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:

  1. 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
  2. By the 2nd Law, F = (Δ p/Δ t). So: (Δ p_A/Δ t) = -(Δ p_B/Δ t)
  3. 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 cosθ and sinθ, 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
  1. X Direction: 2.4 = v_A cos 30^ + v_B cos 40^
  2. Y Direction: 0 = v_A sin 30^ - v_B sin 40^ v_A = v_B (sin 40^/sin 30^)

Substituting (2) into (1) gives v_A = 1.64 m s^-1 and v_B = 1.28 m s^-1.

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