Question

Two identical blocks $\text{A}$ and $\text{B}$ of mass $1\text{kg}$ are joined together with a massless spring and are placed on a smooth surface as shown in the figure. If the block $\text{A}$ moves with an acceleration of $3{\text{m/s}}^2$, then the acceleration of the block $\text{B}$ is:

• A. $4{\text{m/s}}^2$
• B. $1{\text{m/s}}^2$
• C. $3{\text{m/s}}^2$
• D. $2{\text{m/s}}^2$

Answer 1

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

Considering the two blocks $\text{A}$ and $\text{B}$ as a system, $5\text{N}$ force is the external force acting on the system.

From the concept of centre of mass, we can find the acceleration of c.o.m by using the second law of motion,

$a_{com}=\frac{F}{m_{total}}$

where, $m_{total}$ is the total mass of the system.

$\Rightarrow F=m_{\text{total}}\times a_{com}\Rightarrow 5=m_{\text{total}}\times a_{com}......\left(1\right)$
(Given)

If $m_A$ is the mass of block $A$ and $m_b$ is the mass of block $B$ then acceleration $a_{com}$,

$a_{com}=\frac{m_A{a}_A+m_B{a}_B}{m_A+m_B}$

$\Rightarrow m_{\text{total}}\times a_{com}=m_A{a}_A+m_B{a}_B$

$\Rightarrow 5=1\times 3+1\times a_B$ [From $\left(1\right)$]

$\Rightarrow a_B=2{\text{m/s}}^2$

Alternate:
Tension in the spring $=1\times 3=3\text{N}$
Applying Newton's law on body B,
$5-3=1\times a$
$\Rightarrow a=2{\text{m/s}}^2$

$\begin{array}{c}\text{Why this question}?\\ \text{Concept -The concept of centre of mass converts a system}\\ \text{of many particles/bodies into a single particle/body}\\ \text{system. This makes our calculations easier.}\end{array}$