Edexcel A Level (IAL) Physics-1.13 -1.14 Momentum and Conservation of Linear Momentum - Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.13 -1.14 Momentum and Conservation of Linear Momentum- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.13 -1.14 Momentum and Conservation of Linear Momentum- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.13 understand that momentum is defined as p = mv
- 1.14 know the principle of conservation of linear momentum, understand how to relate this to Newton’s laws of motion and understand how to apply this to problems in one dimension
Momentum Defined as \( p = mv \)
Momentum is a fundamental quantity in mechanics that describes the motion of an object in terms of both its mass and its velocity.
Definition: Momentum is the product of an object’s mass and its velocity.
\( p = mv \)
- \( p \) = momentum (kg m s⁻¹)
- \( m \) = mass (kg)
- \( v \) = velocity (m s⁻¹)
Key Points:
- Momentum is a vector quantity (direction is important).
- Momentum increases with either higher mass or higher velocity.
- Objects moving faster or with greater mass have more “impact” when they collide.
- If the velocity changes direction, the momentum direction also changes.
Understanding the Concept
- An object with zero velocity has zero momentum.
- Doubling the mass doubles the momentum.
- Doubling the velocity also doubles the momentum.
- Momentum helps describe motion in collisions and explosions.
How to Calculate Change in Momentum
The change in momentum is defined as the difference between final and initial momentum:
\( \Delta p = p_f – p_i \)
- \( p_i = m u \) = initial momentum
- \( p_f = m v \) = final momentum
- The sign of velocity matters (use negative for opposite direction).
- If an object reverses direction, the change in momentum becomes large because signs differ.
Example (Easy)
A 2 kg object moves with a velocity of \( 3\, \mathrm{m\,s^{-1}} \). Find its momentum.
▶️ Answer / Explanation
\( p = mv = 2 \times 3 = 6\, \mathrm{kg\,m\,s^{-1}} \)
Example (Medium)
A car of mass \( 1200\, \mathrm{kg} \) travels at \( 20\, \mathrm{m\,s^{-1}} \). Calculate its momentum.
▶️ Answer / Explanation
\( p = mv = 1200 \times 20 = 24000\, \mathrm{kg\,m\,s^{-1}} \)
Example (Hard)
A tennis ball of mass \( 0.060\, \mathrm{kg} \) is moving at \( 25\, \mathrm{m\,s^{-1}} \). After hitting a wall, it rebounds in the opposite direction at \( 18\, \mathrm{m\,s^{-1}} \). Find the change in momentum (take rebound direction as negative).
▶️ Answer / Explanation
Initial momentum:
\( p_i = 0.060 \times 25 = 1.5\, \mathrm{kg\,m\,s^{-1}} \)
Final momentum (opposite direction):
\( p_f = 0.060 \times (-18) = -1.08\, \mathrm{kg\,m\,s^{-1}} \)
Change in momentum:
\( \Delta p = p_f – p_i = -1.08 – 1.5 = -2.58\, \mathrm{kg\,m\,s^{-1}} \)
Principle of Conservation of Linear Momentum
Definition: The total linear momentum of a system remains constant provided no external forces act on the system.
\( \text{Total momentum before} = \text{Total momentum after} \)
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Momentum is conserved in all interactions (collisions and explosions).
- Applies only when the net external force on the system is zero.
- Works in one dimension and can be extended to two dimensions.
Connection to Newton’s Laws of Motion
Link to Newton’s Third Law:
- In a collision, the force body A exerts on B is equal and opposite to the force body B exerts on A.
- These forces act for the same time interval.
- Therefore, the impulses are equal and opposite.
- This leads directly to conservation of momentum.
\( F_{AB} \Delta t = -F_{BA} \Delta t \)
This shows that the total momentum change of the two-body system is zero.
Link to Newton’s Second Law:
- Newton’s second law in terms of momentum is \( F = \dfrac{\Delta p}{\Delta t} \).
- If no external force acts, then \( F = 0 \Rightarrow \Delta p = 0 \).
- This means momentum is constant.
Applying Conservation of Momentum in One Dimension
Use the equation:
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
Sign convention: Choose one direction as positive (usually right). Negative sign indicates motion in the opposite direction.
- Collisions (objects come together).
- Explosions (objects move apart).
- Recoil motion (e.g., gun and bullet).
- Elastic or inelastic collisions (momentum always conserved).
Example (Easy)
A 2 kg trolley moving at \( 3\, \mathrm{m\,s^{-1}} \) collides and sticks to a stationary 1 kg trolley. Find their common velocity.
▶️ Answer / Explanation
Total momentum before:
\( p = (2 \times 3) + (1 \times 0) = 6 \)
Total mass after collision = \( 3\, \mathrm{kg} \).
Common velocity:
\( v = \dfrac{6}{3} = 2\, \mathrm{m\,s^{-1}} \)
Example (Medium)
A 1500 kg car moving at \( 18\, \mathrm{m\,s^{-1}} \) hits a 900 kg car moving in the same direction at \( 10\, \mathrm{m\,s^{-1}} \). After collision, they move together. Find their joint velocity.
▶️ Answer / Explanation
Momentum before:
\( p = (1500 \times 18) + (900 \times 10) = 27000 + 9000 = 36000 \)
Total mass = \( 2400\, \mathrm{kg} \).
Velocity after:
\( v = \dfrac{36000}{2400} = 15\, \mathrm{m\,s^{-1}} \)
Example (Hard)
A 0.5 kg ball moving at \( 12\, \mathrm{m\,s^{-1}} \) strikes a 1.2 kg ball moving toward it at \( 4\, \mathrm{m\,s^{-1}} \). After collision, the 0.5 kg ball rebounds at \( -6\, \mathrm{m\,s^{-1}} \). Find the final velocity of the 1.2 kg ball.
▶️ Answer / Explanation
Take the 0.5 kg ball’s initial direction as positive.
Momentum before:
\( p = (0.5 \times 12) + (1.2 \times -4) = 6 – 4.8 = 1.2 \)
Momentum after:
\( 1.2 = (0.5 \times -6) + (1.2 \, v) \)
Solve:
\( 1.2 = -3 + 1.2v \)
\( 1.2v = 4.2 \Rightarrow v = 3.5\, \mathrm{m\,s^{-1}} \)