Question

n atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece A, of mass mA , travels off to the left with speed vA . Piece B, of mass mB , travels off to the right with speed vB . (a) Use conservation of momentum to solve for vB in terms of mA , mB , and vA . (b) Use the results of part (a) to show that KA >KB

Answer 1

Explanation:

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Answer 2

(a) Using the conservation of momentum $v_B=\frac{m_Av_A}{m_B}$.

(b) The ratio of Kinetic energies of piece A to piece B is obtained as

$\frac{KE_A}{KE_B}=\frac{m_B}{m_A}$

Conservation of Momentum

(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 $v_i$ be its velocity. So,

$Mv_i=-m_Av_A+m_Bv_B$

(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,

$v_i = 0\,m/s$

Therefore, the equation becomes;

$0=-m_Av_A+m_Bv_B$

On rearranging the above equation we get;

$v_B=\frac{m_Av_A}{m_B}$

(b) Kinetic energy of piece A,

$KE_A=\frac{1}{2} m_Av_A^2$

The kinetic energy of piece B,

$KE_B=\frac{1}{2} m_Bv_B^2$

But, we got that;

$v_B=\frac{m_Av_A}{m_B}$

Substituting this value in the kinetic energy of B, we get;

$KE_B=\frac{1}{2} m_B(\frac{m_Av_A}{m_B} )^2=\frac{1}{2} (\frac{m_A^2v_A^2}{m_B} )$

Therefore;

$\frac{KE_A}{KE_B}=\frac{\frac{1}{2} m_Av_A^2}{\frac{1}{2} (\frac{m_A^2v_A^2}{m_B} )}=\frac{m_B}{m_A}$

If $m_B>m_A$ then $KE_A>KE_B$.

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