Momentum and relative distance

IIK*JII

1. The problem statement, all variables and given/known data
The attached figure below shows object A (mass,m), which is placed on a table, and wagon B (mass,M), which is in contact with the table. The top of the table and the top of the wagon are at the same height. A is made to slide on the table so that it transfers to the top of B with speed v₀. At that instant, A begins sliding on the top of B, and B begins to move on the floor. Shortly afterwards, A comes to rest with respect to B, and A and B then travel with a constant speed. The coefficient of kinetic friction between A and the top of B is μ'. B moves smoothly on the floor. The size of A is negligible.

What is the distance from the left side of B to the point where A came to rest with respect to B?


2. Relevant equations
Non-conservation energy due to friction between A and B;
ƩE₂-ƩE₁=Work ...(1)

Momentum conserve(non elastic collision)
ƩP₁=ƩP₂ ..(2)

S_A/B=S_A-S_B ..(3)


3. The attempt at a solution

I) When A slides on B surface, I find S_A by using (1)
0-[itex]\frac{1}{2}[/itex]m_av[itex]^{2}_{0}[/itex]=-μ'm_agS_a

∴S_A=[itex]\frac{v^{2}_{0}}{2gμ'}[/itex] ...(4)
II) A and B move together after collision
From (2), I got the velocity(v') after collision
v'=mv₀/m+M

Using (1) again to find displacement of B (B moves by friction of A and B)
Thus, -[itex]\frac{1}{2}[/itex](m+M)([itex]\frac{mv^{2}_{0}}{m+M}[/itex])^2 = -μ'mgS_B

∴S_B=[itex]\frac{mv^{2}_{0}}{2μ'g(m+M)}[/itex]
and S_A/B= S_A-S_B =[itex]\frac{Mv^{2}_{0}}{2μ'g(M+m)}[/itex]

Is it right?? Who can tell me that my procedure to solve this problem is correct??

help is appreciate :))
Thanks a lot

Attached Thumbnails

 

TSny

Can you explain why you set the final kinetic energy of block A equal to 0? Relative to the earth, block A never comes to rest. Seems to me that you need to stick to one frame of reference. So, if you say that the initial velocity of A is v_o, then you are using the earth frame of reference. If you stay in the earth frame of reference, the final velocity of A will not be zero.

Also, I don't understand why you're lumping the masses together when finding the distance B moves. Shouldn't you just use the mass of B alone for M in [itex]\frac{1}{2}[/itex]Mv'²?

However, when I work the problem I get the same final answer as you do! So, maybe your method is ok. But I don't follow it.

IIK*JII

TSny

I think I misunderstood with this problem because I think velocity of A = 0 relative to the Earth so I set final kinetic energy of A = 0. Because of this I understand that 2 masses (A and B) stick together too...

What's the equation did you set???

TSny

In the earth frame block A has a final velocity of

v' = [itex]\frac{mvo}{m+M}[/itex]

So, it seems to me that in your first equation under "The attempt of a solution" your zero should be replaced by the kinetic energy of block A when it has its final velocity. Does that seem right to you?

When setting up the similar equation for block B, I think you should use just the mass M of block B rather than (m+M) when setting up the final kinetic energy of block B: [itex]\frac{1}{2}[/itex]Mv'². Does that also seem right?

If I make these changes and work through the algebra, I get different answers than you did for S_a and S_b. But I get the same answer as you did for the difference: S_a - S_b! I find that kind of amazing.

IIK*JII

In block B: Did you set equation of energy like this

[itex]\frac{1}{2}[/itex]M(v')²=μmgS_B

v'=[itex]\frac{mv₀}{m+M}[/itex]

∴S_B=[itex]\frac{mMv^{2}_{0}}{2μg(m+M)^2}[/itex]
and from your mentioned in #2, I get S_A=v[itex]^{2}_{0}[/itex](mM+M²)/2μg(m+M)²

but when I find S_A/B I can't eliminate M² term....

TSny

I get (2mM + M²) rather than (mM + M²) in the numerator of the expression for S_A. I agree with your expression for S_B.

IIK*JII

now I get S_A like you get :) This is from my wrong calculation

but when find S_A/B, I can't manage the answer as you can :((

IIK*JII

Oh!!! now I can do it!!!

Thank you :)