Graphical Mechanics

Applying conservation of energy

$\frac12mv^2 = mgh$ $\boxed{v = \sqrt{2gh}} \ldots (1)$

Radius of curvature $R$ after falling through height $h$ of $y = -x^4$

$R = \frac{\bigg(1 + (\frac{dy}{dx})^2\bigg)}{\frac{d^2y}{dx^2}}^\frac32$

$\boxed{R = -\frac{(1 + 16x^6)}{12x^2}^\frac32} \ldots (2)$

Slope after falling through height $h$ is

$\tan \theta = \frac{dy}{dx}$ $\tan \theta = -4x^3$ $\boxed{\cos \theta = -\frac{1}{\sqrt{1 + 16x^6}}} \ldots (3)$

FBD after falling through height $h$ if Normal reaction $= N$ is

$mg\cos \theta - N = \frac{mv^2}{R}$

When particle loses contact with wedge, $N = 0$

$mg\cos \theta = \frac{mv^2}{R}$

$g\cos \theta= \frac{2gh}{R}\ldots (\text{by (1)})$

$h = \frac{R\cos \theta}{2}$

$h = \frac{-\frac{(1 + 16x^6)}{12x^2}^\frac{3}{2}.-\frac{1}{\sqrt{1 + 16x^6}}}{2} \ldots (\text{by } (2) \text{ and } (3))$ $h = \frac{1 + 16x^6}{24x^2}$ $\text{ From equation }y = -x^4, \text{ when } y = -h, x = h^{\frac14}$ $h = \frac{1 + 16h^{\frac64}}{24h^{\frac24}}$ $24h^{\frac32} = 16h^{\frac32} + 1$ $8h^{\frac32} = 1$ $h^{\frac32} = \frac18$ $\color{#3D99F6}{\boxed{h = \frac14}}$

$\text{Therefore } a = 1, b = 4, a + b = 5$

Here is the math I used in conjunction with a numerical simulation. Derive an acceleration constraint equation:

$y = -x^4 \\ \dot{y} = -4 x^3 \dot{x} \\ \ddot{y} = -4 x^3 \ddot{x} - 12 x^2 \dot{x}^2$

Let $N$ be the magnitude of the normal force, and let $\vec{u}$ be a unit vector in the direction normal to the curve. $\vec{v}$ is a non-unitized version of $\vec{u}$ .

$\vec{v} = (v_x, v_y) = (4 x^3, 1) \\ \vec{u} = \frac{\vec{v}}{|\vec{v}|}$

Write the Newton's Second Law equations:

$m \ddot{x} = N u_x \\ m \ddot{y} = N u_y - m g$

Substitute these two equations into the acceleration constraint equation, yielding:

$N = \frac{m g - 12 m x^2 \dot{x}^2}{u_y + 4 x^3 u_x}$

Then substitute $N$ back into the Newton's Second Law equations to solve for the accelerations. Numerical integration takes care of the rest. Run the simulation until the normal force becomes negative for the first time. This is the time at which the particle loses contact with the curve. The particle falls through a height $h = \frac{1}{4}$ before this happens.

I have solved it exactly the same way as @Steven Chase has although my nomenclature and terminology is a bit different. One may refer to his solution for details. I am attaching the simulation code here for anyone who is interested.

A plot of the trajectory of the particle is as follows. The simulation runs till the particle loses contact with the wedge. The $-0.25$ y coordinate can be clearly spotted in the plots.

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clear all
clc

% Acceleration due to gravity:
g         = 10;

% Initial conditions position:
x(1)      = 0.05;
y(1)      = -x(1)^4;

% Initial conditions velocity:
dx(1)     = 0;
dy(1)     = 0;

% Time initialisation:
t(1)      = 0;
dt        = 1e-5;

% Reaction force initialisation:
N(1)      = g;

% Loop variable initialisation:
k         = 1;


% While loop until the normal reaction becomes zero:
while N(k) >= 0

    % Sine term in Equation of motion:
    S         = (4*x(k)^3)/(sqrt(1 + 16*x(k)^6));

    % Cosine term in Equation of motion:
    C         = (1)/(sqrt(1 + 16*x(k)^6));

    % Linear system of equations solving for N and acceleration components:
    A         = [1 0 -S;0 1 -C;4*x(k)^3 1 0];
    b         = [0;-g;-12*x(k)^2*dx(k)^2];

    % Matrix inversion:
    S         = inv(A)*b;
    ddx       = S(1);
    ddy       = S(2);
    N(k+1)    = S(3);


    % Numerical integration:
    dx(k+1)   = dx(k) + dt*ddx;
    dy(k+1)   = dy(k) + dt*ddy;
    x(k+1)    = x(k)  + dt*dx(k+1);
    y(k+1)    = y(k)  + dt*dy(k+1);
    t(k+1)    = t(k)  + dt;

    k         = k + 1;
end

Answer = y(end)

The particle's position vector, velocity and acceleration are $\mathbf{r} = \binom{x}{-x^4} \hspace{1cm} \dot{\mathbf{r}} = \dot{x} \binom{1}{-4x^3} \hspace{1cm} \ddot{\mathbf{r}} = \ddot{x}\binom{1}{-4x^3} - \dot{x}^2 \binom{0}{12x^2}$ and so the equation of motion of the particle is $m\ddot{x}\binom{1}{-4x^3} - m\dot{x}^2 \binom{0}{12x^2} \; = \; mR\binom{4x^3}{1} + mg\binom{0}{-1}$ Here $mR\binom{4x^3}{1}$ is the normal reaction. Taking the scalar product of this equation with $\binom{4x^3}{1}$ gives $-12mx^2\dot{x}^2 \; = \; mR(1 + 16x^6) - mg$ and hence $R \; =\; \frac{g - 12x^2 \dot{x}^2}{1 + 16x^2}$ Conservation of energy gives $\tfrac12m(1 + 16x^6)\dot{x}^2 - mgx^4 \; = \; 0$ and hence $R \; =\; \frac{g(1 - 8x^6)}{(1 + 16x^6)^2}$ Thus the particle leaves the surface when $R=0$ , so when $x^6 = \tfrac18$ , or $x^2 = \tfrac12$ , and and hence when $h=\tfrac14$ . This makes the answer $1+4=\boxed{5}$ .