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Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an brutal speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically $5\over6$th of the initial kinetic energy is lost in whole process. What is value of M/m ?

(1) 4 

(2) 5 

(3) 3 

(4) 2

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Ans: (1) 4

Sol: $k_i={1\over2}mv_0^2$

From linear momentum conservation

$mv_0=(2m+M)v_f$

$\implies v_f={mv_0\over2m+M}$

${k_i\over k_f}=6$

$\implies{{1\over2}mv_0^2\over{1\over2}(2m+M)\left({mv_0\over2m+M}\right)^2}=6$

$\implies{2m+M\over m}=6$

$\implies{M\over m}=4$

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Wow! Nice,very satisfying
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How ki by kf is equal to 6?

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