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An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m =...

An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 keV of energy. Suppose the beryllium atom moved in the positive x direction and had a kinetic energy of 44.0 keV. One of the helium atoms is found to be moving in the positive x direction. Find the direction of motion of the second helium, and find the velocity of each of the two helium atoms. Solve this problem in two different ways: by direct application of conservation of momentum and energy; and by applying the results if the original beryllium atom is at rest to a frame of reference moving with the original beryllium atom and then switching to the reference frame in which the beryllium is moving.

(a) by direct application of conservation of momentum and energy

velocity of the first helium (magnitude)     m/s
velocity of the second helium (magnitude)     m/s
direction of motion of the second helium    


(b) by applying the results if the original beryllium atom is at rest to a frame of reference moving with the original beryllium atom and then switching to the reference frame in which the beryllium is moving

velocity of the first helium (magnitude)     m/s
velocity of the second helium (magnitude)     m/s
direction of motion of the second helium  

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Expert Solution

genius_generous answered 1 year ago
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