# Elastic collision

1. May 21, 2008

### Reshma

[SOLVED] Elastic collision

1. The problem statement, all variables and given/known data
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. $\alpha + \beta = {\pi \over 2}$

2. Relevant equations

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

Y-direction:
$$0 = mv'\sin \alpha + Mv'\sin \beta$$

3. The attempt at a solution

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

The momentum conservation equations modify as:
X-direction:
$$mv = mv'\cos \alpha + Mv'\sin \alpha$$

Y-direction:
$$0 = mv'\sin \alpha + Mv'\cos \alpha$$

I am not able to proceed beyond this, I have two unknown quantities on the Right-Side viz. $\alpha$ 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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Last edited: May 21, 2008
2. May 21, 2008

### tiny-tim

Hi Reshma!

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

You can't solve any collision problems without using conservation of energy!

3. May 21, 2008

### Reshma

Thanks for the hint tiny-tim! I will work on it.

4. May 22, 2008

### Reshma

Okay, I figured it out!

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

By conservation of energy:
$${1\over 2}mv^2 = {1\over 2}mv'^2 + {1\over 2}MV'^2$$

On simplification, we get:
$$v^2 = v'^2 + {m\over M}V'^2$$

Conservation of momentum along the X-axis gives:
$$mv = mv'\cos \alpha + MV'\sin \alpha$$

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

We get $v' = v\cos \alpha$ & $V' = v\sin \alpha$, so
$$v^2 = v'^2 + V'^2$$

On comparing the coefficients, we get:
$$m = M$$

5. May 22, 2008

### Reshma

This was an objective type question, however is there a method to solve such a problem with lesser algebra?

6. May 22, 2008

### tiny-tim

Hi Reshma!

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.

7. May 23, 2008