Momentum of a particular particle...?

Question

http://i22.photobucket.com/albums/b338/ravenatic520/Acertainnucleus.jpg
A certain nucleus, at rest, transforms into three particles. Two of them are detected; their masses and velocities are shown in the above link. (a) In unit-vector notation, what is the translational momentum of the third particle, with a mass of 11.7*10^-27 kg?

Answer 1

Here, you have to use conservation of momentum to solve the problem. Since the initial nucleus is at rest before explosion, the sum of the momentums in all directions must equal 0 as well.

Momentum of mass 1: (16.7E-27)(6E6) = 1E-19 kg*m/s
(+x direction)

Momentum of mass 2: (8.35E-27)(-8E6) = -6.68E-20 kg*m/s
(-y direction)

Now you can solve the momentum of mass 3:

x-direction: momentum (mass 1) + momentum (mass 3 in x) = 0
1E-19 kg*m/s + (11.7E-27)(v_x_3) = 0

Therefore, v_x_3 = -8.55E6 m/s (-x direction)

y-direction: momentum (mass 2) + momentum (mass 3 in y) = 0
-6.68E-20 kg*m/s + (11.7E-27)(v_y_3) = 0

Therefore, v_y_3 = 5.71E6 m/s (y-direction)

In unit vector notation, the transitional momentum of mass 3 is:

11.7E-27(-8.55E6 m/s [x] + 5.71E6 [y])

Hope this helps

Answer 2

According to Heisenbergs uncertain ty princinple you can not measure the momentum of a particle. Any atttempt of measurement no matter how accurate and precise the experiment is meant to be =The measurement is just a lucky Guess.
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