AP Physics 1- 4.2 Change in Momentum and Impulse- Study Notes- New Syllabus
AP Physics 1-4.2 Change in Momentum and Impulse – Study Notes
AP Physics 1-4.2 Change in Momentum and Impulse – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Change in Momentum and Impulse
- Net External Force and Momentum
- Newton’s Second Law and Impulse–Momentum Theorem
Change in Momentum
The linear momentum of an object is \( \vec{p} = m \vec{v} \).
The change in momentum is the difference between the final and initial momentum:
\( \Delta \vec{p} = \vec{p}_\text{final} – \vec{p}_\text{initial} \).
Impulse:
Impulse (\( \vec{J} \)) is the product of the average force applied to an object and the time interval over which it acts:
\( \vec{J} = \vec{F}_\text{avg} \Delta t \).
- Impulse is a vector quantity and has the same direction as the applied force.
Impulse-Momentum Theorem: The impulse applied to an object is equal to the change in momentum of the object:
\( \vec{J} = \Delta \vec{p} \) → \( \vec{F}_\text{avg} \Delta t = m \vec{v}_f – m \vec{v}_i \).
- This theorem is useful to analyze situations where forces act over short time intervals (e.g., collisions, strikes, braking).
- Reducing force by increasing the time interval reduces the impact (e.g., airbags, padded surfaces).
Example :
A 0.15 kg baseball moving at 20 m/s is struck by a bat and reverses direction to 30 m/s in 0.01 s. Find the average force exerted by the bat on the ball.
▶️Answer/Explanation
Step 1: Change in momentum → \( \Delta p = m v_f – m v_i = 0.15 \cdot (-30) – 0.15 \cdot 20 = -7.5 \, \text{kg·m/s} \)
Step 2: Impulse → \( J = F_\text{avg} \Delta t = \Delta p \)
Step 3: Solve for \( F_\text{avg} \) → \( F_\text{avg} = \dfrac{\Delta p}{\Delta t} = \dfrac{-7.5}{0.01} = -750 \, \text{N} \)
Final Answer: The average force exerted by the bat is 750 N in the direction opposite to the initial velocity.
Example :
A 1000 kg car moving at 20 m/s comes to a stop in 5 seconds. Find the average braking force applied.
▶️Answer/Explanation
Step 1: Change in momentum → \( \Delta p = m v_f – m v_i = 1000 \cdot 0 – 1000 \cdot 20 = -20{,}000 \, \text{kg·m/s} \)
Step 2: Impulse → \( J = F_\text{avg} \Delta t = \Delta p \)
Step 3: Solve for \( F_\text{avg} \) → \( F_\text{avg} = \dfrac{\Delta p}{\Delta t} = \dfrac{-20{,}000}{5} = -4000 \, \text{N} \)
Final Answer: The average braking force is 4000 N opposite to the motion of the car.
Net External Force and Momentum
The net external force exerted on a system is related to the rate of change of the system’s momentum.
Mathematically:
\( \vec{F}_\text{net} = \dfrac{d\vec{p}}{dt} \),
where \( \vec{p} \) is the momentum of the system.
- Graphically, the slope of a momentum vs. time graph gives the net external force acting on the system.
- The area of a Force vs. time graph gives the momentum of the system.
Internal forces do not affect the total momentum of a system; only external forces change the system’s momentum. This relationship is essentially a restatement of Newton’s Second Law in terms of momentum.
Example:
A 2 kg object has its momentum increase linearly from 0 kg·m/s to 10 kg·m/s over 5 seconds. Find the net external force acting on the object.
▶️Answer/Explanation
Step 1: Slope of momentum vs. time graph → \( F_\text{net} = \dfrac{\Delta p}{\Delta t} \)
Step 2: Substitute values → \( F_\text{net} = \dfrac{10 – 0}{5} = 2 \, \text{N} \)
Final Answer: The net external force on the object is 2 N in the direction of momentum increase.
Newton’s Second Law and Impulse–Momentum Theorem
Newton’s Second Law can be derived from the impulse–momentum theorem for systems with constant mass.
Impulse–momentum theorem:
\( \vec{J} = \Delta \vec{p} \) → \( \vec{F}_\text{net} \Delta t = m \vec{v}_f – m \vec{v}_i \).
Dividing both sides by the time interval \( \Delta t \):
\( \vec{F}_\text{net} = \dfrac{m \vec{v}_f – m \vec{v}_i}{\Delta t} = m \dfrac{\Delta \vec{v}}{\Delta t} = m \vec{a} \).
Thus, Newton’s Second Law, \( \vec{F}_\text{net} = m \vec{a} \), is a direct consequence of the impulse–momentum theorem when mass is constant.
- It shows that the net external force on an object is equal to the rate of change of its momentum.
Example:
A 5 kg cart initially at rest is acted upon by a constant horizontal force of 10 N for 4 seconds. Find the final velocity of the cart using Newton’s Second Law derived from impulse–momentum theorem.
▶️Answer/Explanation
Step 1: Impulse → \( J = F_\text{net} \Delta t = 10 \cdot 4 = 40 \, \text{N·s} \)
Step 2: Impulse = change in momentum → \( J = m \Delta v \)
Step 3: Solve for \( \Delta v \) → \( \Delta v = \dfrac{J}{m} = \dfrac{40}{5} = 8 \, \text{m/s} \)
Final Answer: The final velocity of the cart is 8 m/s in the direction of the force.