Chapter 39, Problem 62. An unstable particle with mass m = 2.81 * 10 ^ -27 kg is

Question

Chapter 39, Problem 62. An unstable particle with mass m = 2.81 * 10 ^ -27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components u1 = 0.987c and u2 = -0.898c. From this information, we wish to determine the masses of fragments 1 and 2. (a) Is the initial system of the unstable particle, which becomes the system of the two fragments, isolated or nonisolated? [i.e. Are there external forces on the system?] (b) Based on your answer to part (a), what two analysis models are appropriate for this situation? [i.e. What two conservation laws are appropriate?] (c) Find the values of y for the two fragments after the decay. (d) Using one of the analysis models in part (b), find a relationship between the masses m1 and m2 of the fragments. (e) Using the second analysis model in part (b), find a second relationship between the masses m1 and m2. (f) Solve the relationships in parts (d) and (e) simultaneously for the masses m1 and m2.

Explanation / Answer

part a) isolated

part b )

Isolated system: conservation of energy, conservation of momentum.

part c )

y = 1/sqrt(1-(v/c)^2))

v = 0.987c

y1 = 1/sqrt(1-(0.987)^2) = 6.22

y2 = 1/sqrt(1-(0.898)^2) = 2.27

part d )

conservation energy gives E1 + E2 = Etotal

y1m1c^2 + y2m2c^2 = mc^2

6.22m1+ 2.27m2 = 2.81 x 10^-27 kg ...(1)

part e )

since the final momentum = 0

p1 = p2

y1m1u1 = y2m2u2

6.22 * 0.987c * m1 = 2.27 * 0.898c*m2

m2 = 3.012m1

put this in eq (1)

6.22m1 + 3.012m1 = 2.81 x 10^-27

m1 = 3.044 x 10^-28 kg

m2 = 9.168 x 10^-28 kg

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