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Problem 22 Easy Difficulty

An atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece $A,$ of mass $m_{A},$ travels off to the left with speed $v_{A}$ . Piece $B$ , of mass $m_{B}$ , travels off to the right with speed $v_{B}$ . (a) Use conservation of momenturn to solve for $v_{B}$ in terms of $m_{A}, m_{B},$ and
$v_{A}-(b)$ Use the results of part (a) to show that $K_{A} / K_{B}=m_{B} / m_{A}$ where $K_{\mathrm{A}}$ and $K_{B}$ are the kinetic energies of the two pieces.

Answer

(a) $v_{B}=\frac{m_{A}}{m_{B}} v_{A}$
(b) First, we write the kinetic energies $K_{A}$ and $K_{B}$ in terms of the mass and speed, substituting for $v_{B}$ from the result of part (a) into $K_{B}$ then using the result, we divide $K_{A}$ by $K_{B}$ so we get $K_{A} / K_{B}=m_{B} / m_{A}$

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Video Transcript

in this question, an atomic nucleus bursts apart and fishing's into two pieces. One piece moves to the right with velocity V B and Mass M B. Another piece moved to the left with velocity V A and Mass M A. In the first item we have to solve for VB in terms off M B M a N v A. For that, we have to use the law of conservation of Momentum. So the net momentum let me call this Puy en before deficient. So P m zero is equals to the net momentum after deficient. So you let me call this Onley. PM. The net momentum before deficient is equal to zero because the atom is just standing still after deficient. The momentum is given by three contributions, one coming from this portion off the atom and another coming from this other person off the atom. Then, in order to properly write this down, we have to choose a reference frame. I'll choose a reference frame that it's pointing to the right so vb it's positive and V A is negative. Then the net momentum after deficient is the following M b times vb minus m. A times v a. Now we solve this equation for VB. We can do that as follows. M a v A is equals to M B v B. The reform VB is given by m a times V eight divided by M. B. And this is the answer to the first item off this question. Now, for the second item, we have to evaluate the ratio between the kinetic energies off the portion A and B off the nucleus on show that it is equals to be divided by M. A. So for that we use the definition off kinetic energy, which is this one. So that ratio is given by the following the kinetic energy off the portion A so one half m a times V a squared divided by one half and me times V b squared. Now, as you can see, we can simplify the factors off one half and we end up with the following That racial is given by M A V a squared divided by m b v b squared. Now we use are less result and substitute vb by this expression that we calculated before by doing that to get the following this is equals to m a Times va squared divided by M b times v B squared. But VB is m a times v a divided by m b squared. Then we get em a times V a squared divided by m b times am a squared times V a squared divided by m b squared. Now we simplify whatever it can so we can simplify v a squared with V a squared. We can also council one factor off m A here and one factor off M b here the reform These results in the following this is one divided by M A, divided by M b, which is equals to m B, divided by M. A as we wanted to show on. This is answer to the second item off this question.

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