Understanding Motion and Collision in the Absence of Gravity

[Nexus99]

Homework Statement

A smooth duct is mounted on a slide with mass M and negligible dimensions, which allows the passage of a ball of mass m, launched towards the slide with initial speed
v0 parallel to the horizontal from the same elevation at a distance d. There
slide is free to move without friction on a vertical track and is let go to
moment of launch.
1. In the absence of gravity, calculate the final sled and ball speeds
2. In the presence of gravity, under which conditions does the ball enter the tube?
3. In the presence of gravity, for which value of v does the slide stop immediately after the impact?

Relevant Equations

Conservation of momentum, conservation of kinetic energy

These are the solution, I'm not understanding something:

In the absence of gravity, the total kinetic energy and vertical momentum are conserved).
So we have

$\frac{1}{2} m v_0 = \frac{1}{2} M \dot{Y}^2 + \frac{1}{2} m \dot{y}^2$
$0 = m \dot{y} + M \dot{y}$

Ok. But, if the ball goes inside the conduct, shouldn't be $v_y = \dot{y} + \dot{Y}$ ?
After some algebra you get:

$\dot{Y} = \pm \sqrt{ \frac{m^2}{M(M+m)}} v_0$
$\dot{y} = \mp \sqrt{ \frac{M}{(M+m)}} v_0$
And the professor writes:
The solution with $\dot{Y}$ < 0, $\dot{y}$ > 0 is clearly not acceptable
Why? Shouldn't the ball and the slide move in the same direction ? (up or down)

 

[PeroK]

And the professor writes:
The solution with $\dot{Y}$ < 0, $\dot{y}$ ̇ > 0 is clearly not acceptable
Why? Shouldn't the ball and the slide move in the same direction ? (up or down)

The assumption is that there are no external forces in the $y$ direction, so momentum in the $y$ direction is conserved. And is initially $0$, of course.

The final y-momenta of the ball and duct must, therefore, be equal and opposite.

The ball is physically constrained to move down the duct, so the final y-momentum of the ball must be downwards - hence the duct must move upwards.

 

[Nexus99]

The assumption is that there are no external forces in the $y$ direction, so momentum in the $y$ direction is conserved. And is initially $0$, of course.

The final y-momenta of the ball and duct must, therefore, be equal and opposite.

The ball is physically constrained to move down the duct, so the final y-momentum of the ball must be downwards - hence the duct must move upwards.

Thanks you're right.
Maybe i understood also my previous doubt:
the term $\dot{y}$ contains yet the velocity of the slide, in fact: $\dot{y} = v_y + V_y$. Am i right?
Is there a collision inside the duct?

 

[PeroK]

Thanks you're right.
Maybe i understood also my previous doubt:
the term $\dot{y}$ contains yet the velocity of the slide, in fact: $\dot{y} = v_y + V_y$. Am i right?
Is there a collision inside the duct?

I assume that $Y$ is the y-coordinate of the duct, so $\dot Y$ is the y-velocity of the duct.

The whole process is a collision!