Coefficient of friction between two blocks shown inFig. 7.203 is mu = 0.4 . The blocks are given velocities of 2 m/s and 8 m/s in the directions shown in figure . The timewhen relative motion between them will stop is ( in sec) . :

Coefficient of friction between two blocks shown inFig. 7.203 is mu =

1.	Coefficient of friction between two blocks shown inFig. 7.203 is mu = 0.4 . The blocks are given velocities of 2 m/s and 8 m/s in the directions shown in figure . The timewhen relative motion between them will stop is ( in sec) . :
Answer» Solution :Using principle of conservation of MOMENTUM: `1 XX (2)+ 2 xx (8) = 3V implies v = 6 m//s ` For 1 kg block , `f = mu m_(1) g= 0.4 XX1 xx 10 = 4 N` `THEREFORE "" a = (f)/(m_(1)) = (4)/(1) = 4 ms^(-2)` Using , `"" v_("rel") = u_("rel") + at ` 6= 2 + (4) t t = 1 second