AP Physics C Mechanics- 4.2 Change in Momentum and Impulse- Study Notes- New Syllabus
AP Physics C Mechanics- 4.2 Change in Momentum and Impulse – Study Notes
AP Physics C Mechanics- 4.2 Change in Momentum and Impulse – Study Notes – per latest Syllabus.
Key Concepts:
- Impulse and the Rate of Change of Momentum
- Graphical Interpretation of Impulse and Force–Momentum Relationship
- Change in Momentum
- Impulse–Momentum Theorem and Its Applications
Impulse and the Rate of Change of Momentum
The relationship between force and momentum is a direct result of Newton’s Second Law. It states that the rate of change of a system’s momentum is equal to the net external force acting on that system.
Relationship Between Force and Momentum
\( \mathrm{\vec{F}_{net} = \dfrac{d\vec{p}}{dt}} \)
- \( \mathrm{\vec{F}_{net}} \): net external force acting on the system
- \( \mathrm{\vec{p}} \): linear momentum (\( \mathrm{m\vec{v}} \))
- This vector equation expresses that a net external force causes a change in the system’s momentum over time.
- If \( \mathrm{\vec{F}_{net} = 0} \), then \( \mathrm{\dfrac{d\vec{p}}{dt} = 0} \) — the momentum of the system remains constant.
Impulse
Impulse is defined as the product of a force and the time interval during which the force acts. More precisely, for a variable force, impulse is the integral of force over time:
\( \mathrm{\vec{J} = \displaystyle \int_{t_1}^{t_2} \vec{F}_{net}(t) \, dt} \)
- \( \mathrm{\vec{J}} \): impulse (unit: \( \mathrm{N \cdot s} \) or \( \mathrm{kg \cdot m/s} \))
- \( \mathrm{\vec{F}_{net}(t)} \): net external force as a function of time
- \( \mathrm{t_1, t_2} \): initial and final times of force application.
Example:
A \( \mathrm{0.5\,kg} \) hockey puck slides on ice and is struck by a stick. The stick applies an average force of \( \mathrm{40\,N} \) to the puck for \( \mathrm{0.05\,s} \). The puck was initially moving at \( \mathrm{2.0\,m/s} \) in the same direction as the applied force. Find
(a) the impulse delivered to the puck and
(b) its final velocity after the hit.
▶️ Answer / Explanation
Step 1: Use the impulse formula for a constant force:
\( \mathrm{\vec{J} = \vec{F}_{net} \, \Delta t} \)
Step 2: Calculate the magnitude of the impulse:
\( \mathrm{J = (40)(0.05) = 2.0\,N \cdot s} \)
The impulse acts in the direction of the force (same as initial motion).
Step 3: Relate impulse to the change in momentum:
\( \mathrm{J = \Delta p = m(v_f – v_i)} \)
Substitute known values:
\( \mathrm{2.0 = (0.5)(v_f – 2.0)} \)
Step 4: Solve for \( \mathrm{v_f} \):
\( \mathrm{v_f – 2.0 = 4.0} \Rightarrow \mathrm{v_f = 6.0\,m/s} \)
Step 5: Interpret the result:
- The impulse of \( \mathrm{2.0\,N \cdot s} \) increases the puck’s momentum.
- The puck’s velocity increases from \( \mathrm{2.0\,m/s} \) to \( \mathrm{6.0\,m/s} \) in the same direction as the applied force.
Graphical Interpretation of Impulse and Force–Momentum Relationship
Impulse is a vector quantity that represents the effect of a net external force acting on a system over a specific time interval. It has both magnitude and direction and is always directed along the same line as the net force applied to the system.
Direction and Nature of Impulse
- Impulse is a vector quantity.
- Its direction is the same as that of the net external force acting on the system.
- Because it equals the change in momentum, the impulse direction also matches the direction of momentum change.
