Answer
See the detailed answer below.
Work Step by Step
We have here three stages,
1- the first stage is when Lisa rides the bobsled just before sliding down; we need to find the system velocity just before sliding down.
2- the second stage is when Lisa rides the bobsled down the incline.
3- the third stage is when Lisa+bobsled hits the spring and stops.
We assume that the system is Lisa+bobsled+Earth+spring. We know that the kinetic energy and the potential energy on the top of the incline will be converted to stored energy in the spring after Lisa hits it and stops.
We assume that the system is isolated and that the surfaces are frictionless.
Thus,
$$E_i=E_f$$
where $E_i$ is at the top of the incline and just before Lisa+bobsled moved down the include, and $E_f$ is when the system hits the spring and stops.
$$K_i+U_{ig}+U_{is}=K_f+U_{fg}+U_{fs}$$
The final kinetic energy is zero since she finally stops, the final gravitational potential energy is zero since it reaches the ground level, and the initial elastic potential energy is zero since the spring is uncompressed.
$$\frac{1}{2}(m_L+m_b)v_{1}^2+ (m_L+m_b)gh+0=0+0+\frac{1}{2}k x ^2 $$
where $m_L$ is Lisa's mass, $m_b$ is the bobsled's mass, $h$ is the height of the incline, and $x$ is the compressed distance of the spring when the whole system stops for a while.
$$\frac{1}{2}(m_L+m_b)v_{1}^2+ (m_L+m_b) g h = \frac{1}{2}k x ^2\tag {*} $$
We can find $h$ by
$$\sin\theta=\dfrac{h}{d}$$ where $d$ is the length of the inclined.
Hence,
$$h=d\sin\theta $$
Thus,
$$\frac{1}{2}(m_L+m_b)v_{1}^2+ (m_L+m_b) g d\sin\theta = \frac{1}{2}k x ^2 \tag 1$$
Now we need to find $v_1$ which is the speed of Lisa+bobsled after the collision.
In the first stage, we can see that the momentum is conserved so
$$p_{ix}=p_{fx}$$
$$m_Lv_L=(m_L+m_b)v_1$$
Thus,
$$v_1=\dfrac{m_Lv_L}{m_L+m_b}$$
Plugging into (1);
$$\frac{1}{2}(m_L+m_b)\left(\dfrac{m_Lv_L}{m_L+m_b}\right)^2+ (m_L+m_b) g d\sin\theta = \frac{1}{2}k x ^2 $$
Solving for $x$;
$$ \left(\dfrac{m_L^2v_L^2}{m_L+m_b} \right)+2 (m_L+m_b) g d\sin\theta = k x ^2 $$
$$ \boxed{ x =\sqrt{\dfrac{ \dfrac{m_L^2v_L^2}{m_L+m_b} +2 (m_L+m_b) g d\sin\theta }{k}}}$$
Plugging the known;
$$ x =\sqrt{\dfrac{ \dfrac{40^2\times 12^2}{40+20} +2 (40+20) (9.8)(50) \sin20^\circ }{2000}}$$
$$x=\color{red}{\bf 3.46}\;\rm m$$
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b) We can see from the boxed formula above, that $d\sin\theta$ is an important factor. But actually, this is just equal to the height since $h=d]sin\theta$. So, the angle here was used just to find the height, and the angle is not important here but the height is.
And we can see in the star formula above that $h$ is there without the angle.
From all the above, if the ramp is frictionless, the angle and the shape of the ramp are not important, it is only the height that matters.
