Question

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 10−28 kg and that of the heavier fragment is 1.62 ✕ 10−27 kg. If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?

Answer 1

The speed of the heavier fragment is 0.335c.

Step-by-step explanation:

Given that,

Mass of the lighter fragment
`M_(l)=2.90*10^(-28)\ kg`

Mass of the heavier fragment
`M_(h)=1.62*10^(-27)\ Kg`

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

`0=\drac{m_(1)v_(1)}{\sqrt{1-(v_(1)^2)/(c^2)}}-\drac{m_(2)v_(2)}{\sqrt{1-(v_(1)^2)/(c^2)}}`

`\drac{m_(1)v_(1)}{\sqrt{1-(v_(1)^2)/(c^2)}}=\drac{m_(2)v_(2)}{\sqrt{1-(v_(1)^2)/(c^2)}}`

`(2.90*10^(-28)*0.893c)/(√(1-(0.893)^2))=\frac{1.62*10^(-27)v_(2)}{\sqrt{1-(v_(2)^2)/(c^2)}}`

`\frac{v_(2)}{\sqrt{1-(v_(2)^2)/(c^2)}}=(2.90*10^(-28)*0.893c)/(1.62*10^(-27)*0.45)`

`\frac{v_(2)}{\sqrt{1-(v_(2)^2)/(c^2)}}=0.355c`

`(v_(2))/(1-(v_(2)^(2))/(c^2))=(0.355c)^2`

`(1-(v_(2)^2)/(c^2))/(v_(2)^2)=(1)/((0.355c))`

`(1)/(v_(2)^2)-(1)/(c^2)=(1)/((0.355c)^2)`

`(1)/(v_(2)^2)=(1)/(c^2)+(1)/(0.126c^2)`

`(1)/(v_(2)^2)=(1)/(c^2)(1+(1)/(0.126))`

`(1)/(v_(2)^2)=(8.93)/(c^2)`

`v_(2)^2=(c^2)/(8.93)`

`v_(2)=0.335c`

Hence, The speed of the heavier fragment is 0.335c.