You are at the controls of a particle accelerator, sending a beam of 3.60×107 m/s protons (mass m) at a gas target of an unknown element. your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. all such protons rebound with a speed of 3.30×107 m/s. assume that the initial speed of the target nucleus is negligible and the collision is elastic.what is the speed of the unknown nucleus immediately after such a collision?

= 7.06 107 m/s

Explanation:

In a crash problem we must define a system, in this case it is formed by the proton and the particle with which it collides. For this system the force during cooking is internal, so the amount of movement is conserved and as it indicates that the shock is elastic, the kinetic energy is also conserved.

      = p₀

      = K₀

Before the crash

     p₀ = + 0

     K₀ = ½ ² + 0

After the crash

      =  + m vf

      = ½ ² + ½ m vf²

Let's write conservation equations

    = + m vf

   ½ ² = ½ ³ + ½ m vf²

    ( -) = m vf

     ( ² -²) = m vf²

We have a System of two equations with two incognitites, so it can be solved, using the relation (a + b) (a-b) = a² -b² and with a little algebra

We calculate the mass of the particle

    = ( / m) ()

     = m ² / m² ()²  

     + = / m  -

   m =  -  /  +    

   m = (3.6 - 3.30) 10 7 / (3.6+ 3.3) 107

   m = 4,248 10-2

We calculate the speed of the particle

    = ( / m)  

    = / (4,248 10-2) (3.6 -3.3) 107

   . = 1 / 4,248 10-2 0.3 107

    = 7.06 107 m / s

its A,C,B i checked it its 100% right