IB DP Physics Newton’s second law in terms of momentum Study Notes- 2025 Syllabus
IB DP Physics Newton’s second law in terms of momentum Study Notes
IB DP Physics Newton’s second law in terms of momentum Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with guiding questions of
- that Newton’s second law in the form F = ma assumes mass is constant where as F = Δp Δt allows for situations where mass is changing
- that linear momentum as given by p = mv
Standard level and higher level: 10 hours
Additional higher level: There is no additional higher level content
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- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Newton’s Second Law: Two Equivalent Forms
Standard Form of Newton’s Second Law: \( \mathbf{F} = m\mathbf{a} \)
This is the most familiar form used in elementary physics. It expresses that the net force \( \mathbf{F} \) acting on an object is equal to its mass \( m \) multiplied by its acceleration \( \mathbf{a} \). This form assumes:
- The mass of the object is constant over time.
- It applies best to simple systems like blocks, cars, or projectiles where mass does not change.
- Acceleration is the direct consequence of the applied force when mass remains fixed.
General (Momentum-Based) Form: \( \mathbf{F} = \dfrac{d\mathbf{p}}{dt} = \dfrac{d}{dt}(m\mathbf{v}) \)
This more fundamental form of Newton’s Second Law expresses force as the time rate of change of momentum \( \mathbf{p} \). It is universally valid, even when mass varies. This form:
- Accounts for both changes in velocity and mass.
- Expands to:
\( \mathbf{F} = \dfrac{d}{dt}(m\mathbf{v}) = m\dfrac{d\mathbf{v}}{dt} + \mathbf{v}\dfrac{dm}{dt} \)
- The second term \( \mathbf{v}\dfrac{dm}{dt} \) becomes significant in systems where mass is not conserved (e.g., rockets, fluid jets, fuel tanks).
When Are They Equivalent?
When mass is constant, the general momentum form simplifies to the standard form:
\( \dfrac{d}{dt}(m\mathbf{v}) = m\dfrac{d\mathbf{v}}{dt} = m\mathbf{a} \)
So: \( \mathbf{F} = m\mathbf{a} \)
When Is the General Form Needed?
In real-world applications where the mass of the system is changing during motion:
- Rocket propulsion – As fuel burns, the rocket’s mass decreases.
- Leaking tanks or carts losing sand – Mass reduces as material exits the system.
- Collisions involving fragmentation – Total system mass may redistribute dynamically.
Example:
A rocket is moving in space by expelling gas at high speed. At a certain moment, the rocket has a mass of \( 1000 \, \text{kg} \), and it ejects fuel at a constant rate of \( 5 \, \text{kg/s} \) with an exhaust velocity of \( 2000 \, \text{m/s} \) relative to the rocket. What is the thrust force acting on the rocket due to the ejected fuel?
▶️Answer/Explanation
Use the general momentum form:
\( F = \frac{dp}{dt} = v_{\text{exhaust}} \cdot \frac{dm}{dt} \)
Insert values:
Given:
- \( v_{\text{exhaust}} = 2000 \, \text{m/s} \)
- \( \frac{dm}{dt} = -5 \, \text{kg/s} \) (mass of rocket is decreasing, so negative)
Calculate thrust (use magnitude only):
\( F = 2000 \times (-5) = -10000 \, \text{N} \Rightarrow \boxed{F = 10000 \, \text{N}} \text{ (in forward direction)} \)
The rocket experiences a forward thrust of \( \boxed{10{,}000 \, \text{N}} \) due to the ejection of fuel.
Linear Momentum
Linear momentum is defined as: \( \mathbf{p} = m\mathbf{v} \), where:
- \( m \) is the mass of the object (kg)
- \( \mathbf{v} \) is the velocity of the object (m/s)
- \( \mathbf{p} \) is the momentum (kg·m/s), a vector quantity
Relation to Newton’s Laws:
- From Newton’s 2nd law in general form: \( \mathbf{F}_{\text{net}} = \dfrac{d\mathbf{p}}{dt} \)
- If \( \mathbf{F}_{\text{net}} = 0 \), then \( \dfrac{d\mathbf{p}}{dt} = 0 \Rightarrow \mathbf{p} = \text{constant} \)
Example:
An ice skater of mass \( 60 \, \text{kg} \) is initially at rest on frictionless ice. She throws a \( 5 \, \text{kg} \) medicine ball horizontally with 100m/s , what is its linear momentum medicine ball ?
▶️Answer/Explanation
System = ice skater + medicine ball.
No external horizontal forces (frictionless ice). So, total horizontal momentum is conserved.
Use conservation of momentum
\( p_{\text{initial}} = (5)(100) \)
\( \boxed{500\, \text{m/s}} \)