What Are the Final Speeds of a Material Point and a Wedge Without Friction?

AI Thread Summary
The discussion focuses on calculating the final speeds of a material point and a wedge without friction, using conservation of energy and momentum principles. The equations derived include kinetic energy and momentum conservation, leading to a final speed expression for the material point. The user also addresses a related problem involving an initial speed for the material point to reach maximum height, applying similar conservation laws. The conclusion suggests that the approach taken is correct, particularly in equating the speeds of the two bodies at maximum height. The overall analysis emphasizes the application of fundamental physics principles in solving the problem.
Karozo
Messages
4
Reaction score
0

Homework Statement



I think that the image is clear.
There isn't friction, not the material point with wedge, not the wedge with the floor.

At time t=0 the material point start to move, I need to find the final speed of the two objects at time t\rightarrow\infty.

Homework Equations



I have used conservation of energy and momentum.

The Attempt at a Solution



So I have two equation:
\frac{1}{2}m{v_m}^2+\frac{1}{2}M{v_M}^2 = mgR

m{v_m}+M{v_M}=0

And the solution is {v_m}=\sqrt{\frac{mgR}{\frac{1}{2}m+\frac{1}{2}\frac{m^2}{M}}}

Am I wrong?
 

Attachments

  • Problem_1-7-2014.png
    Problem_1-7-2014.png
    1.1 KB · Views: 469
Physics news on Phys.org
It can be simplified, but it looks okay. What about the velocity of the other object?
 
Well, if V_m is right is very simple to find V_M, so I haven't written it.
 
I have also a similar problem, you can see the image.

In this the mass m, start with an initial speed v_0, you have to find v_0 so that the material point has maximum height R.

I think that is right to use the two equations:

\frac{1}{2}m{v_m}^2+\frac{1}{2}M{v_M}^2+mgR=\frac{1}{2}m{v_0}^2 energy

m{v_m}+M{v_M}=mv_0 momentum

for the point of maximum height, and then you have {v_m}={v_M} , because the two bodies are in contact.
 

Attachments

  • Problem-2-7-2014.png
    Problem-2-7-2014.png
    1.1 KB · Views: 424
Karozo said:
I have also a similar problem, you can see the image.
Next time, you should put a new problem in a new thread.

In this the mass m, start with an initial speed v_0, you have to find v_0 so that the material point has maximum height R.

I think that is right to use the two equations:

\frac{1}{2}m{v_m}^2+\frac{1}{2}M{v_M}^2+mgR=\frac{1}{2}m{v_0}^2 energy

m{v_m}+M{v_M}=mv_0 momentum

for the point of maximum height, and then you have {v_m}={v_M} , because the two bodies are in contact.
That appears correct.
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top