Physics - Center of Mass Question with Solution | TestHub

For $0\le t<2\text{ s}$, only the block of mass $M=10\text{ kg}$ is present. The acceleration of the COM is $a_{CM}=F/M=30/10=3{\text{ m/s}}^2$. The displacement of COM at $t=2\text{ s}$ is $x_{CM}(2)=\frac{1}{2}{a}_{CM}{t}^2=\frac{1}{2}(3)(2^2)=6\text{ m}$. The velocity of COM at $t=2\text{ s}$ is $v_{CM}(2)=a_{CM}t=3(2)=6\text{ m/s}$. At $t=2\text{ s}$, the object of mass $m=5\text{ kg}$ is added. Since it is dropped vertically, it imparts no horizontal impulse. The total mass becomes $M^{\prime }=M+m=15\text{ kg}$. For $2\text{ s}\le t\le 4\text{ s}$, the acceleration of the COM is $a_{CM}^{\prime }=F/M^{\prime }=30/15=2{\text{ m/s}}^2$. The displacement from $t=2\text{ s}$ to $t=4\text{ s}$ (duration $\Delta t=2\text{ s}$) is $\Delta x^{\prime }=v_{CM}(2)\Delta t+\frac{1}{2}{a}_{CM}^{\prime }(\Delta t{)}^2=6(2)+\frac{1}{2}(2)(2^2)=12+4=16\text{ m}$. The total displacement of the COM from $t=0$ to $t=4\text{ s}$ is $x_{CM}(4)=x_{CM}(2)+\Delta x^{\prime }=6+16=22\text{ m}$. The correct answer is 22.