6.1 Systems, momentum and impulse

Achievable AP Physics 1

6. Linear momentum and collisions

Our AP Physics 1 course is currently in development and is a work-in-progress.

Systems, momentum and impulse

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In this subchapter, you’ll work with these core ideas:

Systems
Centre of mass
Linear momentum
Impulse momentum theorem

What is a system? A system in physics is the collection of objects or particles you choose to analyze together.

We imagine a boundary around the system.
Everything outside that boundary is the environment.

Forces then fall into two categories:

Internal forces act between objects inside the boundary. Internal forces can change how parts of the system move relative to each other, but they can’t change the motion of the system as a whole.
External forces are exerted by the environment on the system. External forces determine the system’s overall motion.

Isolated vs. non-isolated:

An isolated system has no net external force. Its total momentum and the motion of its centre of mass remain unchanged.
A non-isolated system has a net external force, so its centre of mass accelerates.

Centre of mass: The centre of mass (CM) of a system is the unique point where you can treat the system’s mass as if it were concentrated, as far as translational motion is concerned. The CM moves as if all external forces acted at that point.

Definitions

Discrete system: For $N$ point masses $m_{i}$ at position vectors $r_{i}$ , the centre of mass is

$R_{CM} = \frac{1}{M} i = 1 \sum N m_{i} r_{i}, M = i = 1 \sum N m_{i} .$

Continuous distribution: If mass is distributed with density $ρ (r)$ over volume $V$ , then

$R_{CM} = \frac{1}{M} \int_{V} r ρ (r) d V, M = \int_{V} ρ (r) d V .$

Centre of mass for a continuous distribution

Equation of motion of the CM: If you add up all external forces on a system, the centre of mass accelerates according to

$M a_{CM} = \sum F_{ext} .$

So, for translational motion, the system behaves like a single particle of mass $M$ acted on by the net external force.

Example problem 1

Three particles of masses $m_{1} = 1.0$ kg, $m_{2} = 2.0$ kg and $m_{3} = 1.5$ kg lie on the $x$ -axis at $x_{1} = 0$ , $x_{2} = 2.0$ , and $x_{3} = 5.0$ m. Find the centre of mass.

Solution:

(spoiler)

$M = m_{1} + m_{2} + m_{3} = 1.0 + 2.0 + 1.5 = 4.5 kg, x_{CM} = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{M} = \frac{1.0 \cdot 0 + 2.0 \cdot 2.0 + 1.5 \cdot 5.0}{4.5} = \frac{0 + 4.0 + 7.5}{4.5} = 2.56 m .$

Example problem 2

Two point masses, $3 kg$ at $x = 1 m$ and $5 kg$ at $x = 4 m$ , lie on the $x$ -axis. Calculate the position of their CM.

Solution.

(spoiler)

Compute total mass:

$M = 3 + 5 = 8 kg .$

Compute weighted sum of positions:

$\sum m_{i} x_{i} = 3 \cdot 1 + 5 \cdot 4 = 3 + 20 = 23 kg \cdot m .$

Divide to find CM:

$x_{CM} = \frac{\sum m _{i} x _{i}}{M} = \frac{23}{8} = 2.875 m .$

Example problem 3

A thin rod of length $L$ has linear density $λ (x) = λ_{0} (1 + x / L)$ distributed along its length from $x = 0$ to $x = L$ . Show that its centre of mass is at $x$ = $\frac{5}{9} L$ .

Solution.

Total mass:

$M = \int_{0}^{L} λ_{0} (1 + \frac{x}{L}) d x = λ_{0} (L + \frac{L}{2}) = \frac{3}{2} λ_{0} L .$

First moment:

$\int_{0}^{L} x λ (x) d x = λ_{0} (\frac{L ^{2}}{2} + \frac{L ^{2}}{3}) = \frac{5}{6} λ_{0} L^{2} .$

Centre of mass:

$x_{CM} = \frac{\frac{5}{6} λ _{0} L ^{2}}{\frac{3}{2} λ _{0} L} = \frac{5}{9} L .$

Note: This question is for illustration purpose and is not expected to be a part of AP Physics 1 scope.

Example problem 4

Three masses form a right triangle at $(0, 0)$ , $(a, 0)$ , and $(0, a)$ with masses $m$ , $2 m$ , and $3 m$ . Determine the CM coordinates $(x_{CM}, y_{CM})$ .

Solution.

(spoiler)

Total mass: $M = 6 m$ .
$x$ -coordinate:

$x_{CM} = \frac{m \cdot 0 + 2 m \cdot a + 3 m \cdot 0}{6 m} = \frac{1}{3} a .$

$y$ -coordinate:

$y_{CM} = \frac{m \cdot 0 + 2 m \cdot 0 + 3 m \cdot a}{6 m} = \frac{1}{2} a .$

$(x_{CM}, y_{CM}) = (\frac{a}{3}, \frac{a}{2})$ .

