Question

Block B of mass 2kg is placed on smooth
horizontal plane. Block A of mass 1kg is
placed on block B. The coefficient of fric-
tion between A and B is 0.40. The block A
is imparted a velocity 16m/s at t=0. The
time at which momentum of the two blocks
are equal (in second) (g = 10 m/s2) is
1) 4 2) 1 3) 2 4) 8​

Answer 1

Free Body Diagrams of blocks A and B are given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(0,0)(0,-30){2}{\circle*{1}}\multiput(0,0)(0,-30){2}{\vector(0,-1){10}}\multiput(0,0)(0,-30){2}{\vector(0,1){10}}\put(-4,1){\sf{B}}\put(1.5,-34){\sf{A}}\put(0,0){\vector(1,0){10}}\put(0,-30){\vector(-1,0){10}}\put(0,-30){\vector(1,0){20}}\put(2,-10){ \sf{m_Bg+R_b} }\put(1.5,10){ \sf{R_f} }\put(11,-0.5){ \sf{0.4R_b} }\put(2,-42){ \sf{m_Ag} }\put(1.5,-20){ \sf{R_b} }\put(-21,-30){ \sf{0.4R_b} }\put(21,-31){ \sf{16\ m\,s^{-1}} }\end{picture}$

$\sf{R_b}$ and $\sf{R_f}$ are reactions due to the other block and the floor respectively.

From FBD of B,

$\sf{\longrightarrow R_b=m_Bg}$

$\sf{\longrightarrow R_b=10\ N}$

The friction acting between the blocks,

$\sf{\longrightarrow f=0.4\,R_b}$

$\sf{\longrightarrow f=0.4\times10\ N}$

$\sf{\longrightarrow f=4\ N}$

The same frictional force is acting on each block. It is the net horizontal force acting on each block, since no other horizontal force is acting.

So the net horizontal acceleration of block A (opposite to friction),

$\sf{\longrightarrow a_A=-\dfrac{f}{m_A}}$

$\sf{\longrightarrow a_A=-\dfrac{4}{1}}$

$\sf{\longrightarrow a_A=-4\ m\,s^{-2}}$

and that of block B,

$\sf{\longrightarrow a_B=-\dfrac{f}{m_B}}$

$\sf{\longrightarrow a_B=-\dfrac{4}{2}}$

$\sf{\longrightarrow a_B=-2\ m\,s^{-2}}$

By second equation of motion, the velocity of the block A,

$\sf{\longrightarrow v_A=u_A+a_At}$

$\sf{\longrightarrow v_A=16-4t}$

and that of block B,

$\sf{\longrightarrow v_B=u_B+a_Bt}$

$\sf{\longrightarrow v_B=-0+2t}$

$\sf{\longrightarrow v_B=2t}$

Now let us find the time at which the momentum of the blocks become equal.

$\sf{\longrightarrow m_Av_A=m_Bv_B}$

$\sf{\longrightarrow 16-4t=2\times2t}$

$\sf{\longrightarrow 16-4t=4t}$

$\sf{\longrightarrow\underline{\underline{t=2\ s}}}$

Hence (3) is the answer.