Question

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment

Answer 1

Answer: Speed =
`3.10^(-31)` m/s

Step-by-step explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

`p_(f) = p_(i)`

Relativistic momentum is calculated as:

p =
`\frac{mu}{\sqrt{1-(u^(2))/(c^(2)) } }`

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=
`3.10^(8)`m/s)

Initial momentum is zero, then:

`p_(f)` = 0

`p_(1)-p_(2)` = 0

`p_(1) = p_(2)`

To find speed of the heavier fragment:

`\frac{mu_(1)}{\sqrt{1-(u^(2)_(1))/(c^(2)) } }=\frac{mu_(2)}{\sqrt{1-(u^(2)_(2))/(c^(2)) } }`

`\frac{1.86.10^(-27)u_(1)}{\sqrt{1-(u^(2)_(1))/((3.10^(8))^(2)) } }=\frac{3.10^(-28).0.844.3.10^(8)}{\sqrt{1-((0.844c)^(2))/(c^(2)) } }`

`\frac{1.86.10^(-27)u_(1)}{\sqrt{1-(u^(2)_(1))/((3.10^(8))^(2)) } }=1.42.10^(-19)`

`1.86.10^(-27)u_(1) = 1.42.10^(-19).{\sqrt{1-(u^(2)_(1))/((3.10^(8))^(2)) } }`

`(1.86.10^(-27)u_(1))^(2) = (1.42.10^(-19).{\sqrt{1-(u^(2)_(1))/((3.10^(8))^(2)) } })^(2)`

`3.46.10^(-54).u_(1)^(2) = 2.02.10^(-38).(1-(u_(1)^(2))/(9.10^(16)) )`

`3.46.10^(-54).u_(1)^(2) = 2.02.10^(-38) -[2.02.10^(-38)((u_(1)^(2))/(9.10^(16)) )]`

`3.46.10^(-54).u_(1)^(2) = 2.02.10^(-38) -2.24.10^(-23).u^(2)_(1)`

`3.46.10^(-54).u_(1)^(2)+2.24.10^(-23).u^(2)_(1) = 2.02.10^(-38)`

`2.24.10^(-23).u^(2)_(1) = 2.02.10^(-38)`

`u^(2)_(1) = (2.02.10^(-38))/(2.24.10^(-23))`

`u_(1) = \sqrt{9.02.10^(-62)}`

`u_(1) = 3.10^(-31)`

The speed of the heavier fragment is
`u_(1) = 3.10^(-31)`m/s.