JEE Main 20199 JanuaryMorning Shift

JEE Main 2019 · 9 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

Three blocks A, B and C are lying on a smooth horizontal surface,...

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 .

\dfrac{5}{6}\text{t}\text{h}

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

M / m

Options

B. $4$

mv = \left(m + m\right){v}^{′}

\left(m + m\right){v}^{′} = \left(m + m + M\right){v}_{f}

<br /><br />

∴

final velocity

{v}_{f} = \dfrac{m}{2m + M}v

<br /><br />Initial energy

= \dfrac{1}{2}m{v}^{2}

<br /><br />Final energy

= \dfrac{1}{2}\left(2m + M\right){\left(\dfrac{mv}{2m + M}\right)}^{2}

<br /><br />Given that

\dfrac{5}{6}\dfrac{1}{2}m{v}^{2} = \dfrac{1}{2}m{v}^{2} - \dfrac{1}{2}\left(2m + M\right){\left(\dfrac{mv}{2m + M}\right)}^{2}

<br /><br />

\dfrac{1}{12}m{v}^{2} = \dfrac{1}{2}\dfrac{{m}^{2}{v}^{2}}{\left(2m + M\right)}

<br /><br />

6m = 2m + M

<br /><br />

\dfrac{M}{m} = 4