Answer
a) $ v_B =\dfrac{\sqrt{2gd(m+M)^2 + (m+M)\left[ kd^2-2dMg \right] }}{m} $
b) $453\;\rm m/s$
Work Step by Step
We assume that the spring is massless, and the system is the bullet, the block, the spring and the Earth.
This system is isolated which means that the energy is conserved.
$$E_i=E_f$$
where $E_i$ is the energy of the system just after the bullet hits the block and $E_f$ is the energy when the block stops moving up for a while.
$$K_i+U_{gi}+U_{si}=K_f+U_{gf}+U_{sf}$$
The final kinetic energy of the system is zero, the initial gravitational potential energy of the system is $[-(m+M)gy_1]$ (since $y_1$ is down the equilibrium point of the spring), the initial elastic potential energy is given by $\frac{1}{2}ky_1^2$ where $y_1$ is the stretched distance of the spring below due to the weight of the block and its final energy is $\frac{1}{2}ky_2^2$ where $y_2$ is the compressed distance due to the rising of the bullet and the block due to the collision, and the final gravitational potential energy is given by $(m+M)gy_2$; where $d=|y_1|+y_2$ (whereas $d$ is the spring’s maximum compression).
Recall that the two objects, the bullet and the block, will move as one unit after the collision.
$$\frac{1}{2}(m+M)v_{f1}^2-(m+M)gy_1+\frac{1}{2}ky_1^2=0+(m+M)gy_2+\frac{1}{2}ky_2^2$$
where $v_{f1}$ is the speed of the bullet plus the block as one unit just after the collision.
$$\frac{1}{2}(m+M)v_{f1}^2-(m+M)gy_1+\frac{1}{2}ky_1^2=(m+M)gy_2+\frac{1}{2}ky_2^2\tag 1$$
Now we need to use the conservation of momentum law to find $v_{f1}$.
$$p_{iy}=p_{fy}$$
$$mv_B=(m+M)v_{f1}$$
Thus,
$$v_{f1}=\dfrac{mv_B}{m+M}$$
Plugging into (1);
$$\frac{1}{2}(m+M)\left(\dfrac{mv_B}{m+M}\right)^2-(m+M)gy_1+\frac{1}{2}ky_1^2=(m+M)gy_2+\frac{1}{2}ky_2^2 $$
Multiplying both sides by 2;
$$ \dfrac{m^2v_B^2}{ m+M } -2(m+M)gy_1+ ky_1^2=2(m+M)gy_2+ ky_2^2 $$
Solving for $v_B$;
$$ \dfrac{m^2v_B^2}{ m+M } =2(m+M)gy_2+ ky_2^2+2(m+M)gy_1- ky_1^2 $$
$$ \dfrac{m^2v_B^2}{ m+M } =2g(m+M)[y_1+y_2]+ k[y_2^2 - y_1^2] $$
$$ v_B^2 =\dfrac{2gd(m+M)^2 + k(m+M)[y_2^2 - y_1^2] }{m^2}$$
Recall that, $d=y_1+y_2$ , and hence $y_2=d-y_1$.
$$ v_B^2 =\dfrac{2gd(m+M)^2 + k(m+M)[(d-y_1)^2 - y_1^2] }{m^2} $$
$$ v_B^2 =\dfrac{2gd(m+M)^2 + k(m+M)[d^2-2dy_1+\color{red}{\bf\not} y_1^2 - \color{red}{\bf\not} y_1^2] }{m^2} $$
$$ v_B^2 =\dfrac{2gd(m+M)^2 + k(m+M)[d^2-2dy_1 ] }{m^2} \tag 2$$
Now we need to find $y_1$ by applying Newton's second law on the block when it is at attached to the spring and was at rest before the collision (see the middle figure below).
$$\sum F_y=F_{sp}-Mg=Ma_y=M(0)=0$$
Thus,
$$F_{sp}=Mg$$
where $F_{sp}$ is the spring force which is given by Hooke's law, $F_{sp}=-kx$.
Hence,
$$-k(-y_1)=Mg$$
So,
$$y_1=\dfrac{Mg}{k}$$
Plugging into (2);
$$ v_B^2 =\dfrac{2gd(m+M)^2 + k(m+M)[d^2-2d\frac{Mg}{k} ] }{m^2} $$
$$ v_B^2 =\dfrac{2gd(m+M)^2 +\color{red}{\bf\not} k(m+M)\left[ \dfrac{kd^2-2dMg}{\color{red}{\bf\not} k} \right] }{m^2} $$
$$ v_B^2 =\dfrac{2gd(m+M)^2 + (m+M)\left[ kd^2-2dMg \right] }{m^2} $$
Therefroe,
$$ \boxed{ v_B =\dfrac{\sqrt{2gd(m+M)^2 + (m+M)\left[ kd^2-2dMg \right] }}{m} }$$
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b) Here we just need to plug the given into the boxed formula above.
$$ v_B =\dfrac{\sqrt{2\times 9.8\times 0.45(0.01+2)^2 + (0.01+2)\left[ 50(0.45)^2-(2\times 0.45\times 2\times 9.8) \right] }}{0.01} $$
$$v_B=\color{red}{\bf 453}\;\rm m/s$$
