The friction coefficient between the horizont | vedclass

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Question

Retardation of $2 \mathrm{kg}$ block, $\mathrm{a}=\mu \mathrm{g}, 2 \mathrm{m} / \mathrm{s}^{2}$

velocity of $2 \mathrm{kg}$ block before collision

Vedclass · Accepted Answer

$5 $

Vedclass · Answer

Retardation of $2 \mathrm{kg}$ block, $\mathrm{a}=\mu \mathrm{g}, 2 \mathrm{m} / \mathrm{s}^{2}$

velocity of $2 \mathrm{kg}$ block before collision,

$\mathrm{u}_{1}=\sqrt{(1)^{2}-2(2)(0.16)}=0.6 \mathrm{m} / \mathrm{s}$

now $2(0.6)=2 v_{1}+4 v_{2}$ and $v_{2}-v_{1}=0.6$

$	herefore \quad \mathrm{v}_{1}=-0.2 \mathrm{m} / \mathrm{s} \quad$ and $\mathrm{v}_{2}=0.4 \mathrm{m} / \mathrm{s}$

$S_{1}=\frac{(0.2)^{2}}{2 	imes 2}=0.01$

$S_{2}=\frac{(0.4)^{2}}{2 	imes 2}=0.04$

$\mathrm{S}=\mathrm{S}_{1}+\mathrm{S}_{2}=0.05=5 \mathrm{cm}$

The friction coefficient between the horizontal surface and each of the block shown in figure is $0.2.$ The collision between the blocks is perfectly elastic. What is the separation between the blocks when they come to rest :- .............. $\mathrm{cm}$

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