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 will oppose the relative motion. For 1 kg block friction support themotion `&` for 2 kg friction oppose the motion. Let common velocity be v then
for `1kg v=2+a_(1)t` where `a_(1)=(mu(1g))/(2)=(0.4xx10)/(2)=2 ms^(-2)rArr2+4t=8-2trArr6t=5rArrt=1s`
(ii) `v=2+4t=2+4xx1=6ms^(-1)`
(iii) Displacement of 1kg block from initial point `s=ut+(1)/(2)at^(2)rArrs_(1)=2xx1+(1)/(2)xx4xx1^(2)=2+2=4m`
Displacement of 2kg block from intial point `s=ut+(1)/(2)at^(2)rArrs_(2)=8xx1-(1)/(2)xx2xx1^(2)=8-1=7m`