Linear momentum

Linear momentum measures “mass in motion.” For a particle of mass $m$ moving with velocity $v$ ,

$p = m v .$

For a system, the total momentum is the sum of the momenta of its parts:

$P_{sys} = i \sum m_{i} v_{i} = M v_{CM} .$

Impulse-momentum theorem

When an external force acts over a time interval $Δ t$ , it delivers an impulse. For a constant force,

$J = F Δ t$

and that impulse changes the object’s momentum:

$J = Δ p = m v_{f} - m v_{i} .$

Area under the F-t curve gives average/total impulse

Example problem 5 A 2.0 kg puck sliding at 3.0 m/s receives a constant force of 5.0 N in the direction of motion for 0.4 s. Find its final speed.

Solution:

(spoiler)

$J = F Δ t = 5.0 \times 0.4 = 2.0 N \cdot s,$

$p_{i} = 2.0 \times 3.0 = 6.0 kg \cdot m/s,$

$p_{f} = p_{i} + J = 6.0 + 2.0 = 8.0 kg \cdot m/s,$

$v_{f} = \frac{p _{f}}{m} = \frac{8.0}{2.0} = 4.0 m/s .$

Example problem 6 A 0.5 kg toy car moving at 2.0 m/s collides with a spring bumper that exerts an average force of 4.0 N over 0.10 s. Determine the change in the car’s speed and its final velocity.

Solution:

(spoiler)

$p_{i} = m v_{i} = 0.5 \times 2.0 = 1.0 kg \cdot m/s,$

$J = F Δ t = 4.0 \times 0.10 = 0.40 N \cdot s,$

$p_{f} = p_{i} + J = 1.0 + 0.40 = 1.40 kg \cdot m/s,$

$v_{f} = \frac{p _{f}}{m} = \frac{1.40}{0.5} = 2.8 m/s,$

$Δ v = v_{f} - v_{i} = 2.8 - 2.0 = 0.8 m/s .$

Systems

System: collection of objects analyzed together
Internal forces: act within system, don’t change total motion
External forces: from environment, determine system’s overall motion
Isolated system: no net external force, momentum/CM motion unchanged
Non-isolated system: net external force, CM accelerates

Centre of Mass (CM)

CM: point where system’s mass is concentrated for translational motion
Discrete system formula: $R_{CM} = \frac{1}{M} \sum m_{i} r_{i}$
- $M = \sum m_{i}$
Continuous distribution formula: $R_{CM} = \frac{1}{M} \int r ρ (r) d V$
- $M = \int ρ (r) d V$
Equation of motion: $M a_{CM} = \sum F_{ext}$

Linear Momentum

Linear momentum: $p = m v$
System momentum: $P_{sys} = \sum m_{i} v_{i} = M v_{CM}$

Impulse-Momentum Theorem

Impulse: $J = F Δ t$
Change in momentum: $J = Δ p = m v_{f} - m v_{i}$
Area under $F$ - $t$ curve = total impulse

Problem-Solving Habits

Clearly define the system and boundaries
Track directions and sign conventions consistently
Use area under $F$ - $t$ graph for impulse calculations

Next | 6.2 Conservation of momentum and collisions

Systems, momentum and impulse

In this subchapter, you’ll work with these core ideas:

Systems
Centre of mass
Linear momentum
Impulse momentum theorem

What is a system? A system in physics is the collection of objects or particles you choose to analyze together.

We imagine a boundary around the system.
Everything outside that boundary is the environment.

Forces then fall into two categories:

Internal forces act between objects inside the boundary. Internal forces can change how parts of the system move relative to each other, but they can’t change the motion of the system as a whole.
External forces are exerted by the environment on the system. External forces determine the system’s overall motion.

Isolated vs. non-isolated:

An isolated system has no net external force. Its total momentum and the motion of its centre of mass remain unchanged.
A non-isolated system has a net external force, so its centre of mass accelerates.

Definitions

Discrete system: For $N$ point masses $m_{i}$ at position vectors $r_{i}$ , the centre of mass is

$R_{CM} = \frac{1}{M} i = 1 \sum N m_{i} r_{i}, M = i = 1 \sum N m_{i} .$

Continuous distribution: If mass is distributed with density $ρ (r)$ over volume $V$ , then

$R_{CM} = \frac{1}{M} \int_{V} r ρ (r) d V, M = \int_{V} ρ (r) d V .$

Equation of motion of the CM: If you add up all external forces on a system, the centre of mass accelerates according to

$M a_{CM} = \sum F_{ext} .$

So, for translational motion, the system behaves like a single particle of mass $M$ acted on by the net external force.

