Conservation of $\vec{p}$ in Ballistic pendulum So I was thinking about this really simple situation and then I came across a little question. The situation is as follows:
There is a block of wood suspended from a ceiling like support with two strings. The strings are perfectly rigid and light with no friction between various parts of the string. .A bullet is shot horizontally and gets embedded in the block
My question why can we conserve momentum in the horizontal direction in this scenario when there is a component of tension acting in the horizontal direction which is an external force force.
Why do we consider that the bullet block system has a horizontal velocity at max height?
I get that $\frac{(M+m)v^2}{r}=(M+m)g$ And $\vec{F_{net}}=0$ at this point but there is still a velocity component perpendicular to the direction of the string which means the system will rotate about point of suspension further and has a greater height achievable which is a contradiction
In conclusion shouldn't the equation for max height be
$1/2mv^2=(M+m)gh_{max}$
rather than
$1/2\mu v^2=(M+m)gh_{max}$
where $\mu$ is the reduced mass of the system
Please correct my conceptual misunderstandings and shed some light on situation.
Thanks in advance.
 A: In addition to @Bhavay's nice answer I think it's useful to look a just how much energy is lost to permanent deformation, friction, sound, heat and other non-conservables.
As the collision is inelastic, conservation of energy doesn't apply. But without external forces conservation of momentum does:
$$mv_0=(m+M)v_1$$
where $v_0$ is the bullet's velocity and $v_1$ the velocity of bullet and block, immediately after the collision.
$$v_1=\frac{m}{m+M}v_0$$
Now, post-collision, the block plus bullet conserves energy, so that:
$$\frac12 (m+M)v_1^2=(m+M)gh_{max}$$
$$h_{max}=\frac{v_1^2}{2g}$$
$$h_{max}=\Big(\frac{m}{m+M}\Big)^2\frac{v_0^2}{2g}$$
Now assume (erroneously!) that energy was conserved anyway, so that:
$$\frac12mv_0^2=(m+M)gh_{max,2}$$
$$h_{max,2}=\frac{m}{m+M}\frac{v_0^2}{2g}$$
So that $h_{max}$ is a fraction of $h_{max,2}$:
$$h_{max}=\frac{m}{m+M}h_{max,2}$$
A: 
My question why can we conserve momentum in the horizontal direction in this scenario when there is a component of tension acting in the horizontal direction which is an external force force.

There is no external force in the horizontal direction. Tension acts in $\hat j$ direction which is balanced by weight of the block.

In conclusion shouldn't the equation for max height be $$1/2mv^2=(M+m)gh_{max}$$ rather than$$1/2\mu v^2=(M+m)gh_{max}$$

No! The collision is inelastic as "A bullet is shot horizontally and gets embedded in the block." .After the collision the block and bullet behaves as a system. You can't conserve energy directly. You need to apply $\vec p $ conservation first.
