JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Conservation of Momentum

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 initial speed

v

towards B due to which it collides with B perfectly inelastically. The combined mass collides with C , also perfectly inelastically if

\frac{5}{6}

th of the initial kinetic energy is lost in whole process. What is value of

M / m

Answer: (C)

Solution

Work through the two perfectly inelastic collisions in turn. When

A

hits

B

, momentum conservation gives the combined speed

v_{1} = \frac{v}{2}

. When that pair of mass

2 m

then hits

C

(2 m + M) v_{2} = 2 m \cdot \frac{v}{2} = m v

, so

v_{2} = \frac{m v}{2 m + M}

. The fraction of kinetic energy surviving is

\frac{\frac{1}{2} ( 2 m + M ) v _{2}^{2}}{\frac{1}{2} m v ^{2}} = \frac{m}{2 m + M}

. Since

\frac{5}{6}

of the energy is lost, only

\frac{1}{6}

survives, so

\frac{m}{2 m + M} = \frac{1}{6}

, which rearranges to

M = 4 m

and gives

\frac{M}{m} = 4

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option