Center of Mass on a Plane — Practice Problem Solutions
1. Equal masses at the same height: $x_{cm}=\dfrac{(3)(2)+(3)(8)}{6}=\dfrac{30}{6}=5\text{ m}$; $y_{cm}=4\text{ m}$ (both balls share that height). COM $=(5\text{ m},4\text{ m})$.
2. At the geometric center — where the two symmetry lines cross. No calculation is needed because the symmetry guarantees the mass balances equally in every direction from that single point.
3. $x_{cm}=\dfrac{(2)(0)+(4)(6)+(6)(3)}{2+4+6}=\dfrac{0+24+18}{12}=3.5\text{ m}$. $y_{cm}=\dfrac{(2)(0)+(4)(0)+(6)(9)}{12}=\dfrac{54}{12}=4.5\text{ m}$. COM $=(3.5\text{ m},4.5\text{ m})$.
4. The system's center of mass shifts toward the 80 cm mark (the direction the block moves). It must shift by less than 70 cm because the block is only part of the total mass — the system's shift equals $\left(\dfrac{m_{block}}{m_{total}}\right)\times70\text{ cm}$, a fraction of the block's own 70 cm displacement, not the whole thing.
5. In both systems, the two masses are equal and placed symmetrically on opposite sides of the origin — the calculation is identical either way, just with $x$ and $y$ swapped. Both center of mass are at the origin, $(0\text{ m},0\text{ m})$.
6. a) $x_{cm}=\dfrac{(2)(0)+(3)(10)+(5)(5)}{10}=\dfrac{0+30+25}{10}=5.5\text{ m}$. $y_{cm}=\dfrac{(2)(0)+(3)(0)+(5)(20)}{10}=\dfrac{100}{10}=10\text{ m}$. COM $=(5.5\text{ m},10\text{ m})$. b) The classmate took the plain average of the three positions ($x=\frac{0+10+5}{3}=5$, $y=\frac{0+0+20}{3}\approx6.7$) and forgot to weight by mass. Drone C is the heaviest (5 kg) and sits at $y=20\text{ m}$, so it pulls the true $y_{cm}$ well above the naive estimate, up to 10 m.
7. a) Without calculating: the 12 kg sphere is heavier and moves toward lower $x$ (from 10 m to 8 m), so it pulls the system's center of mass to the left (toward lower $x$) as well. b) Before: $x_{cm}=\dfrac{(4)(0)+(12)(10)}{16}=\dfrac{120}{16}=7.5\text{ m}$. After: $x_{cm}=\dfrac{(4)(2)+(12)(8)}{16}=\dfrac{104}{16}=6.5\text{ m}$. The center of mass shifted left, from 7.5 m to 6.5 m — confirming the prediction.
8. $x_{cm}=\dfrac{(3)(0)+(9)(8)}{12}=\dfrac{72}{12}=6\text{ m}$; $y_{cm}=\dfrac{(3)(2)+(9)(6)}{12}=\dfrac{60}{12}=5\text{ m}$. COM $=(6\text{ m},5\text{ m})$ — (B).