# 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.

• "I am imposing various other idealistic conditions as well like the absence of any dissipative forces, absence of any kind of internal friction or any strain in the solid block of wood.A bullet is shot horizontally and gets embedded in the block."... is an oxymoron. Your collision is totally inelastic. Non-conservative forces are needed to permanently deform the block (otherwise the bullet would not be 'sticking' to the block)
– Gert
Dec 11, 2020 at 10:33
• My bad, I copy pasted that from another post with a similar condition.I have fixed it Dec 11, 2020 at 10:38

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 $$h_{max,2}$$:

$$h_{max}=\frac{m}{m+M}h_{max,2}$$

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.

• For some reason I thought that we were conserving momentum when the particle is max height, not immediately after the collision.I get it now! Dec 11, 2020 at 10:43
• "There is no external force in the horizontal direction." If the collision isn't instantaneous then there would be a horizontal force component of tension during the collision actually. Your answer should definitely take this into account. Dec 12, 2020 at 21:45
• @BioPhysicist I agree with you. But then the question becomes unsolvable , doesn't it ? Dec 13, 2020 at 1:22