Blocks are given velocity in direction shown in figure. Co-efficient of friction between two blocks

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4-2.Friction

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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 $g = 10\ m/s^2$

(considering $4\ kg$ block doesn't fall on ground)

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 fstatements is/are correct

A

$i, iii, iv$

B

$ii, iv$

C

$ii, iii, iv$

D

$i, iv$

Solution

$a_{2}=-\frac{20}{4}=-5 \mathrm{m} / \mathrm{s}^{2}$

$a_{1}=\frac{20}{2}=+10 \mathrm{m} / \mathrm{s}^{2}$

Relative motion stopped when

$\mathrm{v}_{1}=\mathrm{v}_{2}$ $-6+10 t=12-5 t$

$t=\frac{18}{15}=1.2 \sec$

common velocity $=-6+10 t$

$=-6+10 \times 1.2=6 \mathrm{m} / \mathrm{s}$

Displacement of $4 \mathrm{kg}$ block

$=\frac{12 \times 6}{5}-\frac{1}{2} \times 5 \times\left(\frac{6}{5}\right)^{2}=\frac{72}{5}-\frac{36}{10}=10.8 \mathrm{m}$

Standard 11

Physics

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