The masses of the blocks A and B are m and M, respectively. Between A and B there is a constant frictional force F, and B can slide frictionlessly on horizontal surface. A is set in motion with velocity v0 while B is at rest. What is the distance moved by A relative to B before they move with the same velocity?
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a
mMv02F(m−M)
b
mMv022F(m−M)
c
mMv02F(m+M)
d
mMv022F(m+M)
answer is D.
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Detailed Solution
Free-body diagrams (figure)Equations of motion: aB=FM( in +x direction )aA=Fm( in −x direction )Relative acceleration of A w.r.t. B : a→A,B=a→A−a→B=−Fm−FM=−Fm+MmM (along −x direction) Initial relative velocity of A w.r.t. B uA,B=v0Final relative velocity of A w.r.t. B = 0Using v2=u2+2as 0=v02−2F(m+M)mMS ⇒ S=Mmv022F(m+M)