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Elastic collision

  • Thread starter Reshma
  • Start date
  • #1
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[SOLVED] Elastic collision

Homework Statement


A proton of mass 'm' collides with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90 degrees with respect to each other. What is the mass of the unknown particle.

Known quantities:
mass of proton = m
velocity of proton = v
mass of unknown particle = M
Angle between M and m after collision is 90 degrees. [itex]\alpha + \beta = {\pi \over 2}[/itex]

Homework Equations



Conservation of Momentum. Momentum must be conserved in x and y-direction:
X-direction:
[tex]mv = mv'\cos \alpha + Mv'\cos \beta[/tex]

Y-direction:
[tex] 0 = mv'\sin \alpha + Mv'\sin \beta [/tex]

The Attempt at a Solution



Since the angle between the particles is known i.e. 90 degrees,
[itex] \beta = {\pi \over 2} - \alpha[/itex] & [itex]\cos ({\pi \over 2} - \alpha) = \sin \alpha[/itex].

The momentum conservation equations modify as:
X-direction:
[tex]mv = mv'\cos \alpha + Mv'\sin \alpha[/tex]

Y-direction:
[tex] 0 = mv'\sin \alpha + Mv'\cos \alpha [/tex]

I am not able to proceed beyond this, I have two unknown quantities on the Right-Side viz. [itex]\alpha[/itex] and the final velocity v'. I have to obtain the unknown mass in terms of m. Am I missing out something here?
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Reshma! :smile:

(btw, don't you need a separate v' and V' ?)

You can't solve any collision problems without using conservation of energy! :smile:
 
  • #3
747
4
Thanks for the hint tiny-tim! I will work on it. :smile:
 
  • #4
747
4
Okay, I figured it out!

Let m and M move at speeds v' and V' respectively.

By conservation of energy:
[tex]{1\over 2}mv^2 = {1\over 2}mv'^2 + {1\over 2}MV'^2[/tex]

On simplification, we get:
[tex]v^2 = v'^2 + {m\over M}V'^2[/tex]

Conservation of momentum along the X-axis gives:
[tex]mv = mv'\cos \alpha + MV'\sin \alpha[/tex]

Multiplying by v on both sides and comparing with the energy conservation equation:
[tex]mv^2 = mv'v\cos \alpha + MV'v\sin \alpha[/tex]

We get [itex]v' = v\cos \alpha[/itex] & [itex]V' = v\sin \alpha[/itex], so
[tex]v^2 = v'^2 + V'^2[/tex]

On comparing the coefficients, we get:
[tex]m = M[/tex]
 
  • #5
747
4
This was an objective type question, however is there a method to solve such a problem with lesser algebra?
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
This was an objective type question, however is there a method to solve such a problem with lesser algebra?
Hi Reshma! :smile:

Actually, there is.

Since we know in this case that the final momentums are perpendicular, it seems a good idea to try solving it using the momentum vectors.

If we call those vectors P Q and R, we know that P = Q + R, and in this case that Q.R = 0.

Try writing the energy equation in terms of the momentum vectors, and then comparing it with the square of P = Q + R. :smile:
 
  • #7
747
4
Thanks for your help tiny-tim! :smile:
 

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