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BITSAT 2015 Paper

Section: Physics

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Question : 9 of 150

Marks: +1, -0

The masses of blocks A and B are m and M respectively. Between A and B, there is a constant frictional force F and B can slide on a smooth horizontal surface. A is set in motion with velocity while B is at rest. What is the distance moved by A relative to B before they move with the same velocity?

$\frac{mMv_0^2}{F(m-M)}$
$\frac{mMv_0^2}{2F(m-M)}$
$\frac{mMv_0^2}{F(m+M)}$
$\frac{mMv_0^2}{2F(m+M)}$

Solution:

For the blocks A and B FBD as shown below

Equations of motion

a_A

=

\frac{F}{M}

(in x direction)

a_B

=

\frac{F}{M}

(in x direction)

Relative acceleration, of A w.r.t. B,

a_{A,B}

=

a_A - a_B

= -

\frac{F}{m} - \frac{F}{M}

= - F

\left(\frac{M+m}{Mm}\right)

(along – x direction)

Initial relative velocity of A w.r.t. B,

u_{AB}

=

v_0

using equation

v^2

=

u^2

+ 2as

0 =

v_0^2

-

\frac{2F(m+M)}{Mm}

⇒ S =

\frac{Mmv_0^2}{2F(m+M)}

i.e., Distance moved by A relative to B

S_{AB}

=

\frac{Mmv_0^2}{2F(M+m)}

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