Conservation of Linear Momentum and Energy in a Two-Block System

[Hamiltonian]

Homework Statement

The figure shows a block A of mass 9m having a smooth semicircular groove of radius R placed on a smooth horizontal surface. A block B of mass m is released from a position in groove where its radius is horizontal. Find the speed of the bigger block when the smaller block reaches its bottom-most position.

Relevant Equations

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since there is no external force in the x-direction linear momentum can be conserved. Hence I get the equation $$0 = mv^2 - 9mV^2$$
where $v$ is the velocity of B towards the right and $V$ is the velocity of A toward the left.
also the conservation of energy gives $$1/2(9m)V^2 + (1/2)mv^2 = mgR$$
solving these two equations yields $V = (\frac {gR}{45})^{1/2}$ which differs quite a bit from the correct answer. Also, I thought maybe I would have to account for the work done by the normal reaction between B and A but shouldn't they get canceled out is that assumption wrong?

 

[LCSphysicist]

Momentum is mv, and not mv²

 

[Hamiltonian]

Momentum is mv, and not mv²

sorry that's a typo my bad
I actually solved it using mv so the rest of the steps should be mathematically correct

 

[LCSphysicist]

sorry that's a typo my bad
I actually solved it using mv so the rest of the steps should be mathematically correct

Did you consider the fact that the little block apparently has a velocity1, but that velocity1 would can be actually a composition of its velocity and the big block velocity? Like a boat in a river?

 

[Hamiltonian]

Did you consider the fact that the little block apparently has a velocity1, but that velocity1 would can be actually a composition of its velocity and the big block velocity? Like a boat in a river?

the $v$ and $V$ that I have used are both with respect to the ground and since I conserve momentum and energy from the ground frame it shouldn't be a problem?

 

[LCSphysicist]

the $v$ and $V$ that I have used are both with respect to the ground and since I conserve momentum and energy from the ground frame it shouldn't be a problem?

Oh yes I got confused in the frame, I also found your answer.

 

[Hamiltonian]

Oh yes I got confused in the frame, I also found your answer.

the correct answer is supposed to be $$V = ( \frac {5}{4} gR)^{1/2}$$
but I don't get understand what I am doing wrong

 

[haruspex]

the correct answer is supposed to be $$V = ( \frac {5}{4} gR)^{1/2}$$
but I don't get understand what I am doing wrong

It's like 1/45 became 1/4 5. Latex error?

 

[Hamiltonian]

It's like 1/45 became 1/4 5. Latex error?

the answer I have got is $$V = (\frac{gR}{45})^{1/2}$$
the correct answer is $$V = (\frac{5gR}{4})^{1/2}$$
so I think it might not be a LaTeX error

 

[Hamiltonian]

this is the only solution I could find. but i don't get what $v_{0}$ is and how they got that equation for $\Delta K_{cm}$

 

[haruspex]

the answer I have got is $$V = (\frac{gR}{45})^{1/2}$$
the correct answer is $$V = (\frac{5gR}{4})^{1/2}$$
so I think it might not be a LaTeX error

@LCSphysicist and I get the same answer as you do.
The book answer is obviously wrong. It would mean the block has KE $\frac{45}8mgR$, when the total energy available is only mgR.

I was trying to understand how the book answer could be 5/4 instead of 1/45. Sometimes it can be explained as a half-updated exercise, where some parameters have been changed but the answer not updated. In the present case, there is no rational ratio of the masses that would give 5/4.
Then it occurred to me that if someone had handwritten 1/45 more like 1/4 5, such that it was misread as $\frac 14 5$ by a typist, it could later have been typeset as $\frac 54$. A longshot, but I couldn't think of another explanation.

 

[Hamiltonian]

this is the only solution I could find. but i don't get what $v_{0}$ is and how they got that equation for $\Delta K_{cm}$

@haruspex I attached the solution in the book above

 

[haruspex]

@haruspex I attached the solution in the book above

The error is clear in the first line. It should read $v_1=9v_2$, not $v_1=\frac{v_0}3$.
With that correction it gives your answer.

 

[Hamiltonian]

I wanted to solve the same problem from the CoM frame and the answer I am getting is different:/

in the Com frame the kinetic energy of a two-block system is given by $$KE = \frac{1}{2} \mu V_{relative}^2$$
where $\mu$ is the reduced mass($\mu = \frac {m1m2}{m1+ m2}$) and $V_{relative}$ is the relative velocity between the two blocks.

by applying conservation of linear momentum we know $v = 9V$
and by applying COE $$\frac{1}{2}\mu(v - V)^2 = mgR$$
solving this gives $$V = \frac {(5gR)^{1/2}}{12}$$

so am I making an error in applying COE from the CoM frame?

 

[haruspex]

I wanted to solve the same problem from the CoM frame and the answer I am getting is different:/

in the Com frame the kinetic energy of a two-block system is given by $$KE = \frac{1}{2} \mu V_{relative}^2$$
where $\mu$ is the reduced mass($\mu = \frac {m1m2}{m1+ m2}$) and $V_{relative}$ is the relative velocity between the two blocks.

by applying conservation of linear momentum we know $v = 9V$
and by applying COE $$\frac{1}{2}\mu(v - V)^2 = mgR$$
solving this gives $$V = \frac {(5gR)^{1/2}}{12}$$

so am I making an error in applying COE from the CoM frame?

The PDF solution had v and V measured with opposite sense, so the relative velocity was V+v. The only error in it was the one I quoted in post #13.