CIE IGCSE Physics (0625) Momentum Study Notes - New Syllabus
CIE IGCSE Physics (0625) Topic 1.6 Momentum Study Notes
Key Concepts:
Core
- Define momentum as mass × velocity; recall and use the equation p = mv
- Define impulse as force × time for which force acts; recall and use the equation impulse = FΔt = Δ(mv)
- Apply the principle of the conservation of momentum to solve simple problems in one dimension
- Define resultant force as the change in momentum per unit time; recall and use the equation F = Δp/Δt
Supplement
- No additional Supplement content
Momentum
Momentum:
- Momentum is a measure of the motion of a moving object.
- It depends on both the mass of the object and its velocity.
- Momentum is a vector quantity — it has both magnitude and direction (same as the direction of velocity).
Formula:
\( p = mv \)
- \( p \) = momentum (kg·m/s)
- \( m \) = mass (kg)
- \( v \) = velocity (m/s)
Key Points:
- Greater mass or greater velocity → larger momentum.
- Stationary objects have zero momentum because \( v = 0 \).
Example:
A 2.5 kg trolley moves at a velocity of 3 m/s. Calculate its momentum.
▶️ Answer/Explanation
Use the formula: \( p = mv \)
\( p = 2.5 \times 3 = \boxed{7.5 \, \text{kg·m/s}} \)
Impulse
Impulse:
- Impulse is the product of the force applied to an object and the time for which the force acts.
- Impulse causes a change in momentum of the object.
- Impulse is a vector quantity and acts in the direction of the applied force.
Formula for Impulse:
\( \text{Impulse} = F \times \Delta t = \Delta (mv) \)
- \( F \) = force (N)
- \( \Delta t \) = time for which force acts (s)
- \( \Delta (mv) \) = change in momentum (kg·m/s)
- Impulse has the same units as momentum: newton-seconds (N·s) or kg·m/s
Key Points:
- If a force acts for a longer time, the change in momentum is larger.
- Impulse is useful in real-life safety applications like airbags, helmets, and cushioned packaging — they increase time and reduce force.
Example:
A cricket ball of mass 0.2 kg is hit by a bat. The velocity changes from 5 m/s to -8 m/s in 0.02 s. Calculate:
- Change in momentum (impulse)
- Average force applied by the bat
▶️ Answer/Explanation
1. Change in momentum:
\( \Delta (mv) = m(v_{\text{final}} – v_{\text{initial}}) = 0.2(-8 – 5) = 0.2 \times (-13) = \boxed{-2.6 \, \text{kg·m/s}} \)
2. Force:
\( F = \dfrac{\text{Impulse}}{\Delta t} = \dfrac{-2.6}{0.02} = \boxed{-130 \, \text{N}} \)
The negative sign indicates that the force is in the opposite direction to the ball’s initial motion.
Law of Conservation of Momentum
Law of Conservation of Momentum:
- The total momentum of a system remains constant, provided no external force acts on it.
- This means the total momentum before a collision or explosion is equal to the total momentum after.
Total momentum before = Total momentum after
\( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \)
- \( m_1, m_2 \) = masses of objects
- \( u_1, u_2 \) = initial velocities
- \( v_1, v_2 \) = final velocities
Conditions:
- This formula assumes motion in a straight line (1D).
- Velocity direction must be considered; opposite directions must be treated as negative.
Example:
A 4 kg trolley moving at 3 m/s collides with a stationary 2 kg trolley. After the collision, the 4 kg trolley slows down to 1.5 m/s. What is the final velocity of the 2 kg trolley? Assume no external forces act.
▶️ Answer/Explanation
Step 1: Use conservation of momentum
Initial momentum: \( (4 \times 3) + (2 \times 0) = 12 \, \text{kg·m/s} \)
Final momentum: \( (4 \times 1.5) + (2 \times v_2) = 6 + 2v_2 \)
Step 2: Set initial = final
\( 12 = 6 + 2v_2 \Rightarrow 2v_2 = 6 \Rightarrow v_2 = \boxed{3 \, \text{m/s}} \)
The 2 kg trolley moves at 3 m/s after the collision.
Resultant Force in Terms of Momentum
Resultant Force in Terms of Momentum:
- The resultant force acting on an object is equal to the rate of change of momentum.
- This is a reformulation of Newton’s Second Law: force causes a change in momentum over time.
Equation:
\( F = \dfrac{\Delta p}{\Delta t} = \dfrac{mv – mu}{t} \)
- \( F \) = resultant force (N)
- \( \Delta p \) = change in momentum (kg·m/s)
- \( m \) = mass (kg)
- \( u \) = initial velocity (m/s)
- \( v \) = final velocity (m/s)
- \( t \) = time (s)
Units: Newton (N) = kg·m/s²
Key Point: If momentum changes quickly, the force is large. If the same momentum change happens over a longer time, the force is smaller (e.g. airbags, crumple zones).
Example:
A 0.1 kg tennis ball is hit, changing its velocity from 10 m/s to -20 m/s in 0.05 s. Calculate the average force applied by the racket.
▶️ Answer/Explanation
Step 1: Calculate change in momentum
\( \Delta p = m(v – u) = 0.1(-20 – 10) = 0.1 \times (-30) = -3.0 \, \text{kg·m/s} \)
Step 2: Use \( F = \dfrac{\Delta p}{\Delta t} \)
\( F = \dfrac{-3.0}{0.05} = \boxed{-60 \, \text{N}} \)
The negative sign means the force is in the opposite direction to the ball’s original motion.