The shape of the lesson
By the end of the lesson, learners can
- 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.
- 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.
- Repeated collisions: With three particles in a line, treat the collisions one at a time.
- Diagram: before and after, masses in kg, every velocity marked.
- Positive direction: velocities the other way are negative.
- Conserve: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\); coalescing uses \((m_1 + m_2)v\).
- Interpret: a negative answer means the particle moves in the negative direction.
- Chains: one collision at a time, then check whether another collision occurs.
One picture every idea returns to
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, \(m_1u_1 + m_2u_2 = (m_1 + m_2)v\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, What goes in, comes out.. Momentum is mass times velocity, and in a collision between two particles the total momentum before equals the total momentum after. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: 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. Project tab 1 of the lab (Momentum lab, momentum and conservation) and let a student drive it while the class predicts. Model the worked example Change of momentum on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: 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. Model the worked example Coalescing (Fig. 1) on the board, one line at a time. 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 \(m\). | Pairs |
| 21 to 30 min | Teach and check: Repeated collisions | With three particles in a line, treat the collisions one at a time. Model the worked example Three particles on the board, one line at a time. | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 20 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Repeated collisions, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| 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. |
| 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. |
| 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. |
Momentum and the conservation of momentum: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Find the magnitude of the momentum of (a) a ball of mass 150 g moving at \(20\ \text{m s}^{-1}\), (b) a van of mass 1200 kg moving at \(15\ \text{m s}^{-1}\).
A ball of mass 0.2 kg moving at \(15\ \text{m s}^{-1}\) hits a wall at right angles and rebounds at \(12\ \text{m s}^{-1}\). Find the magnitude of the change in momentum.
A particle A of mass 4 kg moving at \(6\ \text{m s}^{-1}\) collides with a stationary particle B of mass 2 kg. After the collision A continues in the same direction at \(2\ \text{m s}^{-1}\). Find the speed of B.
Two particles, of masses 0.3 kg and \(m\) kg, move towards each other along a straight line with speeds \(4\ \text{m s}^{-1}\) and \(3\ \text{m s}^{-1}\). They collide and coalesce, and the combined particle is at rest. Find \(m\).
Particles A, B and C, of masses 0.2 kg, 0.3 kg and 0.5 kg, lie at rest in that order on a straight line on a smooth surface. A is projected towards B at \(5\ \text{m s}^{-1}\). After A hits B, A is at rest. B then hits C and they coalesce.
(a) Find the speed of B after the first collision.
(b) Find the speed of the combined particle, and explain why there are no further collisions.
A railway truck of mass 12 tonnes moving at \(3\ \text{m s}^{-1}\) runs into a stationary truck of mass 8 tonnes, and the two couple together.
(a) Find their speed immediately after coupling.
(b) Find the loss of kinetic energy in the collision.
Answer key and marking notes
- (a) \(0.15 \times 20 = 3\ \text{kg m s}^{-1}\).
- (b) \(1200 \times 15 = 18\,000\ \text{kg m s}^{-1}\).
- Towards the wall positive: before \(3\), after \(-2.4\ \text{N s}\).
- Change \(= -2.4 - 3 = -5.4\), magnitude 5.4 N s.
- \(4(6) + 0 = 4(2) + 2v_B\).
- \(24 = 8 + 2v_B\), so \(v_B = 8\ \text{m s}^{-1}\).
- Total momentum after is zero, so total momentum before is zero.
- \(0.3(4) + m(-3) = 0\).
- \(m = 1.2 \div 3 = 0.4\).
- (a) \(0.2(5) = 0.2(0) + 0.3v_B\), so \(v_B = \dfrac{1}{0.3} = 3.33\ \text{m s}^{-1}\).
- (b) \(0.3 \times \dfrac{10}{3} = 0.8v\), so \(1 = 0.8v\) and \(v = 1.25\ \text{m s}^{-1}\).
- A is at rest and the combined particle moves away from it, so they never meet again.
- (a) \(12\,000(3) = 20\,000v\), so \(v = 1.8\ \text{m s}^{-1}\).
- (b) KE before \(= \tfrac{1}{2}(12\,000)(3^2) = 54\,000\) J.
- KE after \(= \tfrac{1}{2}(20\,000)(1.8^2) = 32\,400\) J.
- Loss \(= 21\,600\) J. Momentum is conserved but kinetic energy is not.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A ball of mass 0.15 kg hits a wall at \(20\ \text{m s}^{-1}\) and rebounds along the same line at \(14\ \text{m s}^{-1}\). Find the magnitude of the change in its momentum.
- A5.1 N s
- B0.9 N s
- C2.1 N s
- D510 N s
A particle of mass 4 kg moving at \(6\ \text{m s}^{-1}\) collides with a stationary particle of mass 2 kg. The particles coalesce. Find their common velocity.
