We have the nuclear reaction
"_92^238"U"-> "_90^234"Th"+"_2^4alpha
Let v_alpha and v_"Th" be the velocities of the alpha particle and daughter nuclide respectively.
From conservation of momentum we get
0=m_"Th" xxv_"Th"+m_alphaxxv_alpha
m_"Th" xxv_"Th"=-m_alphaxxv_alpha
Notice that velocity of daughter nuclide is opposite to the velocity of alpha particle.
Squaring both sides
(m_"Th" xxv_"Th")^2=(-m_alphaxxv_alpha)^2
Remembering that Kinetic energy of a body KE=1/2mv^2 and rearranging we get
(m_alphav_alpha^2)/(m_"Th"v_"Th"^2)=m_"Th"/m_alpha
Multiply numerator and denominator of LHS by 1/2
(1/2m_alphav_alpha^2)/(1/2m_"Th"v_"Th"^2)=m_"Th"/m_alpha
=>(KE_alpha)/(KE_"Th")=m_"Th"/m_alpha
Inserting value of masses on the RHS we get
(KE_alpha)/(KE_"Th")=234/4=58.5
