Conservation of Work/Momentum of Puck Sliding Off Plate

[SeventeenForever]

Homework Statement

I saw this problem from a few years ago here on Physics Forum (https://www.physicsforums.com/threads/work-energy-theorem-problem.823990/) and wanted some clarification. Here is the problem:

A curved plate of mass M is placed on the horizontal, frictionless plane as shown. Small puck of mass m is placed at the end of the plate and given an initial horizontal velocity v. How high the puck will shoot up from the plate? Regard radius of curvature to be much smaller than the jump height and consider that the puck leaves the plate vertically.

Homework Equations

K₁ + U₁ = K₂ + U₂
mv₁ = mv₂
K= 0.5mv²
U= mgh

The Attempt at a Solution

[/B]
Based on reading the previous post over and over again, here is my understanding of the problem.

The puck is initially sliding across the horizontal portion of the plate with a velocity, v_o. As it approaches the launch point, some of its kinetic energy is converted to the kinetic energy of the plate. With respect to the ground, the puck still has kinetic energy, since it is on the moving plate. There is also some kinetic energy present that will launch the puck up in the air.

Energy Conservation from Initial to Launch Point: K_{puck, initial} = K_{puck+plate together} + K_puck

Only the kinetic energy of the puck alone (K_puck) gets converted to gravitational potential energy.

Energy Conservation from Launch Point to Max Height: K_puck = U_{max height}

Plugging in the terms:

Energy Conservation from Initial to Launch: 0.5mv_o² = 0.5(m+M)v²_{puck & plate} + 0.5mv²_{puck alone}

Energy Conservation from Launch to Max Height: 0.5mv²_{puck alone} = mgh

To find the unknown variable, v_puck&plate, we need to use conservation of momentum. Since momentum is a vector, it is conserved in each direction. In the horizontal plane, the initial momentum of the puck is conserved as the puck and the plate move together just before launch.

mv_o = (m+M)v_{puck & plate}

I can then solve for v_{puck & plate} in terms of the known variables, m, M, and v_o, and plug this into the Energy Conservation from Initial to Launch Point to find v_{puck alone}. Then I can use the Energy Conservation from Launch Point to Max Height: K_puck = U_{max height} to find the max height and solve the problem.

My biggest questions/lack of understanding is why exactly the plate starts moving. The question (as in the other post) makes no mention of friction. The only thing I can think of is the plate starts moving as the puck is ramping up, but I still haven't quite connected all the dots. Knowing which way the plate moves is also important for the Conservation of Momentum.

Also, kinetic energy isn't a vector. Hence, why is the "horizontal" kinetic energy of the puck (just before the launch point, presumably from its position on the plate, which is moving) not converted to gravitational potential energy. Lastly, if the everything is frictionless, why would the puck even move with plate at the launch point?

Thanks for reading!

 

[.Scott]

The force of the puck on the plate is normal to the surface of the plate.
Where that surface is horizontal, the force of m on M will be entirely vertical and there is no change in M's velocity.
Where that surface is not horizontal, there will be a horizontal component to the m on M force. And that component will cause M to accelerate horizontally.

 

[SeventeenForever]

That seems to make sense. The problem as written doesn't focus on the ramping so I neglected that.

Does my work seem correct to you?

Thanks for responding so quickly!

 

[haruspex]

Hence, why is the "horizontal" kinetic energy of the puck (just before the launch point, presumably from its position on the plate, which is moving) not converted to gravitational potential energy

Although energy is not a vector, force, acceleration and velocity are. Increasing GPE requires upward movement. With no initial vertical velocity that will require an upward force.
In the set-up, the plate is simply not able to supply the necessary forces to convert all horizontal KE to vertical KE. Indeed, the conservation of momentum means some KE has to remain in the horizontal motion.

Does my work seem correct to you?

Yes.

 

[SeventeenForever]

Thanks guys! It makes more sense now