Question

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An atomic nucleus suddenly bursts apart (fission) into two pieces. Piece A with mass mA travels to the left with a speed of vA. Piece B with mass mB travels to the right with speed vB. Show the velocity of piece B in terms of mA, mB and VA.

Solution:

Consider that the nucleus is not acted by an external force. Thus, momentum is conserved, so: pBf + pAf = 0

Substituting the expression for momentum results to

mBv (_______) + (_________) mAv (______) = 0

Deriving the expression for the velocity of piece B results to

v(_____) =(mA/m ______)vA

Answer 1

Answer:

$v_{B} = \frac{m_{A}v_{A}}{m_{B}}$

Explanation:

We can apply the law of conservation of momentum on the nucleus in its initial and final state of nucleus:

Initial Momentum = Final Momentum

momentum of nucleus before bursting = momentum of piece A + momentum of Piece B

$(m_{nucleus})(velocity of nucleus) = m_{A}v_{A} + m_{B}v_{B}$

since, nucleus was initially at rest. Therefore,

velocity of nucleus = 0 m/s

and due to opposite direction of forces:

Vb = - Vb

Therefore,

$(m_{nucleus})(0) = m_{A}v_{A} - m_{B}v_{B}\\\\m_{A}v_{A} = m_{B}v_{B}\\\\v_{B} = \frac{m_{A}v_{A}}{m_{B}}$