- #1
IIK*JII
- 21
- 0
Homework Statement
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?
Homework 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)
The Attempt at a Solution
I) When A slides on B surface, I find SA by using (1)
0-[itex]\frac{1}{2}[/itex]mav[itex]^{2}_{0}[/itex]=-μ'magSa
∴SA=[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'mgSB
∴SB=[itex]\frac{mv^{2}_{0}}{2μ'g(m+M)}[/itex]
and SA/B= SA-SB =[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