Systems and Center of Mass Practice

Sample questions

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1. Sample 1difficulty 2/5

   Three masses on the x-axis: $1~\text{kg}$ at $x = 0$, $2~\text{kg}$ at $x = 4~\text{m}$, $1~\text{kg}$ at $x = 8~\text{m}$. The center of mass is at

   • A

     $6~\text{m}$

   • B

     $2~\text{m}$

   • C

     $8/3~\text{m}$

   • D

     $4~\text{m}$

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   Why

   $x_\text{cm} = (1 \cdot 0 + 2 \cdot 4 + 1 \cdot 8)/4 = 16/4 = 4~\text{m}$.

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2. Sample 2difficulty 2/5

   Masses $m_1 = 1~\text{kg}$ at $x_1 = 0$ and $m_2 = 3~\text{kg}$ at $x_2 = 4~\text{m}$. Where is the center of mass?

   • A

     $4~\text{m}$

   • B

     $2~\text{m}$

   • C

     $1~\text{m}$

   • D

     $3~\text{m}$

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   Why

   $x_{\text{cm}} = (m_1 x_1 + m_2 x_2)/(m_1 + m_2) = (0 + 12)/4 = 3~\text{m}$.

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3. Sample 3difficulty 3/5

   Two masses on a frictionless surface push apart via an internal spring. The center of mass of the two-mass system

   • A

     Accelerates

   • B

     Moves at constant velocity

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   • C

     Cannot be determined

   • D

     Stops

   Why

   Internal forces cannot change the system's momentum. The CM moves at its initial constant velocity (zero if the system was at rest).

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4. Sample 4difficulty 3/5

   A $60~\text{kg}$ person stands at one end of a $120~\text{kg}$, $4.0~\text{m}$ long boat at rest in still water. He walks to the other end. Ignoring water resistance, the boat shifts (relative to water) by

   • A

     $1.0~\text{m}$ opposite to walking direction

   • B

     $2.0~\text{m}$ opposite

   • C

     $1.3~\text{m}$ opposite to walking direction

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   • D

     $0$

   Why

   Center of mass doesn't move. Person shifts $+\Delta x_p$, boat shifts $-\Delta x_b$. With $60(\Delta x_p) = 120(\Delta x_b)$ and $\Delta x_p + \Delta x_b = 4.0$: $\Delta x_b = 4.0 \cdot 60/(60+120) \approx 1.33~\text{m}$.

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5. Sample 5difficulty 4/5

   Two stars of masses $3m$ and $m$ are separated by a distance $d$. How far is the center of mass from the heavier star?

   • A

     $d/4$

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   • B

     $d/2$

   • C

     $d/3$

   • D

     $3d/4$

   Why

   $x_{cm} = (m \cdot d)/(3m + m) = d/4$ from the $3m$ star.

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