\( \mathrm{\vec{J} = \displaystyle \int_{t_1}^{t_2} \vec{F}_{net}(t)\,dt = \Delta \vec{p}} \)
Impulse as the Area Under the Force–Time Graph
The magnitude of impulse delivered to a system equals the area under the curve of a graph of the net external force as a function of time:
\( \mathrm{J = \text{Area under } F_{net}\text{ vs. } t \text{ curve}} \)
- For constant force: \( \mathrm{J = F_{net} \Delta t} \)
- For variable force: \( \mathrm{J = \displaystyle \int_{t_1}^{t_2} F_{net}(t)\,dt} \)
- The total area (including sign) gives the net impulse on the system.
- If the graph lies partly above and below the time axis, the areas must be added algebraically.
Force as the Rate of Change of Momentum
The net external force on a system equals the slope of the momentum–time graph:
\( \mathrm{\vec{F}_{net} = \dfrac{d\vec{p}}{dt}} \)
- The steeper the slope of the \( \mathrm{p\text{–}t} \) graph, the larger the net force acting on the system.
- A zero slope (horizontal line) indicates no net force — momentum remains constant.
- A negative slope represents a force acting opposite to the direction of motion.
Example:
The graph below shows the net force acting on a \( \mathrm{2.0\,kg} \) object as a function of time.
- From \( \mathrm{t = 0\,s} \) to \( \mathrm{t = 3\,s} \), \( \mathrm{F_{net} = 6\,N} \).
- From \( \mathrm{t = 3\,s} \) to \( \mathrm{t = 5\,s} \), \( \mathrm{F_{net} = -3\,N} \).
Determine the total impulse delivered to the object and its change in momentum.
▶️ Answer / Explanation
Step 1: Calculate impulse for each region (area under \( \mathrm{F–t} \) graph).
- Region 1: \( \mathrm{J_1 = F \Delta t = (6)(3) = +18\,N \cdot s} \)
- Region 2: \( \mathrm{J_2 = F \Delta t = (-3)(2) = -6\,N \cdot s} \)
Step 2: Find total impulse:
\( \mathrm{J_{total} = J_1 + J_2 = 18 – 6 = +12\,N \cdot s} \)
Step 3: Relate to change in momentum:
\( \mathrm{\Delta p = J_{total} = +12\,kg \cdot m/s} \)
Interpretation: The object’s momentum increases by \( \mathrm{12\,kg \cdot m/s} \) in the direction of the net applied force.
Change in Momentum
The change in momentum of a system represents the difference between its final momentum and initial momentum. It quantifies how much and in what direction the momentum of a system changes due to applied forces.
Relevant Equation:
\( \mathrm{\Delta \vec{p} = \vec{p}_f – \vec{p}_i} \)
- \( \mathrm{\Delta \vec{p}} \): change in momentum (vector)
- \( \mathrm{\vec{p}_i} \): initial momentum
- \( \mathrm{\vec{p}_f} \): final momentum
Nature of Momentum Change:
- \( \mathrm{\Delta \vec{p}} \) is a vector quantity with both magnitude and direction.
- Its direction is the same as that of the net force or impulse applied to the system.
Relation to Impulse:
The change in momentum of a system equals the impulse delivered to it:
\( \mathrm{\vec{J} = \Delta \vec{p}} \)
- This is the basis of the Impulse–Momentum Theorem.
- A greater impulse results in a larger change in momentum.
Example:
A \( \mathrm{0.2\,kg} \) ball is moving horizontally to the right at \( \mathrm{5\,m/s} \). After being struck by a wall, it rebounds to the left with a velocity of \( \mathrm{3\,m/s} \). Calculate the change in momentum of the ball.
▶️ Answer / Explanation
Step 1: Write known values.
- Mass: \( \mathrm{m = 0.2\,kg} \)
- Initial velocity: \( \mathrm{v_i = +5\,m/s} \) (to the right)
- Final velocity: \( \mathrm{v_f = -3\,m/s} \) (to the left)
Step 2: Use the change in momentum equation.
