# Momentum and relative distance

1. Jun 17, 2012

### 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)

SA/B=SA-SB ..(3)

3. The attempt at a solution

I) When A slides on B surface, I find SA by using (1)
0-$\frac{1}{2}$mav$^{2}_{0}$=-μ'magSa

∴SA=$\frac{v^{2}_{0}}{2gμ'}$ ...(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, -$\frac{1}{2}$(m+M)($\frac{mv^{2}_{0}}{m+M}$)^2 = -μ'mgSB

∴SB=$\frac{mv^{2}_{0}}{2μ'g(m+M)}$
and SA/B= SA-SB =$\frac{Mv^{2}_{0}}{2μ'g(M+m)}$

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

help is appreciate
Thanks a lot

#### Attached Files:

• ###### momentum2.JPG
File size:
13.5 KB
Views:
60
2. Jun 18, 2012

### 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 vo, 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 $\frac{1}{2}$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.

3. Jun 18, 2012

### 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???

4. Jun 18, 2012

### TSny

In the earth frame block A has a final velocity of

v' = $\frac{mvo}{m+M}$

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: $\frac{1}{2}$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 Sa and Sb. But I get the same answer as you did for the difference: Sa - Sb! I find that kind of amazing.

5. Jun 18, 2012

### IIK*JII

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

$\frac{1}{2}$M(v')2=μmgSB

v'=$\frac{mv0}{m+M}$

∴SB=$\frac{mMv^{2}_{0}}{2μg(m+M)^2}$
and from your mentioned in #2, I get SA=v$^{2}_{0}$(mM+M2)/2μg(m+M)2

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

6. Jun 18, 2012

### TSny

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

7. Jun 19, 2012

### IIK*JII

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

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

8. Jun 19, 2012

### IIK*JII

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

Thank you :)