Conservation of momentum
Section: Dynamics | Syllabus: Cambridge AS Level Physics 9702
The Principle of Conservation of Momentum Principle of Conservation of Momentum In a closed system (where no external forces act), the total momentum remains constant during any interaction. Derivation from Newton's Third Law Conservation of momentum follows directly from Newton's third law.
If two balls A and B collide: By the 3rd Law, the force on A by B (F_A) is equal and opposite to the force on B by A (F_B): F_A = -F_B By the 2nd Law, F = fraction. So: fraction = -fraction The interaction time Δ t is identical for both: Δ p_A = -Δ p_B This means the momentum gained by one object is exactly lost by the other.
The total momentum before collision equals the total momentum after. Interactions in One Dimension When objects interact in a straight line, we use the scalar form of the conservation equation: m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 Always assign one direction as positive (e.g., right) and the other as negative (left).
Diagram: 1D Collision (Before and After) Show two trolleys moving towards each other (labeled with masses and velocities) in the 'Before' state, and moving together or apart in the 'After' state. Worked Example: Snooker Ball Collision Scenario: A white ball (m=0.16 kg) moving at 0.5 m s^-1 hits a stationary red ball of equal mass.
If the white ball stops dead, what is the final velocity of the red ball? Solution Total initial momentum = 0.16 × 0.5 + 0.16 × 0 = 0.08 kg m s^-1 Total final momentum = 0.16 × 0 + 0.16 × v = 0.16v 0.16v = 0.08 v = 0.5 m s^-1 Worked Example: Railway Wagons Joining Question: A wagon (8.1 × 10^4 kg) at 2.4 m s^-1 hits another (4.5 × 10^4 kg) moving at 0.75 m s^-1 in the opposite direction.
They join together. Find their final velocity. Solution Let Right be positive (+): (8.1 × 10^4 × 2.4) + (4.5 × 10^4 × -0.75) = (8.1 × 10^4 + 4.5 × 10^4)v 194,400 - 33,750 = 126,000v 160,650 = 126,000v v = 1.28 m s^-1 (moving right) Explosions and Recoil An "explosion" in physics refers to any interaction where objects separate from each other.
The total initial momentum is zero. Worked Example: Cannon Recoil Question: A 100 kg cannon fires a 5 kg ball at 80 m s^-1. Calculate the recoil velocity of the cannon. Calculation Initial p = 0. After firing: 0 = m_c v_c + m_b v_b 0 = (100 × v_c) + (5 × 80) 100v_c = -400 v_c = -4 m s^-1 (opposite to ball) Interactions in Two Dimensions Momentum is a vector, so it must be conserved independently in the X and Y directions.
Strategy: Simultaneous Equations Often you will resolve velocities into components using θ and θ, creating two equations to solve for two unknowns. Diagram: Vector Resolution for 2D Collision Show two particles colliding at an angle.
Draw X and Y axes and show the velocity vectors being resolved into v sin heta and v cos heta components. Worked Example: 2D Tennis Ball Collision Scenario: Ball A hits stationary Ball B. A moves at 30^ to its path, B at 40^ to the other side.
Initial u_A = 2.4 m s^-1. Masses are equal. Solution X Direction: 2.4 = v_A 30^ + v_B 40^ Y Direction: 0 = v_A 30^ - v_B 40^ v_A = v_B fraction Substituting (2) into (1) gives v_A = 1.64 m s^-1 and v_B = 1.28 m s^-1.
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