Interesting+Fact

=__**Interesting Fact!-(common misconception too)**__=

1)Which would be more damaging: driving into a massive concrete wall, or driving at the same speed into a head-on collision with an identical car traveling toward you at the same speed?

 * __Ans: Both would be equally damaging!__**

Explanation: Lets take the car driving into the wall as car A and the car driving into another identical car as car B1 and B2.

For Car A, assuming that the concrete wall is solid and so massive that it is immovable, when the car collides with the wall, Newton's 2nd law, F=ma can be applied, taking m=mass, a=deceleration. After calculations, this would be the force applied onto the wall. Based on **__Newton's 3rd law, there is an equal and opposite force applied for every action,__** the wall being static and unbreakable, exerts that much force back onto the car, causing damage to the car. This force would be the force generated by a deceleration from 'x' km/h

For car B, assuming both cars which would collide head-on are equal in every aspects(mass,speed) except for their direction by the **__law of momentum conservation__**, for a collision occurring between objects in an isolated system, __**the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision**__. That is, the **__momentum lost by object 1 is equal to the momentum gained by object 2__**, we can tell that both sustain **__equal damage__** as we know that both cars come to a stop, so the two cars' total momenta would equate to, taking direction into considerations:

(Note that momentum is the force of the object as defined in "Physics Concepts behind Car collisions")

__Before:__ Car B1=__**M1(force from 'x' km/h)**__, Car B2=**__M2(force from '-x' km/h__**)

__After:__ Car B1=__**M1*(at 0 km/h)**__, Car B2=**__M2*(at 0 km/h)__ __M1* and M2* are both 0__**

Hence __**M1 = -M2**__, **__M1*__****__-M2*__** and **__M1 + M2 = M1* + M2*= 0__**

Since the loss in momentum by Car B1 is a M1, the momentum gained by B2 is also M1: ( M2+M1=M2* ) Similarly,the loss in momentum by Car B2 is M2, the momentum gained by B1 is also M2: ( M1+M2=M1* ) Thus we can conclude that both Cars sustain the same damage as both had the same amount of momentum gained although in different directions, the force sustained by the car would be equal. Take Car A's kinetic energy as 'K', so Cars B1 and B2 would have 2'K'. Since Car B1 and B2 take the same amount of damage, each would take 1'K' worth of damage, which is equal to the 1'K' worth of damage from Car A.