## scattering

$$\alpha$$ 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.
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