Ballistic pendulum

we can apply law of conservation of momentum and law of conservation f energy, let the bullet be m1, the block be m2, and the velocity before and after the bullet embeds are v1 and v2. thus we have.(the block dont have momentum before the bullet embeds because it has no initial velocity)

$m_{1}v_{1}=(m_{1}+m_{2})v{2}$ $0.02\times 100=(0.02+1)v_{2}$ $\frac{2}{1.02}=v_{2}$

now we will apply law of conservation of energy

$KE=EP$ $\frac{1}{2}m(v_{2})^{2}=mgh$ $\frac{1}{2}(v_{2})^{2}=gh$ $\frac{1}{2}(\frac{2}{1.02})^{2}=9.8\times h$ $0.1961568941302011=h$

$\sin\theta=\frac{1-0.1961568941302011}{1}$ $\sin\theta=0.8038431058697989$ $\theta=53.498675106806$ \theta is maximum inclination to the horizontal, so the inclination to the vertical is $90-53.498675106806=36.501324893194$

Let $v_A$ be the entrance speed given: 100 m/s. Let $v_B$ be speed at which the block-bullet system sets off. We know, from conservation of energy, that

$v_B^2 = 2gh,$ where $h$ is the height above the original position of the block to which the system rises. So, from conservation of momentum, if $m$ is the mass of the bullet and $M$ is the mass of the block, we have:

$mv_A = (m+M) v_B \implies v_B = \dfrac {m}{m+M} \cdot v_A$

So, we have: $h = \dfrac {v_B^2}{2g} = \dfrac {\left( \dfrac {m}{m+M} \cdot v_A \right)^2}{2g}$ .

Our goal is to find $\theta$ , the rise angle. Using some trigonometry, we see that

$\cos \left( \theta \right) = 1 - \dfrac {h}{\ell},$

where $\ell$ is the length of the string. Substituting the given numbers, we have $\theta = \boxed {36.5^\circ}$ .

When the bullet is embedding itself into the block you can conserve momentum of the bullet-block system because we can neglect the impulse of gravity and tension in the string(Both have much less magnitude than the normal force)

So we have $20g * 100 m/s = 1020g *v$ $v=\frac{2}{1.02} m/s$

Note: Considering $v\approx 2$ gives the correct answer of 37 degrees, otherwise the angle rounds off to 36 and not 37 degrees. It should have been made clear in the question to neglect the mass of the bullet, or they could have provided multiple choice options.

Then using conservation of mechanical energy we get:- $\frac{v^2}{2}=lg(1-\cos\theta)$ $\cos\theta\approx \frac 45$ Thus $\theta\approx 37$

By "conservation law of momentum" we get the velocity of the two bodies after the impact ------> M1V1 + M2 V2 = Vc( M1+M2 ) ------> (0.02 X 100) + (1 X 0) = Vc ( 0.02 + 1 ) ------> Vc = 1.9607 ------> Vc = sqrt ( 2 X g X h ) ------> 1.9607 = sqrt (2 X 9.8 X h ) ------> h = 0.196 ------> h = L( 1 - cos x ) ------> 0.196 = 1(1-cos x ) ------> then, angle ------> x = 36.5