Coefficient of friction between two blocks shown in figure is `mu=0.4` The blocks are given velocities of `2ms^(-1)` and `8ms^(-1)` in the directions shown in figure. Find
(i) The time when relative motion between them will stop.
(ii) The common velocities of blocks upto that instant.
(iii) Displacement of blocks upto that instant `(g=10ms^(2))`

Explanation

(i) Frictional force between two blocks eill oppose the relative motion. For 1 kg block friction support the motion & for 2 kg friction oppose the motion. Let common velocity by v then for `1kg v=2+a_(1)t"where"a_(1)=(mu(1g))/(1)=(0.4xx10)/(1)=4ms^(2)`
for `2kgv=8-a_(2)t"where"a_(2)=(mu(1g))/(2)=(0.4xx10)/(2)=2ms^(-2)implies2+4t=8 0 2timplies6t=6impliest=1s`
(ii) `v=2+4t=2+4xx1=6ms^(-1)`
(iii) Displacement of 1 kg block from rest
`s=ut+1/2at^(2)impliess_(2)=8xx1-1/2xx2xx1^(2)=8-1=7m`