Respuesta :
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment [tex]M_{l}=2.90\times10^{-28}\ kg[/tex]
Mass of the heavier fragment [tex]M_{h}=1.62\times10^{-27}\ Kg[/tex]
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
[tex]0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}[/tex]
[tex]\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}[/tex]
[tex]\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}[/tex]
[tex]\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}[/tex]
[tex]\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c[/tex]
[tex]\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2[/tex]
[tex]\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}[/tex]
[tex]\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}[/tex]
[tex]\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}[/tex]
[tex]\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})[/tex]
[tex]\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}[/tex]
[tex]v_{2}^2=\dfrac{c^2}{8.93}[/tex]
[tex]v_{2}=0.335c[/tex]
Hence, The speed of the heavier fragment is 0.335c.
