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(B) The frictional force between the blocks is $f = \mu N = 0.5 	imes 4 	imes 10 = 20\ N$.For the $4\ kg$ block (moving right at $12\ m/s$): $a_2 =

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$ii, iv$

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(B) The frictional force between the blocks is $f = \mu N = 0.5 	imes 4 	imes 10 = 20\ N$.For the $4\ kg$ block (moving right at $12\ m/s$): $a_2 = -f/m_2 = -20/4 = -5\ m/s^2$.For the $2\ kg$ block (moving left at $6\ m/s$): $a_1 = +f/m_1 = +20/2 = +10\ m/s^2$.Relative motion stops when $v_1 = v_2$:$-6 + 10t = 12 - 5t$$15t = 18 \implies t = 1.2\ s$. (Statement $ii$ is correct).Common velocity $v = 12 - 5(1.2) = 6\ m/s$ (towards the right). (Statement $iii$ is incorrect).Displacement of $4\ kg$ block: $s = ut + 0.5at^2 = 12(1.2) + 0.5(-5)(1.2)^2 = 14.4 - 3.6 = 10.8\ m$. (Statement $iv$ is correct).Thus,statements $ii$ and $iv$ are correct.

Similar Questions

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