1) Applying momentum conservation,
(234 x 0) + (1 x 2.5 x 10^8)i = (231 x 9.2 x 10^5)(cos(30)i +
sin30)j) + (4 v)(cos(theta)i + sin(theta)j)
along j,
(231 x 9.2 x 10^5)sin30 + (4 v sin(theta)) = 90
4 v sin(theta) = - 1.063 x 10^8 ... (i)
along i,
2.5 x 10^8 = 1.841x 10^8 + 4 v cos(theta)
4 v cos(theta) = 0.659 x 10^8 .... (ii)
(i)/(ii) = tan(theta) = -1.613
theta = - 58 deg .....Ans
v = 0.312 x 10^8 m/s
Introductory Physics I Group Problem 1) A Uranium-234 nucleus is initially stationary, but it is hit...
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A uranium-238 atom can break up into a thorium-234atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10−27 kg). When a uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.2 × 105 m/s. Please give the answer in Joules
A uranium-238 atom can break up into a thorium-234atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.1 × 105 m/s. How much of the uranium atom's internal energy is released in the breakup?...
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