(a) Using the conservation of momentum [tex]v_B=\frac{m_Av_A}{m_B}[/tex].
(b) The ratio of Kinetic energies of piece A to piece B is obtained as
[tex]\frac{KE_A}{KE_B}=\frac{m_B}{m_A}[/tex]
(a) According to the conservation of momentum;
The momentum before the collision will be equal to the momentum after the collision.
Let 'M' be the mass of the parent nucleus and [tex]v_i[/tex] be its velocity. So,
[tex]Mv_i=-m_Av_A+m_Bv_B[/tex]
(Left is taken to be negative by convention)
The atomic nucleus is considered to be at rest initially. So, the parent nucleus has a velocity,
[tex]v_i = 0\,m/s[/tex]
Therefore, the equation becomes;
[tex]0=-m_Av_A+m_Bv_B[/tex]
On rearranging the above equation we get;
[tex]v_B=\frac{m_Av_A}{m_B}[/tex]
(b) Kinetic energy of piece A,
[tex]KE_A=\frac{1}{2} m_Av_A^2[/tex]
The kinetic energy of piece B,
[tex]KE_B=\frac{1}{2} m_Bv_B^2[/tex]
But, we got that;
[tex]v_B=\frac{m_Av_A}{m_B}[/tex]
Substituting this value in the kinetic energy of B, we get;
[tex]KE_B=\frac{1}{2} m_B(\frac{m_Av_A}{m_B} )^2=\frac{1}{2} (\frac{m_A^2v_A^2}{m_B} )[/tex]
Therefore;
[tex]\frac{KE_A}{KE_B}=\frac{\frac{1}{2} m_Av_A^2}{\frac{1}{2} (\frac{m_A^2v_A^2}{m_B} )}=\frac{m_B}{m_A}[/tex]
If [tex]m_B>m_A[/tex] then [tex]KE_A>KE_B[/tex].
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