Weird Blocks

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After the blocks come in contact the net force would be in the direction of the $10 N$ force. So, the net force will be $(10-5)N = 5N$ .

After doing some geometry, you will find that the normal reaction on the block on the left hand side by the block on the right hand side makes an angle of $60°$ .After both the blocks come in contact we can treat the blocks as one system so the acceleration of both the blocks becomes $\frac{ { F }_{ net }}{m_{in motion}} = \frac{10-5}{1+1} = \frac{5}{2}$ As the net force on the block on the left hand side is $5N$ in the direction of the $10N$ force hence, we get the equation as follows

$Ncos60° - 5 = ma$

$\Longrightarrow\ Ncos60° - 5 = \frac{5}{2}$

$\Longrightarrow\ Ncos60° = \frac{15}{2}$

$\Longrightarrow\ \frac{N}{2} = \frac{15}{2}$

$\Longrightarrow\ N = \boxed{15}$

Similarly for the other block, the equation formed will be

$10 - Ncos60° = \frac{5}{2}$

And on solving it we will get

$N = \boxed{15}$