Question

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

Answer 1

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 $v$:

$\overrightarrow{v_1'}=\overrightarrow{v_1}+\overrightarrow{v}\Rarr \overrightarrow{v_1}=\overrightarrow{v_1'}-\overrightarrow{v},\\ \overrightarrow{v_2'}=\overrightarrow{v_2}+\overrightarrow{v}\Rarr \overrightarrow{v_2}=\overrightarrow{v_2'}-\overrightarrow{v},\\ \overrightarrow{u_1'}=\overrightarrow{u_1}+\overrightarrow{v}\Rarr \overrightarrow{u_1}=\overrightarrow{u_1'}-\overrightarrow{v},\\ \overrightarrow{u_2'}=\overrightarrow{u_2}+\overrightarrow{v}\Rarr \overrightarrow{u_2}=\overrightarrow{u_2'}-\overrightarrow{v}$

Let's insert this values into the equation (1):

$(\overrightarrow{v_1'}-\overrightarrow{v})+(\overrightarrow{v_2'}-\overrightarrow{v})=(\overrightarrow{u_1'}-\overrightarrow{v})+(\overrightarrow{u_2'}-\overrightarrow{v})\Rarr\\ \Rarr\overrightarrow{v_1'}+\overrightarrow{v_2'}=\overrightarrow{u_1'}+\overrightarrow{u_2'},$

so, we can see that momentum is also conserved in the ground reference system.