Block Sliding Down a Ramp that is Free to Slide

AI Thread Summary
A block slides down a frictionless ramp, which is also free to slide on a horizontal surface, and the goal is to determine the block's speed after it leaves the ramp. The discussion includes analyzing forces acting on both the block and the ramp, with emphasis on free body diagrams and the conservation of mechanical energy and momentum. The final speed of the block is derived using energy conservation principles, leading to the equation v = sqrt(2gh). The scenario is likened to a collision problem due to the conservation of mechanical energy in the block-ramp system, despite the complexities introduced by the normal forces. The conversation also touches on the differences in energy conservation when transitioning from the ramp to the flat surface.
PokemonMaster
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Homework Statement


A block with mass m rests on a smooth, frictionless ramp with mass M and height h. The ramp itself sits on a frictionless horizontal surface in which it is free to slide. The block slides smoothly down the ramp from rest. We want to find the speed of the block after it has left the ramp as it is moving horizontally on the flat surface.

a. Draw a clear and complete free body diagram for the block AND the ramp at an instant when it is sliding at an arbitrary location on the ramp.
b. SEE TABLE - Determine the 1: Average Net Force (Direction Arrow), 2: Net Work Done By All Forces (+,-, 0), 3: Work done by Non-Conservative Forces (+,-, 0), 4. Kinetic Energy Conserved? (C, NC), 5. Mechanical Energy Conserved? (C, NC), and 6: Momentum (x,y) Conserved (C, NC)? for EACH of the 3 systems: A. Block, B. Ramp, and C. Block-Ramp. Assuming that the total mechanical energy of the Block-Ramp system is conserved.
c. Find the final speed of the block once it has reached the horizontal surface.
d. How can we treat this situation the same way we treat collision problems?

Homework Equations


There are many, including:

Wc + Wnc = delta K
Wnc = delta Emech

The Attempt at a Solution


a. the block: "mg" down, "n1" perpendicular to ramp, and "mg cos theta" parallel to ramp
the ramp**: "Mg" down, "n1" going into the ramp, and a large "n2" coming up perpendicular to the ground
I am less sure about the ramp diagram.
b. I attached a photo including my table.
c.
Ui = Kf
mgh = 1/2 mv^2
2gh = v^2
v = sqrt (2gh)
d. This situation is similar to a collision problem because mechanical energy of the block-ramp system is conserved and therefore the momentum of the block-ramp system is conserved. The normal force that the ramp exerts on the block causes the block to move to the right and the reactive force that the block exerts on the ramp causes the ramp to move to the left, so the normal force can be equated to a collision force where the two objects move in opposite directions with different velocities depending on their masses like in an elastic collision.
 

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I believe that the conservative work on both the block and the ramp is done by gravity while the nonconservative work on both is by the normal force. I have updated the filled in chart I posted for part b and now think that the column for the block is: down and to the right arrow, +, + , NC, C, x: NC and y: NC. Then for the ramp: right arrow, +, +, NC, NC, x: NC, y: C. Lastly, for the block-ramp system I believe that there are no external forces acting on the system so the values would be: 0, 0, 0, C, C (given), x: C and y: C.
 
Do you know how the terminal velocity of the block on this ramp compares to a fixed ramp?
 
PokemonMaster said:
This situation is similar to a collision problem because mechanical energy of the block-ramp system is conserved
Mechanical energy is typically not conserved in a collision. What is conserved?

Also, consider the transition from sliding down the ramp to sliding on the horizontal surface. Will mechanical energy be conserved then?
 
Sashimoo the necromancer strikes again!

PokémonMaster has not been online in about 6 months it would appear.
 
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