A block of mass `m` is pushed with a velocity `v_(0)` along the surface of a trolley car of mass `M`. If the horizontal ground is smooth and the coefficient of kinetic friction between the block of plank is `mu`. Find the minimum distance of relative sliding between the block and plank.

Explanation

Method 1: Using ground frame method
decide system: block and plank
Observation: No external force acting on system nn horizontal direction
Conclusion linear momentum of the system will be conserved.
As friction in kinetic nature and system.
when relative sliding between block and troley stops, velocity `V`.
Problem solving: Using conservation of linear momentum in horizontal direction.
` mv_(0)+0=(m+M)VimpliesV=(mv_(0))/((m+M)`.........i
now using WE theorem from ground frame.
`W_(ext)+w_(int)=/_\K`
`W_(ext)=0`
`W_(int)=W_("friction")=-fx=-mumgx`
`0+(-mumgx)=(1/2(m+M)V^(2)-1/2mv_(0)^(2))`..........ii
From i and ii relative distance of sliding
`x=v_(0)^(2)/(2(1+m/M)mug)`
Method 2: work energy thoerem from centre of mass frame
Considering the system from `CM` frame
`W_(ext)+W_("int") =(/_\K)_(cm)=(k_(f)-k_(i))_(cm)`
Initial kinetic energy of system w.r.t `CM`
`(K_(i))_(cm)=1/2(mM)/((m+M) v_(0)^(2)`
Final kinetic energy of sytem w.r.t `CM`
`(K_(f))_(cm)=0` or `|(V_("rel"))f|=0`
hence `0+(-mumgx)=0-1/2(mM)/((m+M))v_(0)^(2)`
`impliesx=v_(0)^(2)/(2(1+m/M)mug`