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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.

Sagot :

The velocity of the piece b is expressed as vb = (ma×va)/mb.

The Law of Momentum conservation states that the Momentum of particle before collision will be equal to momentum of particles after collision. If we assume that the mass of parent nuclei is M and its velocity is V, then according to law of momentum conservation

MV = - ma×va + mb×vb ( we use negative sign because part a travels left)

Now, the initial velocity of parent nuclei is zero because it was at rest that is V = 0. So, the equation will be

0 = - ma×va + mb×vb

ma×va = mb×vb

From this we get

vb = (ma×va)/mb.

Learn more about Momentum conservation at:

brainly.com/question/7538238

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