Help me 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

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

since, nucleus was initially at rest. Therefore,

velocity of nucleus = 0 m/s

and due to opposite direction of forces:

Vb = - Vb

Therefore,

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

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