Question

1. An unstable particle with mass 2.81 x 10-27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components u, 0.987c and u, --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 γ for the two fragments after the decay. (d) Using one of the analysis models in part (b), find a relationship between the masses m, and m, of the fragments. (e) Using the second analysis model in part (b), find a second relationship between the masses m, and m2 (f) Solve the relationships in parts (d) and (e) simultaneously for the masses m, and m,.

Answer 1

(A) There is not any force is acting on the particles so the system is isolated. (b) For the analysis of this situation we can use 1. law of conservation of momentum and 2. law of conservation of energy. (c) gamma_1 = 1/squareroot 1 - u^2_1/c^2 here u_1 = 0.987 c, put this value in above equation we will get gamma_1 = 6.22 similarly gamma_2 = 1/squareroot 1 - u^2_2/c_2 u_2 = -0.898c we get gamma_2 = 2.27 \r\n (d) By using conservation of momentum Total momentum before decay = Total momentum after decay Here momentum before decay = 0[initially particle at rest] momentum after decay = m_1 u_1 + m_2 u_2 so 0 = m_1 u_1 + m_2 u_2 = > m_1 u_1 = -m_2 u_2 [m_1/m_2 = -u_2/u_1] here u_2 = -0.898 c u_1 = 0.987 c so m_1/m_2 = -(-0.898 c)/0.987 c = 0.9098 [m_1 = 0.91 m_2] approximately where m_1 = r_1 m_10 m_2 = r_2 m_20 m_1 = 6.22 m_10 m_2

At a time,4 parts can be solved.Still these will help you in others too.Comment regarding any query .Thanks...

Answer 2

(a) There is not any force is acting on the particles so the system is isolated.

(b) For the analysis of this situation we can use 1. law of conservation of momentum and 2. law of conservation of energy.

(c)  

  $\gamma_1=\frac{1}{\sqrt{1-\frac{u_1^2}{c^2}}}$

here u_1=0.987 c , put this value in above equation we will get

$\gamma_1=6.22$

similarly

$\gamma_2=\frac{1}{\sqrt{1-\frac{u_2^2}{c^2}}}$

u_2=-0.898c

we get

$\gamma_2=2.27$   

(d)