Three blocks $\mathrm { A } , \mathrm { B }$ 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 initial speed $v$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with $C$, also perfectly inelastically . $\frac { 5 } { 6 }$ th of the initial kinetic energy is lost in the whole process. What is the value of $M / m$ ?
(1) 3
(2) 4
(3) 5
(4) 2

Blocks of masses $\mathrm { m } , 2 \mathrm {~m} , 4 \mathrm {~m}$ and 8 m are arranged in a line of a frictionless floor. Another block of mass m , moving with speed $v$ along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8 m starts moving the total energy loss is $\mathrm { p } \%$ of the original energy. Value of 'p' is close to:
(1) 77
(2) 94
(3) 37
(4) 87

Three objects $A , B$ and $C$ are kept in a straight line on a frictionless horizontal surface. The masses of $A , B$ and $C$ are $m , 2 m$ and $2 m$ respectively. $A$ moves towards $B$ with a speed of $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and makes an elastic collision with it. Thereafter $B$ makes a completely inelastic collision with $C$. All motions occur along the same straight line. The final speed of $C$ is:
(1) $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(3) $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$