys04;165920 wrote: That's not true.
If the bikes weighed the same and their velocities were equal, their momentum would be equal, and in an "ideal" collision, both bikes would stop dead in their tracks. For example, 400lb bikes travelling at 80mph, their individual moment equals 32,000 lb*m/sec (not a normal unit for momentum, but you get the idea). However, when you take into account the vector of each bike, if colliding perfectly at 180degrees, 32,000-32,000=0, which means dead stop.
Now look at car vs. bike. 400lb bike at 80mph vs a 6000lb car at 35mph. 32,000-210,000= -178,000 in the car's favor. Notice that at less than half of the velocity of the bike, the car still has almost a 6x advantage in a head-on collision.
However, the scenario you are talking about is referring to mph, rather than momentum. In this case, it's not about vectors and such, but more about referrence points. If you are travelling down the free way in a car at 65mph (car A), and someone in the same model car rear ends you at 70mph (car B), the collision would be the same as if you were parked and someone rear ended you at 5mph. With car A as the reference point, car B is travelling 5mph faster than A. With B as the reference point, A is travelling 5mph slower than B. If you pick a road sign as a reference point, A (at 65mph) - B (at 70mph) = -5mph in B's favor.
Same thing goes with head on collisions, but you add a negative sign. Bike A (at 80mph) - (-) bike B (at 80mph) = 160mph. (2 negative signs yields a postitive)
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