Momentum: Nucleus decays into 3

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The discussion revolves around calculating the momentum of a third particle resulting from the decay of a nucleus at rest into three particles. The user initially attempted to break down the momentum of the first two particles into components but found it incorrect. It is confirmed that the total momentum should equal zero since the nucleus started from rest. Participants suggest summing the x and y components of the momenta of the first two particles and setting them to zero to find the momentum of the third particle. The conversation emphasizes the importance of correctly applying momentum conservation principles in this scenario.
sweetpete28
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Can someone please help with this one? I'm not sure what to do here. I broke momentum for particles one and two into components (found m1v1cosθ1 + m2v2cosθ2 and m1v1sinθ1 + m2v2sinθ2) but this is not correct and I really don't know what to do. Please help!


A certain nucleus at rest suddenly decays into three particles, two of which are charged and can be easily detected. The data gathered for these two particles is:

Particle 1 has mass m1 = 3.00 x 10^-20 kg moving at speed v1 = 5.25 x 10^5 m/s at θ1 = 51.9o.

Particle 2 has mass m2 = 2.11 x 10^-25 kg moving at speed v2 = 1.10 x 10^7 m/s at θ2 = 220o.

Find p3, the momentum of the third piece in i-j notation
 
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Can anyone help? Since it started from rest should the total momentum be 0?
 
sweetpete28 said:
Can anyone help? Since it started from rest should the total momentum be 0?

Yes. Write up the sum of all the x and y components in separate equations and equate them to zero.

ehild
 
Thank you!
 
sweetpete28 said:
Thank you!

You are welcome. Have you got the correct solution?


ehild
 
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