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 Energy and Work Energy and Work Physics Help Forum Apr 14th 2008, 08:27 PM #1 Junior Member   Join Date: Apr 2008 Posts: 7 Am I right or am I missing something? I am dealing with this problem. We have two bodies. One with mass of m1 and one with m2. The second one is moving with a constant velocity u and its height is h. The ground and the sides of both of the bodies are frictionless. The question is how much is the minimum velocity u in order for the first body m1 to climb up the m2 body all the way to the top, and not to stop but just to start moving the other way down the second body. (I Have attached a sample image to the post) My solution is that in order to happen that the kinetic energy of the second body must transform in potential energy for the first body. Therefore $\displaystyle m_1gh=\frac{1}{2}m_2u^2~\rightarrow~u=\sqrt{\frac{ 2m_1gh}{m_2}}$ That means that when the m1 body reaches the top the second one must stop moving because it lost all it's potential energy. So the minimun velocity of the second body is $\displaystyle u\geq \sqrt{\frac{2m_1gh}{m_2}}$ Is this solution ok? Am I missing something? Thanks a lot in advance. Akis   Apr 14th 2008, 09:46 PM #2 Junior Member   Join Date: Apr 2008 Location: Pittsburgh, PA Posts: 11 It looks good to me...only question I have is...Is m1 not moving? m2 is the only object moving in this situation? It seems weird that the "hill" m1 has to climb is the thing moving, but I don't know. The math is definitely right, and as long as m1 is stationary, then you should have everything you need.   Apr 15th 2008, 02:41 AM #3 Junior Member   Join Date: Apr 2008 Posts: 7 You 're right. Sorry that I forgot to mention that m1 is not moving and m2 is the only moving body with velocity u. In other words it is like the body m1 is still, and the ground beneath it, is moving. I hope that helps. Thanks a lot again   Apr 15th 2008, 04:58 AM   #4
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 Originally Posted by zenctheo You 're right. Sorry that I forgot to mention that m1 is not moving and m2 is the only moving body with velocity u. In other words it is like the body m1 is still, and the ground beneath it, is moving. I hope that helps. Thanks a lot again
That still doesn't quite answer the question. If the ground is frictionless then any kind of impact between the masses will push m1 to the left.

Since m2 does have a slope I think it is reasonable to conclude that m1 will climb up it to a certain degree at least. And since there is no friction, energy will be conserved.

The problem is that momentum is not: The vertical component of the momentum of m1 is increasing as the object is raised into the air.

It's been a while since I've done one of these: I'll have to ponder it and get back to you.

-Dan
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See the forum rules here.   Apr 15th 2008, 05:49 AM   #5
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 The problem is that momentum is not: The vertical component of the momentum of m1 is increasing as the object is raised into the air.
I agree. When the m2 body reaches m1 the weight of m1 will act as a deceleration force to m2. Not a constant one because the component of weight which decelerates the body is always changing because m1 is following the slope.
It will begin with zero just before it will start climbing m2, it will be increase until it reaches the top of m2 where it will be zero again.
Since there is no collision happening between the bodies, I don't thing that momentum is conserved.
Therefore I can only study the system using the conservation of energy.
However I am not sure if this is the right way.   Apr 15th 2008, 06:22 AM #6 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,781 Okay, here we go. When m1 is at the top of m2 we know that the y component of the velocity of m1 is 0 m/s. At this point in time the horizontal component of m1's velocity must be the same as the velocity of m2, because m1 is about to change direction relative to the motion of m2. Now, m1's momentum is not conserved in the y direction but it is conserved in the x, so we have at this point in time: $\displaystyle m_2u = m_1 v + m_2 v$ (The velocity of m1 is all in the x direction at this point, so I'm just calling the x component of v1, v.) We also know that energy is conserved, so when m1 reaches it's max height we have that $\displaystyle \frac{1}{2}m_2u^2 = \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 + m_1gh$ So using the momentum equation we get $\displaystyle v = \frac{m_2u}{m_1 + m_2}$ and plugging this into the energy equation: $\displaystyle \frac{1}{2}m_2u^2 = \frac{1}{2}(m_1 + m_2) \left ( \frac{m_2u}{m_1 + m_2} \right ) ^2 + m_1gh$ $\displaystyle \frac{1}{2}m_2u^2 = \frac{1}{2} \left ( \frac{m_2^2}{m_1 + m_2} \right ) u^2 + m_1gh$ which you can solve for u from there. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.   Apr 15th 2008, 06:26 AM   #7
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 Originally Posted by zenctheo Since there is no collision happening between the bodies, I don't thing that momentum is conserved.
The two objects are interacting in some way, so we can consider this to be a collision. (You can even do this when the objects never touch.) We know that the momentum (in the horizontal direction) is conserved because there is no net force (x component) acting on the system composed of both masses.

-Dan
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