Faster and faster we go

0.103 J = m g y + 0.5 * m (V)^2 At x = 1m, y = 1 - 1 1 => y = 0m Here, m = 10 g = 0.01 kg (V)^2 = 20.6 (m/s)^2 ......(i)

y = 1 - x^2 Taking derivatives both sides,

Vy = - 2* Vx (V)^2 = (Vy)^2 + (Vx)^2 (V)^2 = 5 * (Vx)^2 .............(ii)

(Vx)^2 = 4.12 (m/s)^2 ;

Vx = 2.029778 m/s

At x=1m, y=0 m. So, potential energy at x=1m is zero. So, the whole of total energy initially present will be converted into Kinetic Energy when the coin reaches x=1m. So, K.E. (at x=1m) = 0.103 Joules. Now, K.E. = $(1/2)mv^2$ . On solving, v=4.539 m/s This is the velocity at x=1m but this will be at the angle θ = arctan(dy/dx) = arctan(-2) to the x axis. The horizontal component of this velocity will be $v cosθ = v/ \sqrt(5) = 2.03$ m/s. Hence, the answer 2.03 m/s.

It is clear that, Final Potential Energy = $0$

Now, since Energy is conserved,

Final Kinetic Energy = Total Energy

$\Rightarrow \frac{1}{2} m v^2 = 0.103J$

$\Rightarrow v \approx 4.538 m/s$

Now,

$y = 1-x^2$

$\Rightarrow \frac{\mathrm d}{\mathrm d x}y = -2x$

Finally, $x = 1m$

Therefore,

$\Rightarrow \frac{\mathrm d}{\mathrm d x}y = -2$

$\Rightarrow \tan\theta = -2$

$\Rightarrow \cos\theta = \frac{1}{\sqrt{5}}$

Drawing a vector diagram, clearly,

$\cos\theta = \frac{v_x}{v}$ , where $v_x$ represents the horizontal component of the final velocity

Hence,

$v_x \approx 2.029 m/s$

By conservation of energy, $Initial energy=Final energy$ . Hence, $0.103=\frac{1}{2}mv^2$ . Solving, for velocity v, v=4.54 m/s. Since the velocity vector is always tangent to the path, $y=1-x^2$ , the slope of the velocity vector is equal to $\frac{dy}{dx} at x=1$ which is -2. This implies that the velocity vector points to the direction 1 horizontal and 2 vertical (downward). So, the horizontal component of velocity is $\frac{1}{\sqrt5}v=2.03m/s$ .

When the coin has reached $x=1m$ , the height is $y=0m$ . By using the law of conservation of mechanical energy, the initial energy is equal to the kinetic energy of the coin at $y=0m$ , particularly: $E_k = 0.103 J$ From the kinetic energy, we can calculate the velocity of the coin: $v=\sqrt{\frac{2E_k}{m}}$ The velocity of the coin is parallel to the tangent of the hill at $x=1m$ , where the slope is: $y'=\tan \theta = -2x = -2$ Finally, the horizontal component of the coin's velocity is: $v_x = v\cos \theta = \sqrt{\frac{2E_k}{m}} \sqrt{\frac{1}{1+\tan^2 \theta}} = \sqrt{\frac{2\times0.103}{0.01}} \sqrt{\frac{1}{1+(-2)^2}} = 2.030 (m/s)$

2.03

Energy conservation -> at x=1; y=0, the potential energy = 0
=> Kinetic energy there is $0.103 = mv^2/2$
$y ' = -2x$ so $y '(1)=-2$
Horizontal component of the coin's velocity : $v_h = v/ \sqrt{2^2+1^2}*1$
$v_h = 2.029778313$

$$ By using the law of conservation of energy we obtain $$ $\begin{aligned} E_{initial}&=E_p+E_k\\ 0.103&=mgy+\frac{1}{2}m\left(v_x^2+v_y^2\right)\\ &=mg(1-x^2)+\frac{1}{2}m\left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2\right)\\ &=mg(1-x^2)+\frac{1}{2}m\left(\left(\frac{dx}{dt}\right)^2+\left(-2x\frac{dx}{dt}\right)^2\right)\\ &=mg(1-1^2)+\frac{1}{2}m\left(\left(\frac{dx}{dt}\right)^2+\left(-2(1)\frac{dx}{dt}\right)^2\right)\\ &=\frac{5}{2}m\left(\frac{dx}{dt}\right)^2\\ \frac{dx}{dt}&=\sqrt{\frac{2\times0.103}{5m}}\\ &\approx\boxed{2.03\;\text{m/s}} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$