Question

Two particles are shown in figure. At $t=0$, a constant force $F=6$N starts acting on the 3 kg mass. Find the velocity of the centre of mass of these particles at $t=5s$.

• A. $5m/s$
• B. $4m/s$
• C. $6m/s$
• D. $3m/s$

Answer 1

Answer: (C)

$6m/s$

We know that the total linear momentum of a system of partilas is equal to linar momentumm of center of man of the system.

$\Rightarrow (m_1+m_2){\vartheta }_{cm}=m_1{\vartheta }_1+m_2{\vartheta }_2$

At t= 5 sec,

${\vartheta }_1=0m/s$

${\vartheta }_2=u_2+at$

${\vartheta }_2=2\times 5=10m/s(\because u_2=0$ and $a=\frac{F}{m_2}=2m/s^2)$

$\Rightarrow 5\times {\vartheta }_{cm}=0+3\times 10$

$\Rightarrow {\vartheta }_{cm}=6m/s$