Question

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

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

Answer 1

The correct option is C $6$ $m/s$

Acceleration of $3$kg:
$a_2=\frac{6}{3}=2m/s^2$
Velocity of $3$kg at $t=5s$: $v_2=u_2+a_{2}t=0+2\times 5=10$ m/s
$v_{CM(t-5s)}=\frac{2\times 0+3\times 10}{2+3}=6$m/s.
Method-$2$:
$a_{CM}=\frac{F}{m_1+m_2}=\frac{6}{2+3}=1.2m/s^2$
$v_{CM}=u_{CM}+a_{CM}t=0+1.2\times 5=6$m/s