What is the Kinetic Energy of the Daughter Nucleus?

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The discussion revolves around a radioactive decay process where a parent nucleus transforms into a daughter nucleus by emitting an electron and a neutrino. The problem involves calculating the linear momentum of the daughter nucleus and its kinetic energy, given the momenta of the emitted particles.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and momentum, with some attempting to derive kinetic energy from momentum using formulas. Questions arise regarding the correct application of these relationships.

Discussion Status

Participants have explored the connection between momentum and kinetic energy, with one providing a formula derived from the definitions. There is acknowledgment of a successful understanding of the relationship, although the exact calculations remain unverified.

Contextual Notes

There is a focus on ensuring the correct application of physics formulas, particularly in the context of kinetic energy and momentum, while some participants express uncertainty about their calculations.

mb85
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A certain radioactive (parent) nucleus transforms to a different (daughter) nucleus by emitting an electron and a neutrino. The parent was at rest at the origin of an xy coordinate system. The electron moves away from the origin with linear momentum (-3.3 x 10-22 kg m/s) ; the neutrino moves away from the origin with linear momentum (-2.2 x 10-23 kg m/s) . What are (a) the magnitude and (b) angle (from the +x axis) of the linear momentum of the daughter nucleus? (c) If the daughter nucleus has a mass of 1.8 x 10-26 kg, what is its kinetic energy?

i know part a and b i did are correct:
part a. i got 3.3x10^-22 kg m/s
part b the angle is 3.81 degrees

but i can't figure out part c? what formula am i supposed to use.
 
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Do you know how to write down the kinectic energy of a particle in terms of its momentum?
 
p = mv?

the KE = (m -mi)V^2

?
 
mb85 said:
p = mv?

the KE = (m -mi)V^2

?
K.E = 1/2 mv^2 . Multiply and divide RHS by m. K.E = (mv)^2/2m
 
neutrino said:
K.E = 1/2 mv^2 . Multiply and divide RHS by m. K.E = (mv)^2/2m

what is RHS?
K.E = (mv)^2/2m
i get 9.8x10-70...but that doesn't look right.
 
mb85 said:
what is RHS?
Right hand side.

You start with K.E. = \frac{1}{2}mv^2

Then multiply both sides by m, but to keep them the same you must divide by m, i.e.:
K.E. = \frac{1}{2}mv^2 . \frac{m}{m}
K.E. = \frac{m^2 v^2}{2m}
K.E. = \frac{(mv)^2}{2m}
Then use the fact that p = mv to get:
K.E. = \frac{p^2}{2m}

Can you use that formula to get the answer?
 
thanks. i got it now. :blushing:
 

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