Speed of Particle: Conservation of Momentum & Energy

[mike217]

Can you please give me a few pointers on this problem.

A particle of rest mass m is accelerated to a kinetic energy K in a nuclear reactor. This particle is incident on a stationary target particle, also of rest mass m.

a) Show that the speed of the centre of mass (that is the speed of the frame in which the total momentum is zero) is $$\gamma v/(\gamma +1)$$ where v is the speed of the incident particle and $$\gamma =(1-v^2/c^2)^-1/2$$. Verify that this expression reduces to the usual one in the non relativistic case when v<<c.

b) If the collision is perfectly inelastic-show by using the conservation of energy and momentum that the mass M of the resulting composite is $$M=m\sqrt{2(\gamma +1)}$$ Verify that this reduces to the usual value in the non relativistic case v<<c.

For part 'a' I get by writing the conservation of momentum:

$$Pbefore=mv/\sqrt{1-v^2/c^2}$$ and $$Pafter=MV/\sqrt{1-V^2/c^2}$$ after that I am stuck on what to do.

For part 'b' I get,

$$mc^2/\sqrt{(1-v^2/c^2)} + mc^2=Mc^2/\sqrt{(1-V^2/c^2)}$$

Thank you.

 

[mike217]

Hi,

Please you don't have to solve this question. I just need a few tips on how to proceed, i.e. which particle to take as one having zero momentum etc.

Thank you.

 

[wais]

Sure, here are some pointers to help you with this problem:

a) To start, you can use the conservation of momentum in the x-direction to set up an equation:

Pbefore = Pafter

where Pbefore is the momentum of the incident particle and Pafter is the momentum of the resulting composite. You can also use the definition of momentum P = mv/\sqrt{1-v^2/c^2} to rewrite this equation as:

mv/\sqrt{1-v^2/c^2} = MV/\sqrt{1-V^2/c^2}

Then, you can solve for V in terms of v and M by rearranging the equation. Once you have V in terms of v and M, you can use the definition of \gamma to simplify the expression and get the desired result.

b) For this part, you can use the conservation of energy and momentum to set up two equations:

Energy before = Energy after
Momentum before = Momentum after

Using the definitions of energy E = \sqrt{m^2c^4 + p^2c^2} and momentum P = mv/\sqrt{1-v^2/c^2}, you can rewrite these equations in terms of v and V. Then, you can solve for V in terms of v and M, and use the definition of \gamma to simplify the expression and verify the result.

Hope this helps! Let me know if you have any other questions.