Conservation of Energy problem

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SUMMARY

The discussion centers on a conservation of energy problem involving two blocks, M1 and M2, on a frictionless circular path with radius R. The user successfully derived the velocity of mass M2 as it exits the circular path using the conservation of energy principle, resulting in the equation v(21) = SQRT(2gR). The momentum conservation equation was also applied, leading to the final velocity of M2, V2, expressed as V2 = M1[SQRT(2gR)]/(M1 + M2). The user noted a discrepancy when M1 equals M2, where the expected result is V2 = SQRT(gR).

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  • Familiarity with momentum conservation equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of energy conservation and momentum in mechanical systems.

a9211l
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Homework Statement



If there is a frictionless circular path with radius R cut into a block of mass M1 and a block of mass M2 slides down the path, what is the velocity of mass M2 as it leaves the block?

There is no friction between M1 and the ground.

M2=mass of falling block
M1=mass of
v(21)=velocity of M2 with respect to M1
v(1)=velocity of M1 with respect to the earth
v(2)=velocity of M2 with respect to the earth

Homework Equations



I know that this is a conservation of momentum problem, so I used the equations. E(initial)=E(final) and P(initial)=P(final)

The Attempt at a Solution



By using conservation of energy, I found the velocity of the mass M2 with respect to M1 as it leaves the circular path. Since it falls a distance R, I set

(M2)gR=m[v(21)]^2/2

And found v(21)=SQRT(2gR)

Since the system began with no momentum and momentum must be conserved:

0=(M2)(V2)+(M1)(V1)

What I'm trying to find is V2, I know that V2=V21+V1, so substituting V1=V2-V21 into the momentum equation...

0=(M2)(V2)+(M1)(V2-V21)=(M2)(V2)+(M1)(V2-[SQRT(2gR)])

Now the final answer I got was that V2=M1[SQRT(2gR)]/(M1+M2).

But the answer hint was that if M1=M2, that V2=SQRT(gR) which wasn't what my answer would get.
 
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just kidding, I got it.
 

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