Coefficient of friction between two blocks shown in Fig. 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 time when relative motion between them will stop is ( in sec) . :

Answer

3

Explanation

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 xx 1 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