Question

A proton, moving with a velocity of v_iî, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 1.40 times the speed of the proton initially at rest, find the following.

(a) the speed of each proton after the collision in terms of v_i

initially moving proton

initially at rest proton

(b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction)

initially moving proton

initially at rest proton

Answer 1

Ans:

   1)    the mass of a proton is given mp=1.672·10^-27kg and because collision is elastic energy is conserved in the system we have,

           Ki=         .........................(1) vi is initial speed of moving proton

          Kf=             .....(2)    Vf1 is final velocity of initially moving proton

                                                            Vf2 is final velocity of initially rested proton.

            we have been given that ,

                   Vf1=1.40Vf2   ....3

          by conservation of energy Kf=Ki

                  =       

     put 3 in this equation and simplifying we get

         Vf1=                  final velocity of initially moving proton

       

           Vf2=           from (3)

               Vf2 = Vi/0.5812          Final velocity after collision of initially rested proton

             total Vf=Vf1+Vf2    =1.5894Vi      ........(4)

2) Let Vf1 be the final velocity for the incident proton, and Vf2 be the final velocity for the proton initially at rest. Conserving momentum in the y direction

                Piy =0= Pf y=mpVf1y+mp.Vf2y          for y component initially proton was in x direction thats why momemtum in y direction is Piy=0

                vf1y=−1.40*vf2y putting values of eq 3

the magnitude of their speeds in the x direction should also be equal   |vf1x|=1.40*|vf2x|. Conserving momentum in the   x    direction.

      Pix=mp*Vi= Pfx   =   mp*Vf1x+   mp*Vf2x   =1.40mp*Vf2x    by using eq3

            V_f2x =Vi/1.40    and V_f1x =V_i /1.714

          Vfx=Vf1x+Vf2x=0.5834Vi..................(5)

     total velocity magnitude given by Vf

        V²_f =Vfx² + Vfy²

       by putting value from 4 and 5 we get

          Vfy=0.5834Vi=Vfx

So because the x and y components of Vf are the same, both protons are deflected away at an angle of    θ=45◦   from the x   direction, with   opposite y   components (so the angle between   Vf1   and Vf2    is 90◦)