Question

Two blocks with masses $m_1=3\text{kg}$ and $m_2=4\text{kg}$ are touching each other on a fricionless surface. If a force of $5\text{N}$ is applied on $m_1$ as shown in figure. The acceleration of each block is

• A. $a_1=\frac{5}{7}{\text{m/s}}^2,a_2=\frac{5}{4}{\text{m/s}}^2$
• B. $a_1=\frac{5}{3}{\text{m/s}}^2,a_2=\frac{5}{4}{\text{m/s}}^2$
• C. $a_1=\frac{5}{4}{\text{m/s}}^2,a_2=\frac{5}{3}{\text{m/s}}^2$
• D. $a_1=\frac{5}{7}{\text{m/s}}^2,a_2=\frac{5}{7}{\text{m/s}}^2$

Answer 1

The correct option is D $a_1=\frac{5}{7}{\text{m/s}}^2,a_2=\frac{5}{7}{\text{m/s}}^2$

By the FBD of $m_1$:
$5-R=m_{1}a...\left(i\right)$ where $R$ is the reaction on the block $m_1$ by the block $m_2$

By the FBD of $m_2$:
$R=m_{2}a....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$ we get and $R_1$ and $R_2$ does not contribute to horizontal accelerations.
$5=(m_1+m_2)a$
$\Rightarrow a=\frac{5}{3+4}=\frac{5}{7}{\text{m/s}}^2$ both the blocks move together with the same acceleration

Hence option D is the correct answer