\( \mathrm{\Delta \vec{p} = m(v_f – v_i)} \)
\( \mathrm{\Delta \vec{p} = 0.2(-3 – 5) = 0.2(-8) = -1.6\,kg \cdot m/s} \)
Step 3: Interpret the result.
- The negative sign indicates the change in momentum is toward the left, opposite the initial direction of motion.
- Magnitude of momentum change: \( \mathrm{1.6\,kg \cdot m/s} \)
Impulse–Momentum Theorem and Its Applications
The Impulse–Momentum Theorem states that the impulse delivered to an object is equal to the change in its momentum. This theorem connects the concepts of force, time, and motion, and serves as a time-integrated form of Newton’s Second Law.
Impulse Equals Change in Momentum
The impulse exerted on an object is equal to the object’s change in momentum. This relationship defines how a force acting over a time interval changes the motion of a system.
\( \mathrm{\vec{J} = \int_{t_1}^{t_2} \vec{F}_{net}(t)\,dt = \Delta \vec{p}} \)
- \( \mathrm{\vec{J}} \): Impulse (unit: \( \mathrm{N \cdot s} \) or \( \mathrm{kg \cdot m/s} \))
- \( \mathrm{\vec{F}_{net}} \): Net external force acting on the system
- \( \mathrm{\Delta \vec{p}} \): Change in momentum (\( \mathrm{\vec{p}_f – \vec{p}_i} \))
- Impulse and momentum are vector quantities that point in the same direction as the net external force.
- This equation is valid for both constant and variable forces.
Newton’s Second Law from the Impulse–Momentum Theorem (Constant Mass Systems)
Newton’s Second Law can be expressed as the time rate of change of momentum. For systems where the mass remains constant, the impulse–momentum theorem directly leads to the familiar form of the law:
\( \mathrm{\vec{F}_{net} = \dfrac{d\vec{p}}{dt}} \)
Substituting \( \mathrm{\vec{p} = m\vec{v}} \) (constant \( \mathrm{m} \)):
\( \mathrm{\vec{F}_{net} = m \dfrac{d\vec{v}}{dt} = m\vec{a}} \)
- For constant-mass systems, the net external force is proportional to the rate of change of velocity (acceleration).
- This form applies to most rigid-body and particle motion problems in classical mechanics.
Systems with Variable Mass
The impulse–momentum theorem can also describe systems in which the mass changes with time — such as rockets, flowing fluids, or systems that lose or gain material.
\( \mathrm{\vec{F}_{net} = \dfrac{d\vec{p}}{dt} = \dfrac{dm}{dt} \vec{v}} \)
- Here, the velocity of the system may remain constant, but the mass changes with respect to time.
- In such systems, changes in momentum arise due to varying mass rather than acceleration.
- This extended form of Newton’s second law is crucial in understanding propulsion systems such as rockets or jets.
Example:
A \( \mathrm{0.2\,kg} \) ball is hit by a bat. The force on the ball varies with time according to the graph below (force acts for \( \mathrm{0.04\,s} \)). The maximum force is \( \mathrm{100\,N} \), and the average force is \( \mathrm{50\,N} \). Find
(a) the impulse and
(b) the change in velocity of the ball.
▶️ Answer / Explanation
Step 1: Use impulse formula:
\( \mathrm{J = F_{avg} \, \Delta t = (50)(0.04) = 2.0\,N \cdot s} \)
Step 2: Relate impulse to change in momentum:
\( \mathrm{J = \Delta p = m(v_f – v_i)} \)
If the ball was initially at rest (\( \mathrm{v_i = 0} \)):
\( \mathrm{2.0 = (0.2)(v_f)} \Rightarrow \mathrm{v_f = 10\,m/s}} \)
Result: Impulse \( \mathrm{J = 2.0\,N \cdot s} \) → Change in velocity \( \mathrm{\Delta v = 10\,m/s} \) in the direction of the applied force.