The blocks are given velocity in the direction shown in figure. Co-efficient of friction between two blocks shown in figure is $μ = 0.5$ . Take $g = 10 m / s 2$

Consider the following statements-
(i) Time when relative motion between them is stopped is 1.4 second.
(ii) Time when relative motion between them is stopped in 1.2 second.
(iii) The common velocity of the two blocks is 8 m/s, towards right.
(iv) The displacement of the 4 kg block when relative motion stopped is 10.8 m.
Which of the following option is correct-

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i, iii, iv

ii, iv

ii, iii, iv

i, iv

answer is B.

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Detailed Solution

$a 2 = − 204 = − 5 m / s 2$

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$a 1 = 202 = + 10 m / s 2$
Relative motion stopped when $V 1 = V 2$
$− 6 + 10 t = 12 − 5 t t = 1815 = 1.2 sec$
common velocity $= − 6 + 10 t$
$= − 6 + 10 × 1.2 = 6 m / s$
Displacement of 4kg block
$= 12 × 6 5 − 12 × 5 × 652 = 725 − 3610 = 10.8 m$