Question

A small block of mass $m=2kg$ is projected on a larger block of $20kg$ and of length $20\text{metres}$, with a velocity of $vm/s$ as shown in the figure. The coefficient of friction between the two blocks is $1.2$, while between the lower block and ground is $0.1$. The time taken by the small block to fall off the larger block is (Take $g=10m/s^2$)

• A. $2.5s$
• B. $4s$
• C. $3s$
• D. $2s$

Answer 1

The correct option is D $2s$

FBDs of the blocks are as shown.
Here, $R_1$ and $R_2$ are normal reaction forces on the respective blocks.


From the FBDs, we have,
$R_1=2g=20\text{N}$
$f_1={\mu }_1{R}_1=1.2\times 20=24\text{N}$

$R_2=R_1+20g=20+200=220\text{N}$
$f_2={\mu }_2{R}_2=0.1\times 220=22\text{N}$

$\because f_1>f_2$ so, upper block will retard and lower block will accelerate.

Retardation of upper block is
$a_1=\frac{f_1}{m}=\frac{24}{2}=12m/s^2$
and acceleration of lower block is
$a_2=\frac{f_1-f_2}{m}=\frac{24-22}{20}=0.1m/s^2$

$\therefore$ Relative retardation of upper block w.r.t lower block is
$a_{12}=a_1+a_2=12.1m/s^2$ (constant acceleration)

For minimum value of velocity, upper block will just come to rest as it falls from lower block
So, using 3rd law of motion, we can write
$v_{12}^2=u_{12}^2-2a_{12}{s}_{12}$
$\Rightarrow 0=v^2-2\times 12.1\times 20$
$\Rightarrow v^2=484\Rightarrow v=22m/s$

To find time taken, use 1st law of motion,
$v_{12}=u_{12}-a_{12}t$
$\Rightarrow 0=v-at=22-12.1t$
$\Rightarrow t=\frac{22}{12.1}=1.82s\approx 2s$