Category : JEE Main & Advanced
When two masses compressed towards each other and suddenly released then energy acquired by each block will be dissipated against friction and finally block comes to rest
i.e., F × S = E [Where F = Friction, S = Distance covered by block, E = Initial kinetic energy of the block]
\[\Rightarrow \] \[F\times S=\frac{{{P}^{2}}}{2m}\] [Where P = momentum of block]
\[\Rightarrow \] \[\mu mg\times S=\frac{{{P}^{2}}}{2m}\] [As F = m mg]
\[\Rightarrow \] \[S=\frac{{{P}^{2}}}{2\mu {{m}^{2}}g}\]
In the given condition P and \[\mu \] are same for both the blocks.
So, \[S\propto \frac{1}{{{m}^{2}}}\]; \[\therefore \] \[\frac{{{S}_{1}}}{{{S}_{2}}}={{\left[ \frac{{{m}_{2}}}{{{m}_{1}}} \right]}^{2}}\]
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