momentum is conserved in a collision of two objects as measured by an observer on a uniformly moving train. show that momentum is also conserved for a ground observed
Momentum conserving in the train reference system:
"\\overrightarrow{v_1}+\\overrightarrow{v_2}=\\overrightarrow{u_1}+\\overrightarrow{u_2},~~~(1)"
where "\\overrightarrow{v_1},~\\overrightarrow{v_2}" - velocities of the first and second object before collision, "\\overrightarrow{u_1},~\\overrightarrow{u_2}" - after collision.
In the ground reference system each of those velocities will be increased with velocity of train "v":
"\\overrightarrow{v_1'}=\\overrightarrow{v_1}+\\overrightarrow{v}\\Rarr \\overrightarrow{v_1}=\\overrightarrow{v_1'}-\\overrightarrow{v},\\\\\n\\overrightarrow{v_2'}=\\overrightarrow{v_2}+\\overrightarrow{v}\\Rarr \\overrightarrow{v_2}=\\overrightarrow{v_2'}-\\overrightarrow{v},\\\\\n\\overrightarrow{u_1'}=\\overrightarrow{u_1}+\\overrightarrow{v}\\Rarr \\overrightarrow{u_1}=\\overrightarrow{u_1'}-\\overrightarrow{v},\\\\\n\\overrightarrow{u_2'}=\\overrightarrow{u_2}+\\overrightarrow{v}\\Rarr \\overrightarrow{u_2}=\\overrightarrow{u_2'}-\\overrightarrow{v}"
Let's insert this values into the equation (1):
"(\\overrightarrow{v_1'}-\\overrightarrow{v})+(\\overrightarrow{v_2'}-\\overrightarrow{v})=(\\overrightarrow{u_1'}-\\overrightarrow{v})+(\\overrightarrow{u_2'}-\\overrightarrow{v})\\Rarr\\\\\n\\Rarr\\overrightarrow{v_1'}+\\overrightarrow{v_2'}=\\overrightarrow{u_1'}+\\overrightarrow{u_2'},"
so, we can see that momentum is also conserved in the ground reference system.
Comments
Leave a comment