Question

For a spring block system as shown in figure, a time varying force $F=5t\text{N}$ is applied on mass $2\text{kg}$. After $10\text{seconds}$, velocity of $3\text{kg}$ mass is $30\text{m/s}$. Find the velocity of $2\text{kg}$ mass at this instant.

• A. $40\text{m/s}$
• B. $80\text{m/s}$
• C. $\frac{80}{3}\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is B $80\text{m/s}$

From the fundamental definition:
$F_{\text{ext}}=m{a}_{\text{com}}....\left(i\right)$
Considering (blocks+spring) as system $F_{\text{ext}}=5t\text{N}$.
Spring force is an internal force for the system, hence velocity of centre of mass at that instant can be given as,
$5t=\left(m_{\text{sys}}\right)\frac{d{v}_{\text{com}}}{dt}$
Integrating w.r.t time with limits $t=0$ to $t=10\text{s}$
$\left(m_{\text{sys}}\right){\int }_0^{v_{\text{com}}}d{v}_{\text{com}}={\int }_0^{10}5tdt$
$\Rightarrow (3+2)v_{\text{com}}={\left[\frac{5t^2}{2}\right]}_0^{10}$
$\therefore v_{\text{com}}=\frac{5}{10}\times ({10}^2-0)$
$=50\text{m/s}...\left(ii\right)$

Considering both blocks to be moving along direction of force $F$, at $t=10\text{s}$, velocity of $3\text{kg}$ mass, $v_1=30\text{m/s}$ and let velocity of $2\text{kg}$ mass be $v_2\text{m/s}$.
$\Rightarrow v_{\text{com}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}...\left(iii\right)$
From Eq. $\left(ii\right)$ and $\left(iii\right)$,
$50=\frac{(3\times 30)+(2\times v_2)}{3+2}$
$\therefore v_2=\frac{250-90}{2}=80\text{m/s}$