Coefficient of friction between two block shown in figure is `mu = 0.4`. The blocks are given velocities of `2 m//s` and `8 m//s` in the directions figure. Find

(a) The time when relative motion between them will stop
(b) the common velocities of blocks upto that instant. (c) Displacement of `1 kg` and `2 kg` block upto that instant `(g = 10 m//s^(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`