Momentum is mass times velocity, and in a collision between two particles the total momentum before equals the total momentum after. With a sign for each direction, one equation solves every Paper 4 collision.
Momentum \(= mv\), measured in \(\text{kg m s}^{-1}\) (or N s), with a sign showing its direction. In a direct collision: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\). If the particles coalesce, \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\).
Fig. 1: two particles moving towards each other coalesce. With right as positive, the 2 kg particle has velocity −2 m s⁻¹, so the combined particle moves right at 2.2 m s⁻¹
Section 01
Momentum is a vector.
In the lab, give A and B velocities in opposite directions and watch the total momentum readout before and after the collision. Change the masses: the individual momenta change, but the total stays the same.
Momentum \(= mv\), in \(\text{kg m s}^{-1}\) or N s. Masses in grams must be changed to kg.
Direction matters: choose a positive direction; a velocity the other way is negative, and so is its momentum.
Conservation: in a collision the particles push on each other with equal and opposite forces (Newton's third law), so the total momentum does not change.
Impulse and the coefficient of restitution are not part of this syllabus.
Worked example Change of momentum
A ball of mass 200 g hits a wall at \(15\ \text{m s}^{-1}\) and rebounds along the same line at \(12\ \text{m s}^{-1}\). Find the change in its momentum.
Mass \(= 0.2\) kg. Take the direction towards the wall as positive.
Subtracting the speeds gives \(0.2(15 - 12) = 0.6\) N s, which is wrong. The velocity changed sign, so the sizes ADD: \(0.2(15 + 12) = 5.4\) N s.
Section 02
Conservation in a collision.
Draw a before-and-after diagram with every velocity marked, choose a positive direction, and write one equation: total momentum before \(=\) total momentum after.
Worked example Coalescing (Fig. 1)
A particle of mass 3 kg moving at \(5\ \text{m s}^{-1}\) collides directly with a particle of mass 2 kg moving at \(2\ \text{m s}^{-1}\) in the opposite direction. They coalesce. Find their common velocity.
Positive in the direction of the 3 kg particle.
Before: \(3(5) + 2(-2) = 15 - 4 = 11\).
After: \((3 + 2)v = 5v\).
\(5v = 11\), so \(v = 2.2\).
Answer: \(2.2\ \text{m s}^{-1}\) in the original direction of the 3 kg particle.
Worked example One particle continues
A particle A of mass 4 kg moving at \(6\ \text{m s}^{-1}\) hits a stationary particle B of mass 2 kg. After the collision A moves on in the same direction at \(2\ \text{m s}^{-1}\). Find the velocity of B.
Before: \(4(6) + 2(0) = 24\).
After: \(4(2) + 2v_B = 8 + 2v_B\).
\(8 + 2v_B = 24\), so \(v_B = 8\).
Answer: \(8\ \text{m s}^{-1}\) in the direction A was moving.
Worked example A particle reverses
Particles A (0.5 kg) and B (0.3 kg) move towards each other with speeds \(4\ \text{m s}^{-1}\) and \(6\ \text{m s}^{-1}\). After colliding, B moves at \(3\ \text{m s}^{-1}\) in the direction A was originally moving. Find the velocity of A after the collision.
Positive in A's original direction. Before: \(0.5(4) + 0.3(-6) = 2 - 1.8 = 0.2\).
After: \(0.5v_A + 0.3(3) = 0.5v_A + 0.9\).
\(0.5v_A = 0.2 - 0.9 = -0.7\), so \(v_A = -1.4\).
Answer: A moves at \(1.4\ \text{m s}^{-1}\) in the opposite direction to its original motion.
Your turn
A particle of mass \(m\) kg moving at \(5\ \text{m s}^{-1}\) hits a stationary particle of mass 3 kg. They coalesce and move at \(2\ \text{m s}^{-1}\). Find \(m\).
\(5m = 2(m + 3)\), so \(3m = 6\) and \(m = 2\).
Examiner trap
Every unknown velocity after the collision should be written as a signed variable, and the sign of the answer tells you the direction. A negative answer is not an error; it means the particle moves in the negative direction.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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Find the momentum of a 0.4 kg ball moving at \(5\ \text{m s}^{-1}\).
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Which unit is equivalent to \(\text{kg m s}^{-1}\)?
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A 2 kg particle moving at \(3\ \text{m s}^{-1}\) hits a stationary 1 kg particle and they coalesce. Find their common speed.
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Why is momentum conserved when two particles collide?
Section 03
Repeated collisions.
With three particles in a line, treat the collisions one at a time. The velocity after the first collision becomes the "before" velocity for the second. At the end, compare velocities to decide whether any more collisions happen.
Worked example Three particles
Particles A (2 kg), B (1 kg) and C (3 kg) lie in a straight line. A moves towards B at \(6\ \text{m s}^{-1}\); B is at rest. A and B coalesce, and the combined particle then hits C, which is moving towards it at \(1\ \text{m s}^{-1}\). All three coalesce. Find the final velocity.
First collision: \(2(6) + 1(0) = 3v_1\), so \(v_1 = 4\ \text{m s}^{-1}\).
Second collision, positive in A's direction: \(3(4) + 3(-1) = 6v_2\).
\(6v_2 = 9\), so \(v_2 = 1.5\).
Answer: \(1.5\ \text{m s}^{-1}\) in A's original direction.
Examiner trap
Do not do both collisions in one equation unless all the particles really do end up moving together. If A and B separate after the first collision, each must be tracked on its own.
Stage 2 · Exam
Exam-style questions
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A 3 kg particle moving at \(5\ \text{m s}^{-1}\) and a 2 kg particle moving at \(2\ \text{m s}^{-1}\) in the opposite direction collide and coalesce. Find their common speed.
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A (0.5 kg) and B (0.3 kg) approach each other at \(4\ \text{m s}^{-1}\) and \(6\ \text{m s}^{-1}\). After the collision B moves at \(3\ \text{m s}^{-1}\) in A's original direction. Find A's velocity after the collision.
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A (2 kg, \(6\ \text{m s}^{-1}\)) hits stationary B (1 kg) and they coalesce. The combined particle then hits C (3 kg), moving towards it at \(1\ \text{m s}^{-1}\), and all three coalesce. Find the final speed.
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P, of mass \(m\) kg, moving at \(5\ \text{m s}^{-1}\), catches up with Q, of mass 3 kg, moving at \(1\ \text{m s}^{-1}\) in the same direction. They coalesce and move at \(2.5\ \text{m s}^{-1}\). Find \(m\).
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Momentum conserved.
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