Momentum of a System (College Board AP® Physics 1: Algebra-Based): Revision Note

A rocket of mass $M$ and shuttle of mass $5M$ are initially traveling together to the right with velocity $v$, as shown in the diagram.

The shuttle separates from the rocket and continues to travel to the right with velocity $v_s$ while the rocket continues to travel to the right with velocity $v_r$, where $v_r>v$, as shown in the following diagram.

Which of the following is true for the velocity $v_{cm}$ of the center of mass of the system after the shuttle and rocket separate?

A      $v_{cm}=v_s$

B      $v_{cm}=v$

C      $v_{cm}<v$

D      $v<v_{cm}<v_r$

The correct answer is B

Answer:

Step 1: Analyze the scenario

• When traveling together, the velocity of the center of mass of the rocket and shuttle is

$v_{cm}=\frac{{\displaystyle \sum _i\left(m_i{\overrightarrow{v}}_i\right)}}{\sum m_i}=\frac{\left(M+5M\right)v}{\left(M+5M\right)}$

$v_{cm}=v$

• During the separation, no external forces act on the system, so there is no impulse on the system

Step 2: Eliminate incorrect options

• Since no external forces act on the system, the total momentum of the system must be constant

$\overrightarrow{p}=\left(M+\hspace{0.17em}5M\right){\overrightarrow{v}}_{cm}$

• The velocity of the center of mass was $v$ before the separation

• Therefore, it cannot be greater than $v$ after the separation

  • This eliminates option D

• It also cannot be less than $v$ after the separation

  • This eliminates option C

• After the separation, the velocity of the rocket $v_r$ was greater than $v$

• Therefore, the velocity of the shuttle $v_s$ must be less than $v$

  • This eliminates option A

Step 3: Deduce the correct option

• As the center of mass was moving with velocity $v$ before the separation, it will continue to move with velocity $v$

  • Therefore, option B is correct

The masses and velocities of three objects in a system are given in the table below.

The positions of the three objects in an xy plane are shown in the diagram.

For the system of three objects, determine:

(A) the position of the center of mass, ${\overrightarrow{x}}_{cm}$.

(B) the velocity of the center of mass, ${\overrightarrow{v}}_{cm}$.

Answer:

Part (A)

Step 1: Determine the xy coordinates of the positions of each object

• Object 1 $\left(m_1=1\mathrm{kg}\right)$ has coordinates $\left(4,4\right)$, so $x_1=4\mathrm{m}$ and $y_1=4\mathrm{m}$

• Object 2 $\left(m_2=3\mathrm{kg}\right)$ has coordinates $\left(2,2\right)$, so $x_2=2\mathrm{m}$ and $y_2=2\mathrm{m}$

• Object 3 $\left(m_3=2\mathrm{kg}\right)$ has coordinates $\left(4,1\right)$, so $x_3=4\mathrm{m}$ and $y_3=1\mathrm{m}$

Step 2: Determine the x-component of the position of the center of mass

$x_{cm}=\frac{{\displaystyle m_1{x}_1+m_2{x}_2+m_3{x}_3}}{m_1+m_2+m_3}$

$x_{cm}=\frac{{\displaystyle \left(1\times 4\right)+\left(3\times 2\right)+\left(2\times 4\right)}}{1+3+2}=3\mathrm{m}$

Step 3: Determine the y-component of the position of the center of mass

$y_{cm}=\frac{{\displaystyle m_1{y}_1+m_2{y}_2+m_3{y}_3}}{m_1+m_2+m_3}$

$y_{cm}=\frac{{\displaystyle \left(1\times 4\right)+\left(3\times 2\right)+\left(2\times 1\right)}}{1+3+2}=2\mathrm{m}$

Step 4: Determine the position of the center of mass

• The xy coordinate of the position of the center of mass is ${\overrightarrow{x}}_{cm}=\left(3,2\right)$

Part (B)

Step 1: Determine the xy velocity components of each object

• The velocity vectors of $m_1$ and $m_2$ are in opposite directions

• Using their position coordinates, we can determine the angle between their velocity vectors and the x-axis

• This angle can be found using $\tan \theta =\frac{y}{x}$

• The positions of $m_1$ and $m_2$ are $\left(4,4\right)$ and $\left(2,2\right)$ respectively, therefore:

