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#