A proton of mass m undergoes a head on collision with a stationary atom of mass 16m.  If the initial speed of the proton is 470 m/s, find the speed of the proton after the collision?

I'm using the equation m1v1 + m2v2= m1v3 +m2v4 and I get so far as#m(470) = m1(v3) +16m(v4)# and then I solved for v3 to get:

#((470m-16v4)/m)=v3#

And I can plug it back into the equation but I don't know how to solve it from there when I still don't know m, v3, or v4.

1 Answer
Morgan
Aug 14, 2017

The final speed of the proton is #3525" m"//"s"#.

Explanation:

This is easily solvable if you make the assumption that the collision is perfectly elastic, which is an approximation often made for atomic collisions.

Derived from equations of both momentum and energy conservation:

#color(darkblue)(v_(1f)=((m_1-m_2)/(m_1+m_2))v_(1i)+((2m_2)/(m_1+m_2))v_(2i))#

We have the following information:

  • #|->m_1=m#
  • #|->m_2=16m#
  • #|->v_(1i)=470" m"//"s"#
  • #|->v_(2i)=0#

Substituting into the above equation:

#=>v_(1f)=((cancelm-16cancelm)/(2cancelm))v_(1i)#

#=>v_(1f)=-15/2(470"m"//"s")#

#=>color(darkblue)(v_(1f)=-3525"m"//"s")#

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