Scattering angle in the CM frame

[indigojoker]

$$\alpha$$ particle of mass m is scattered by a nucleus of mass M. $$\Theta$$ is the scattering angle of the $$\alpha$$ particle in the LAB reference frame, and $$\theta$$ is the scattering angle in the CM frame.

What is the relation between $$\Theta$$ and $$\theta$$ using conservation of energy and momentum?
I am suing v1 as velocity of m and v2 as velocity of M

$$v_{CM} = \frac{m v_1}{m+M}$$

in the CM frame (denoted by v'):
$$v'_2 = v_1 - v_{CM} = \frac{M v_1}{m+M}$$
$$v'_2 = v_{CM}=\frac{mv_1}{m+M}$$

for elastic scattering:

$$v cos \theta -v_{CM} = v'_1 cos\Theta$$
$$v cos \theta = v'_1 cos\Theta +v_{CM}$$

we also know:
$$v \sin \theta = v'_1 \sin \Theta$$

dividing the two expressions, we get:

$$\tan \theta = \frac{sin\Theta}{\cos \Theta + \frac{m}{M}}$$

i was wondering this was the correct logic to solving this problem? I didnt use conservation of energy, and was wondering if there was something that i missed.

 

[Jhero]

Yes, your logic is correct. You have used conservation of momentum to derive the relation between \Theta and \theta. This is because in elastic scattering, the total momentum before and after the collision remains the same. However, you have not used conservation of energy, which is also important in this problem.

To incorporate conservation of energy, you can use the fact that the total kinetic energy before and after the collision must also remain the same. This means that:

\frac{1}{2}mv^2 = \frac{1}{2}m{v'_1}^2 + \frac{1}{2}M{v'_2}^2

Substituting the expressions for v'_1 and v'_2 from the CM frame, we get:

\frac{1}{2}mv^2 = \frac{1}{2}m{v'_1}^2 + \frac{1}{2}M{v_{CM}}^2

Solving for v'_1, we get:

v'_1 = \sqrt{v^2 - \frac{M^2}{(m+M)^2}v^2}

Using this expression for v'_1 in the relation you derived using conservation of momentum, we get:

\tan \theta = \frac{\sin \Theta}{\cos \Theta + \frac{m}{M}\sqrt{1-\frac{M^2}{(m+M)^2}\sin^2 \Theta}}

This takes into account conservation of both momentum and energy.