Collision between a block and a curved up-ramp surface

[rashida564]

Homework Statement

A block of mass mb sits at rest on a frictionless table; the block has a circular surface
of radius R as shown in the figure. A small cube of mass mc < mb and speed vc,0 is
incident upon the block; the cube moves without friction on the table and the block.
What is the maximum height h above the table that is reached by the cube?
Assume vc,0 is small enough so that h < R. Express your answer using some or
all of the following variables: vc,0, g, mb and mc

Relevant Equations

Pi=Pf
Ei=Ef

I tried to solve this question using conservation of momentum. The momentum of the system is mc(vc)=mcvc'+mbvf'. But after that I have no idea I want to use the conservation of kinetic energy but the question doesn't say it's elastic collision, I need to find the velocity of the mb after the collision so I can find the maximum height

 

[PeroK]

Conservation of momentum in the x-direction is the key. You can't use conservation of kinetic energy.

What is the critical time after the collision?

 

[Cutter Ketch]

A couple of things. The two objects are changing state all the time, but the problem asks about one very specific moment in time. What can you say about the velocities at the moment the cube reaches its highest point? Also, for equations you mention conservation of momentum and conservation of energy. So far you have only thought about conservation of momentum. I bet you can guess what you should think about next ;)

 

[rashida564]

Conservation of energy. At the highest point the velocity of the block will be equal to zero with respect to the block, so the only velocity will be the velocity of the block and the conservation is conserved through out the whole journey for the x-axis because no net force.
So mcgh+1/2 (mc+mb)vf^2=1/2 mc vi^2.
mc vi= (mc+mb)vf

 

[PeroK]

Conservation of energy. At the highest point the velocity of the block will be equal to zero with respect to the block, so the only velocity will be the velocity of the block and the conservation is conserved through out the whole journey for the x-axis because no net force.
So mcgh+1/2 (mc+mb)vf^2=1/2 mc vi^2.
mc vi= (mc+mb)vf

That looks good. If you are not using Latex, maybe m, M are better than mc, mb. And, u, v instead of vi, vf.

 

[rashida564]

Mate how can I write the way you write with symbols and fractions. @PeroK

 

[PeroK]

Mate how can I write the way you write with symbols and fractions. @PeroK

Help is here:

https://www.physicsforums.com/help/latexhelp/

And, when you reply you get some Latex which you can copy and paste. E.g.

$m_cgh + \frac 1 2 (m_c + m_b) v_f^2 = \frac 1 2 m_c v_i^2$

 

[Cutter Ketch]

Conservation of energy. At the highest point the velocity of the block will be equal to zero with respect to the block, so the only velocity will be the velocity of the block and the conservation is conserved through out the whole journey for the x-axis because no net force.
So mcgh+1/2 (mc+mb)vf^2=1/2 mc vi^2.
mc vi= (mc+mb)vf

That’s it! Now solve these two equations for h.

 

[haruspex]

Mate how can I write the way you write with symbols and fractions. @PeroK

You should try to use LaTeX, but if you find that too bothersome at least use the pulldowns at the top of the editing area. The ... pulldown gives subscript and superscript, while the √x pulldown gives lots of special symbols and characters.

 

[rashida564]

@haruspex can I use word then snapshot my working?

 

[haruspex]

@haruspex can I use word then snapshot my working?

We discourage that because it makes it hard for us to quote individual lines in our posts. Maybe ok if you number all your equations.

 

[neilparker62]

Alternate method:

At the moment the small block reaches its maximum height its velocity relative to the large block is zero. This is typically what happens in an inelastic or "collide and coalesce" collision and the energy loss in such a case is given by the expression: $$ ΔE = ½ μ Δv^2 $$ where: $$μ=\frac{m_cm_b}{m_c+m_b}$$ ... the 'reduced mass' of the colliding bodies and Δv is their (initial) relative velocity $v_c$ in this instance. But here the collision energy is not lost - it is simply converted to potential energy and therefore you can equate the above energy expression to the standard expression for PE.