  • $\tan \theta =\frac{2}{2}=\frac{4}{4}=1$

  • $\theta ={\tan }^{-1}\left(1\right)=45°$

• The x- and y- velocity components of object 1 $\left(m_1=1\mathrm{kg}\right)$ are:

  • $v_{1x}=1\cos 45°$

  • $v_{1y}=1\sin 45°$

• The x- and y- velocity components of object 2 $\left(m_2=3\mathrm{kg}\right)$ are:

  • $v_{2x}=-2\cos 45°$

  • $v_{2y}=-2\sin 45°$

• The x- and y- velocity components of object 3 $\left(m_3=2\mathrm{kg}\right)$ are:

  • $v_{3x}=0$

  • $v_{3y}=3$

Step 2: Determine the x-component of the velocity of the center of mass

$v_{x,cm}=\frac{{\displaystyle m_1{v}_{x1}+m_2{v}_{x2}+m_3{v}_{x3}}}{m_1+m_2+m_3}$

$v_{x,cm}=\frac{{\displaystyle \left(1\times 1\cos 45°\right)-\left(3\times 2\cos 45°\right)+0}}{1+3+2}=-0.589$

Step 3: Determine the y-component of the velocity of the center of mass

$v_{y,cm}=\frac{{\displaystyle m_1{v}_{y1}+m_2{v}_{y2}+m_3{v}_{y3}}}{m_1+m_2+m_3}$

$v_{y,cm}=\frac{{\displaystyle \left(1\times 1\sin 45°\right)-\left(3\times 2\sin 45°\right)+\left(2\times 3\right)}}{1+3+2}=0.411$

Step 4: Determine the velocity of the center of mass

• Using Pythagoras theorem, the magnitude of the velocity is:

${\overrightarrow{v}}_{cm}=\sqrt{{\left(v_{x,cm}\right)}^2+{\left(v_{y,cm}\right)}^2}$

${\overrightarrow{v}}_{cm}=\sqrt{0.{589}^2+0.{411}^2}=0.72\mathrm{m}/\mathrm{s}$

• Using trigonometry, the direction of the velocity is:

${\theta }_{cm}={\tan }^{-1}\left(\frac{v_{y,cm}}{v_{x,cm}}\right)$

${\theta }_{cm}={\tan }^{-1}\left(-\frac{0.411}{0.589}\right)=-35°$

• The magnitude and direction of the velocity of the system's center of mass can be represented on the xy plane as:

Make sure you understand the position and velocity of a center of mass:

• is not necessarily at the geometric center of a system

• does not change regardless of what the objects in a system do

A dart of mass $m$ moving horizontally with constant speed becomes embedded in a block of mass $M$, as shown in the diagram. The dart and block move together with speed $v$ immediately after the collision. The block is suspended from two light strings and swings up to a maximum height $h$ above the block’s initial position.

In which of the following systems is linear momentum conserved?

A      The dart-block system in the horizontal direction

B      The dart-block-string system

C      The dart-block-Earth system

D      The dart-block-string-Earth system

The correct answer is A

Answer:

Step 1: Analyze the scenario

• This scenario, known as the Ballistic Pendulum, can be analyzed in three stages

  • Stage 1: before the collision

  • Stage 2: after the collision

  • Stage 3: at the maximum height

• According to the Impulse-Momentum Theorem, external forces cause an impulse to act which results in a change in momentum

• Therefore, linear momentum will only be conserved in a system where there are no external forces acting

Step 2: Eliminate incorrect options

• The Earth exerts an external (gravitational) force when there is a change in height, so conservation of momentum cannot be used

  • This eliminates options C and D

• The strings are described as light, so they can be ignored in calculations of momentum

  • This eliminates option B

Step 3: Deduce the correct option

• Before the collision, the dart is moving horizontally, and just after the collision, the dart and the block continue to move horizontally momentarily

• Therefore, the conservation of momentum principle can be applied to the dart-block system in the horizontal direction only

  • Therefore, option A is correct

You must not perform conservation of momentum calculations when the net external force on a system is nonzero, as the conservation of momentum principle is not valid in this situation. However, you will be expected to select systems appropriately so that the net external force on the system is zero. Often in AP Physics 1, the system will be chosen for you, so you must read the question carefully and analyze the system you are given.