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IB DP Physics Conservation of momentum Study Notes

IB DP Physics Conservation of momentum Study Notes - 2025 Syllabus

IB DP Physics Conservation of momentum Study Notes

IB DP Physics Conservation 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

  • Conservation of Momentum and Newton’s Second Law Forms

Standard level and higher level: 10  hours
Additional higher level: There is no additional higher level content

IB DP Physics 2025 -Study Notes -All Topics

Conservation of Momentum and Newton’s Second Law Forms

Newton’s Second Law can be written in two equivalent but context-specific forms:

  • Form 1 (constant mass): \( F = ma \)
  • Form 2 (changing or constant mass): \( F = \frac{\Delta p}{\Delta t} \)

1. \( F = ma \): The Standard Form (Constant Mass Only)

  • Derived from Newton’s second law assuming mass is constant.
  • Since momentum is \( p = mv \), differentiating gives:

\( \frac{dp}{dt} = \frac{d(mv)}{dt} = m \frac{dv}{dt} = ma \)

  • This is applicable for:
    • Rigid bodies where the mass doesn’t change.
    • Everyday forces on cars, balls, blocks, etc.

2. \( F = \frac{\Delta p}{\Delta t} \): General Form (Mass May Vary)

  • This is the more general form of Newton’s Second Law.
  • It works even if mass is changing, as in:
    • Rockets losing mass as fuel burns
    • Rain accumulating in a moving cart
    • Jet propulsion
  • Momentum is defined as \( p = mv \). The total force is the rate of change of momentum:

\( F = \frac{\Delta p}{\Delta t} = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt} \)

  • This accounts for cases where mass changes over time and can even include variable ejection speeds (as in a rocket).

3. Momentum Conservation Principle

  • If there is no external force acting on a system, then from:

\( F = \frac{\Delta p}{\Delta t} \Rightarrow 0 = \frac{\Delta p}{\Delta t} \Rightarrow \Delta p = 0 \)

  • This implies that momentum remains constant:

\( p_{\text{initial}} = p_{\text{final}} \)

  • Momentum conservation is valid in all types of interactions (e.g., explosions, collisions), provided there’s no net external force.

Example

A 1000 kg rocket in deep space expels 10 kg of gas at 500 m/s. What is the recoil velocity of the rocket immediately after the expulsion?

▶️ Answer/Explanation

Initial momentum = 0 (rocket and gas at rest)

Let recoil velocity of rocket be \( v \)

By conservation of momentum:

\( 0 = m_{\text{rocket}} \cdot v + m_{\text{gas}} \cdot (-500) \)

\( 1000v = 5000 \Rightarrow v = \boxed{5 \, \text{m/s}} \)

Example

A baseball of mass \( 0.15 \, \text{kg} \) is hit by a bat, changing its velocity from \( -20 \, \text{m/s} \) to \( +30 \, \text{m/s} \) in \( 0.01 \, \text{s} \). Find the impulse and average force exerted by the bat.

▶️ Answer/Explanation

Impulse \( J = \Delta p = m(v_f – v_i) \)

\( J = 0.15(30 – (-20)) = 0.15(50) = 7.5 \, \text{Ns} \)

Average force: \( F = \frac{J}{\Delta t} = \frac{7.5}{0.01} = \boxed{750 \, \text{N}} \)

Example

A truck of mass \( 5000 \, \text{kg} \) moves at \( 10 \, \text{m/s} \). It leaks sand at a constant rate of \( 100 \, \text{kg/s} \). What is the rate of change of momentum after 5 seconds?

▶️ Answer/Explanation

At \( t = 5 \, \text{s} \), mass lost = \( 100 \times 5 = 500 \, \text{kg} \)

Remaining mass = \( 5000 – 500 = 4500 \, \text{kg} \)

Momentum = \( p = mv = 4500 \times 10 = 45000 \, \text{kg·m/s} \)

Initial momentum = \( 5000 \times 10 = 50000 \, \text{kg·m/s} \)

Change in momentum = \( \Delta p = 45000 – 50000 = -5000 \, \text{kg·m/s} \)

Rate = \( \frac{-5000}{5} = \boxed{-1000 \, \text{N}} \) (acting backward)

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