Conservation of momentum of a bullet question

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SUMMARY

The discussion focuses on the conservation of momentum in a bullet-wood collision scenario. It establishes that the horizontal momentum of the bullet-wood system remains conserved during the collision despite the tension forces from the strings. The derived expression for the speed of the bullet and wood immediately after the collision is given by \( v' = \frac{m v}{m + M} \). Additionally, the law of conservation of energy is identified as the principle that relates the maximum vertical height to the speed just after the collision.

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Homework Statement


A bullet is shot horizontally into a block suspended by two strings. The bullet remains embedded in the wood and the wood and bullet together suspend upward.
a) Explain why the horizontal momentum of the bullet-wood system is conserved during the collision even though the strings exert tension forces on the wood.
b) If the bullet and wood have masses of m and M respectively, and the bullet has an initial speed of v, derive an algebraic expression for the speed of the bullet and wood immediately after the collision before they swing upward in terms of m, M and v.
c) What law of nature can be used to relate the max vertical height to speed just after the collision?
d) Derive an expression for the max vertical height in terms of m, M, v and g.

The Attempt at a Solution


a) I think it is because as the bullet strikes the wood, there is a force of the bullet acting on the wood and a force of the wood acting on the bullet and those 2 forces together equal the initial force of the bullet so momentum is conserved?

b)P = P'
mvm + MvM = mvm’ + MvM ‘
mvm + MvM = (m + M)v
(mvm + MvM)/(m + M) = v

c) I don't know what law of nature they're talking about here.
d)
 
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Law of conservation of energy.
 

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