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postnew [5]
3 years ago
11

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

tes into three particles. One particle has a mass of 5.0010-27kg and velocity of 6.00106 jm/s, and a second particle has a mass of 8.4010-27kg and velocity of 4.00106 im/s. Calculate (a) the velocity of the third particle and (b) the total increase in kinetic energy as a result of the decay. (Note that this kinetic energy actually comes from the mass of the original nucleus, according to E = ∆mc2, but only a tiny amount of mass is turned into energy, so we don’t have to worry about it for this problem.)
Physics
1 answer:
slega [8]3 years ago
8 0

Answer:

Part a)

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

E = 4.4 \times 10^{-13} J

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

m_1v_1 + m_2v_2 + m_3v_3 = 0

(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0

(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

By equation of kinetic energy we have

E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2

E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}

E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}

E = 4.4 \times 10^{-13} J

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