Question

In a nuclear collision, an alpha particle A of mass 4 unit is incident with velocity v on a stationary helium nucleus B of 4 mass unit. After collision, A moves in the direction BC with velocity v/2, where BC makes angle 60 with initial direction AB and the helium nucleus moves along BD. Calculate the velocity of rebound of the helium nucleus along BD and angle made with the direction AB.

Answer 1

common brother :P .. ALLTEHMAFFS ...

Answer 2

conservation of momentum has to apply in both directions.

\[ p_{ix}=p_{yx}\]
\[m_{A}v_{Aix}+\cancel{m_{B}v_{Bix}}^0=m_{A}v_{Afx}+m_{B}v_{Bfx}\]
\[m_{Ai}v_{Ai}=m_{A}v_{Aix}-m_{B}v_{Bx} \]
\[m_A=m_B \quad ; \quad v_{Aix} = v \quad ; \quad v_{Bfx}=\frac{v}{2}\cos60=\frac{v}{4}\]
\[v_{Afx}=v-\frac{v}{4}\]

Then the same setup for y, except there's no initial momentum in the y direction, so

\[v_{Afy}=-v_{Bfy}\]
\[v_{Afy}=v \sin 60º=-v\frac{\sqrt{3}}{2}\]

Now you know the x and y components of vA, so tan phi = Vfay/Vfax

^_^