# Scattering angle in the CM frame

by indigojoker
Tags: scattering
 P: 247 $$\alpha$$ 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.