Source: High school physics (Chinese)
Problem Sets:
Problem
An $\alpha$ particle collides with an initially stationary oxygen nucleus. After the collision, the $\alpha$ particle is scattered at an angle of $64^\circ$ with its initial path. The oxygen nucleus recoils at an angle of $51^\circ$ on the other side of the initial path. The mass of the oxygen nucleus is four times that of the $\alpha$ particle.
Let $m$ be the mass of the $\alpha$ particle, so the mass of the oxygen nucleus is $4m$. Let the initial direction of the $\alpha$ particle be the x-axis. The initial momentum of the system in the y-direction is zero. Since there are no external forces, the total momentum is conserved. We can apply conservation of momentum for the y-component.
Let $v_\alpha$ and $v_O$ be the final speeds of the $\alpha$ particle and the oxygen nucleus, respectively. The y-component of the final momentum is:
$$p_{fy} = m v_\alpha \sin(64^\circ) + (4m) v_O \sin(-51^\circ) = m v_\alpha \sin(64^\circ) - 4m v_O \sin(51^\circ)$$By conservation of momentum in the y-direction, $p_{fy} = p_{iy} = 0$:
$$m v_\alpha \sin(64^\circ) - 4m v_O \sin(51^\circ) = 0$$ $$v_\alpha \sin(64^\circ) = 4 v_O \sin(51^\circ)$$The ratio of the speeds is:
$$\frac{v_\alpha}{v_O} = \frac{4 \sin(51^\circ)}{\sin(64^\circ)}$$