In Fig. 9-21, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{64}\mathbf{.}\mathbf{0}\mathbf{°}$and the oxygen nucleus recoils with speed localid="1661254577203" $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and at angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{51}\mathbf{.}\mathbf{0}\mathbf{°}$.In atomic mass units, the mass of the alpha particle is 4.00 uand the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle?

Mass of particle 2 is,$m_2=16amu$

Mass of particle 1 is,$m_2=4amu$

Velocity of particle 2 is,$v_2=1.2\times {10}^{5}m/s$

Velocity of particle 1 is,$v_1=4.15\times {10}^{5}m/s$

Scattering angle${\theta }_2=51°$

Scattering angle${\theta }_1=64°$

Consider particle 1 as alpha particle and particle 2 as oxygen. When they scatter, they may conserve energy as well as momentum. By using the linear momentum equation, find the solution for the initial and final velocities.According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formula is as follow:

Initial momentum = Final momentum

Conservation of linear momentum can be expressed as,

$$\begin{gathered} m_1{v}_{1initial}=m_1{v}_{1final}\cos {\theta }_1+m_2{v}_{2final}\cos {\theta }_2...........\left(1\right) \\ 0=m_1{v}_{1final}\sin {\theta }_1-m_2{v}_{2final}\sin {\theta }_2...........\left(2\right) \end{gathered}$$

From equation (2), to calculatethefinal speed of alpha particle,

$v_{1final}=\frac{m_2{v}_{2final}\sin {\theta }_2}{m_1\sin {\theta }_1}$

From equation (1), to calculate the initial speed of alpha particle,

$v_{1initial}=\frac{m_1{v}_{1final}\cos {\theta }_1+m_2{v}_{2final}\cos {\theta }_2}{m_1}$

Calculating final speed of a alpha particle,

$$\begin{gathered} v_{1final}=\frac{m_2{v}_{2final}\sin {\theta }_2}{m_1\sin {\theta }_1} \\ {v}_{1final}=\frac{\left(16\right)\left(1.2\times {10}^5\right)\sin 51°}{\left(4\right)\sin 64°} \\ {v}_{1final}=4.15\times {10}^{5}m/s \end{gathered}$$

Hence, final speed of alpha particle 1 is $4.15\times {10}^{5}m/s$.

Calculating initial speed of a alpha particle,

$$\begin{gathered} v_{1initial}=\frac{m_1{v}_{1final}\cos {\theta }_1+m_2{v}_{2final}\cos {\theta }_2}{m_1} \\ {v}_{1initial}=\frac{\left(4\right)\times \left(4.15\times {10}^5\right)\cos 64°+\left(16\right)\left(1.2\times {10}^5\right)\cos 51°}{4} \\ {v}_{1initial}=4.84\times {10}^{5}m/s \end{gathered}$$

Hence,initial speed of particle 1 is$4.84\times {10}^{5}m/s$.

Therefore, using conservation of linear momentum for two particles scattered together, its initial and final velocity can be found.