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
v1=2.52e6
v2 = ?
| velocity of the second helium (magnitude) | m/s |
| Incorrect Answer = -0.4599e6 m/s |
(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
v1= 2.52e6
v2 = ?
|
velocity of the second helium (magnitude) Incorrect Answer = -0.4599e6 m/s |
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An atom of beryllium (m = 8.00 u) splits into two atoms
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An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m =...
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