Example problem 1

Three particles of masses $m_{1} = 1.0$ kg, $m_{2} = 2.0$ kg and $m_{3} = 1.5$ kg lie on the $x$ -axis at $x_{1} = 0$ , $x_{2} = 2.0$ , and $x_{3} = 5.0$ m. Find the centre of mass.

Solution:

(spoiler)

Example problem 2

Two point masses, $3 kg$ at $x = 1 m$ and $5 kg$ at $x = 4 m$ , lie on the $x$ -axis. Calculate the position of their CM.

Solution.

(spoiler)

Compute total mass:

$M = 3 + 5 = 8 kg .$

Compute weighted sum of positions:

$\sum m_{i} x_{i} = 3 \cdot 1 + 5 \cdot 4 = 3 + 20 = 23 kg \cdot m .$

Divide to find CM:

$x_{CM} = \frac{\sum m _{i} x _{i}}{M} = \frac{23}{8} = 2.875 m .$

Example problem 3

A thin rod of length $L$ has linear density $λ (x) = λ_{0} (1 + x / L)$ distributed along its length from $x = 0$ to $x = L$ . Show that its centre of mass is at $x$ = $\frac{5}{9} L$ .

Solution.

Total mass:

$M = \int_{0}^{L} λ_{0} (1 + \frac{x}{L}) d x = λ_{0} (L + \frac{L}{2}) = \frac{3}{2} λ_{0} L .$

First moment:

$\int_{0}^{L} x λ (x) d x = λ_{0} (\frac{L ^{2}}{2} + \frac{L ^{2}}{3}) = \frac{5}{6} λ_{0} L^{2} .$

Centre of mass:

$x_{CM} = \frac{\frac{5}{6} λ _{0} L ^{2}}{\frac{3}{2} λ _{0} L} = \frac{5}{9} L .$

Note: This question is for illustration purpose and is not expected to be a part of AP Physics 1 scope.

Example problem 4

Three masses form a right triangle at $(0, 0)$ , $(a, 0)$ , and $(0, a)$ with masses $m$ , $2 m$ , and $3 m$ . Determine the CM coordinates $(x_{CM}, y_{CM})$ .

Solution.

(spoiler)

Total mass: $M = 6 m$ .
$x$ -coordinate:

$x_{CM} = \frac{m \cdot 0 + 2 m \cdot a + 3 m \cdot 0}{6 m} = \frac{1}{3} a .$

$y$ -coordinate:

$y_{CM} = \frac{m \cdot 0 + 2 m \cdot 0 + 3 m \cdot a}{6 m} = \frac{1}{2} a .$

$(x_{CM}, y_{CM}) = (\frac{a}{3}, \frac{a}{2})$ .

Linear momentum

Linear momentum measures “mass in motion.” For a particle of mass $m$ moving with velocity $v$ ,

$p = m v .$

For a system, the total momentum is the sum of the momenta of its parts:

$P_{sys} = i \sum m_{i} v_{i} = M v_{CM} .$

Impulse-momentum theorem

When an external force acts over a time interval $Δ t$ , it delivers an impulse. For a constant force,

$J = F Δ t$

and that impulse changes the object’s momentum:

$J = Δ p = m v_{f} - m v_{i} .$

Example problem 5 A 2.0 kg puck sliding at 3.0 m/s receives a constant force of 5.0 N in the direction of motion for 0.4 s. Find its final speed.

Solution:

(spoiler)

$J = F Δ t = 5.0 \times 0.4 = 2.0 N \cdot s,$

$p_{i} = 2.0 \times 3.0 = 6.0 kg \cdot m/s,$

$p_{f} = p_{i} + J = 6.0 + 2.0 = 8.0 kg \cdot m/s,$

$v_{f} = \frac{p _{f}}{m} = \frac{8.0}{2.0} = 4.0 m/s .$

Solution:

(spoiler)

$p_{i} = m v_{i} = 0.5 \times 2.0 = 1.0 kg \cdot m/s,$

$J = F Δ t = 4.0 \times 0.10 = 0.40 N \cdot s,$

$p_{f} = p_{i} + J = 1.0 + 0.40 = 1.40 kg \cdot m/s,$

$v_{f} = \frac{p _{f}}{m} = \frac{1.40}{0.5} = 2.8 m/s,$

$Δ v = v_{f} - v_{i} = 2.8 - 2.0 = 0.8 m/s .$

Achievable AP Physics 1

6. Linear momentum and collisions

Our AP Physics 1 course is currently in development and is a work-in-progress.