- #1
Nexus99
- 103
- 9
- 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)
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