- #1
alizeid
- 14
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A particle of mass m comes with the speed 0.6c and collides with another particle of mass m which is at rest. In the collision melts the particles together and form a particle. What is the mass and velocity of the particle is formed?
solution:
The momentum and the total energy is conserved
[tex]E_{before}=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}+mc^2,v=0.6c \\ \Rightarrow E_{before}=2.25mc^2 \\ [/tex]
[tex]E_{after}=\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}\\ [/tex] , where xm is the new mass and z is the velocity
Now, the energy is conserved
[tex]\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}=2.25mc^2\Leftrightarrow X^2=2.25^2(1-\frac{z^2}{c^2})\\ [/tex]
Now we do the same for momentum:
[tex]P_{befor}=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}, v=0.6c\Rightarrow P_{befor}=0,75mc[/tex]
momentum after the collision:
[tex]P_{after}=\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}[/tex]
the momentum is conserved:
[tex]\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}=0.75mc\Leftrightarrow xz=0.75c\cdot\sqrt{1-\frac{z^2}{c^2}}\\ \Leftrightarrow x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})[/tex]
we have from the first equation that [tex]x^2=2.25^2\cdot(1-\frac{z^2}{c^2})[/tex]:
[tex] x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ \Leftrightarrow 2.25^2\cdot(1-\frac{z^2}{c^2})z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ z^2=\frac{0.75^2}{2.25^2}c^2\\ \Leftrightarrow z=0.33c[/tex]
Now we have z, and then we can solve X. I get x = 2.12. Thus the mass to be 2.12m and the speed 0.33c. Is this true? Thanks for the help
solution:
The momentum and the total energy is conserved
[tex]E_{before}=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}+mc^2,v=0.6c \\ \Rightarrow E_{before}=2.25mc^2 \\ [/tex]
[tex]E_{after}=\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}\\ [/tex] , where xm is the new mass and z is the velocity
Now, the energy is conserved
[tex]\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}=2.25mc^2\Leftrightarrow X^2=2.25^2(1-\frac{z^2}{c^2})\\ [/tex]
Now we do the same for momentum:
[tex]P_{befor}=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}, v=0.6c\Rightarrow P_{befor}=0,75mc[/tex]
momentum after the collision:
[tex]P_{after}=\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}[/tex]
the momentum is conserved:
[tex]\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}=0.75mc\Leftrightarrow xz=0.75c\cdot\sqrt{1-\frac{z^2}{c^2}}\\ \Leftrightarrow x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})[/tex]
we have from the first equation that [tex]x^2=2.25^2\cdot(1-\frac{z^2}{c^2})[/tex]:
[tex] x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ \Leftrightarrow 2.25^2\cdot(1-\frac{z^2}{c^2})z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ z^2=\frac{0.75^2}{2.25^2}c^2\\ \Leftrightarrow z=0.33c[/tex]
Now we have z, and then we can solve X. I get x = 2.12. Thus the mass to be 2.12m and the speed 0.33c. Is this true? Thanks for the help