Question

Both the blocks as shown in the given arrangement are given together a horizontal velocity towards right. If $a_{cm}$ is the subsequent acceleration of the centre of mass of the system of blocks, then $a_{cm}$ equals [Take $g=10{\text{m/s}}^2$]

• A. $1{\text{m/s}}^2$
• B. $2{\text{m/s}}^2$
• C. $5{\text{m/s}}^2$
• D. $0.5{\text{m/s}}^2$

Answer 1

The correct option is A $1{\text{m/s}}^2$

Assuming the blocks move together as a system, we know that, acceleration of centre of mass

$a_{cm}=\frac{\text{Net external force}}{\text{Total mass}}$

Here, only external force on the system of blocks is friction from the ground.
Coefficient of friction between ground and the blocks be $\mu$.

Hence, $a_{cm}=\frac{\mu \times (2+1)g}{(2+1)}$

$=\mu g=1{\text{m/s}}^2$

Then, required friction force on mass $2\text{kg}=2\times 1=2\text{N}$ which is less than the maximum possible friction between the blocks ($={\mu }^{\prime }\times 2g=0.2\times 2g=4\text{N}$
Hence, blocks will move as a system and acceleration of COM of blocks, $a_{CM}=1{\text{m/s}}^2$.