- A\(4\ \text{m s}^{-1}\)
- B\(12\ \text{m s}^{-1}\)
- C\(3\ \text{m s}^{-1}\)
- D\(6\ \text{m s}^{-1}\)
Particles of mass 3 kg and 5 kg move towards each other along a straight line with speeds \(4\ \text{m s}^{-1}\) and \(2\ \text{m s}^{-1}\) respectively. They collide and coalesce. Find the velocity of the combined particle.
- A\(0.25\ \text{m s}^{-1}\) in the original direction of the 3 kg particle
- B\(0.25\ \text{m s}^{-1}\) in the original direction of the 5 kg particle
- C\(2.75\ \text{m s}^{-1}\) in the original direction of the 3 kg particle
- D\(1\ \text{m s}^{-1}\) in the original direction of the 3 kg particle
Which of these units is equivalent to the unit of momentum, \(\text{kg m s}^{-1}\)?
- AN s
- BN
- CJ
- DW
Particle A of mass 0.4 kg moving at \(5\ \text{m s}^{-1}\) and particle B of mass 0.6 kg moving at \(3\ \text{m s}^{-1}\) approach each other along a straight line. After the collision, A moves back along its original path at \(2\ \text{m s}^{-1}\), as shown. Find the velocity \(v\) of B after the collision.
- A\(1.67\ \text{m s}^{-1}\) in A's original direction
- B\(1\ \text{m s}^{-1}\) in B's original direction
- C\(7.67\ \text{m s}^{-1}\) in A's original direction
- D\(1.67\ \text{m s}^{-1}\) in B's original direction
Particles A (3 kg), B (2 kg) and C (4 kg) lie in that order on a straight line. A moves towards B at \(5\ \text{m s}^{-1}\); B is at rest; C moves away from B at \(1\ \text{m s}^{-1}\). A and B collide and coalesce, then the combined particle collides with C and they coalesce. Find the final velocity.
- A\(2.11\ \text{m s}^{-1}\)
- B\(1.22\ \text{m s}^{-1}\)
- C\(1.67\ \text{m s}^{-1}\)
- D\(3\ \text{m s}^{-1}\)
Particle A (2 kg) moving at \(6\ \text{m s}^{-1}\) hits particle B (3 kg), which is at rest. After the collision A moves on in the same direction at \(0.6\ \text{m s}^{-1}\). B then hits particle C (5 kg), which is at rest, and B and C coalesce. Find the velocity of the combined particle and decide whether A collides with it again.
- A\(1.35\ \text{m s}^{-1}\); A does not collide again
- B\(1.35\ \text{m s}^{-1}\); A collides with it again
- C\(1.5\ \text{m s}^{-1}\); A does not collide again
- D\(2.16\ \text{m s}^{-1}\); A does not collide again
Particle P of mass \(m\) kg moving at \(6\ \text{m s}^{-1}\) collides directly with particle Q of mass 2 kg moving towards it at \(1.5\ \text{m s}^{-1}\). After the collision P is at rest and Q moves at \(3\ \text{m s}^{-1}\) in P's original direction. Find \(m\).
- A1.5
- B0.5
- C1
- D0.75
Quiz key
Change \(= 0.15 \times 14 - 0.15 \times (-20) = 2.1 + 3 = 5.1\) N s. The tempting 0.9 N s is \(0.15(20 - 14)\), which treats the rebound as if the ball carried on in the same direction; because the velocity reverses, the sizes add. 2.1 N s is the momentum after the rebound only, and 510 uses the mass in grams.
\(4 \times 6 + 2 \times 0 = 6v\), so \(v = 4\ \text{m s}^{-1}\). The tempting 12 divides the momentum by the 2 kg mass only, forgetting that the 4 kg particle moves on too. 3 averages the two velocities without weighting by mass, and 6 assumes the speed is unchanged, which would create momentum.
\(3 \times 4 + 5 \times (-2) = 8v\), so \(2 = 8v\) and \(v = 0.25\ \text{m s}^{-1}\). It is positive, so the combined particle moves in the 3 kg particle's original direction (that particle had more momentum: 12 against 10). The tempting 2.75 adds the momenta (\(22 \div 8\)), ignoring the opposite directions. 1 divides the net momentum by 2 instead of the total mass 8.
\(1\ \text{N s} = 1\ \text{kg m s}^{-2} \times \text{s} = 1\ \text{kg m s}^{-1}\), so momentum (and change in momentum, the impulse) can be given in N s. The newton alone is \(\text{kg m s}^{-2}\), the unit of force. The joule is \(\text{kg m}^2\,\text{s}^{-2}\) (energy) and the watt is \(\text{J s}^{-1}\) (power).
Before: \(0.4 \times 5 + 0.6 \times (-3) = 0.2\). After: \(0.4 \times (-2) + 0.6v = 0.2\), so \(0.6v = 1\) and \(v = 1.67\ \text{m s}^{-1}\), positive, so in A's original direction. The tempting \(1\ \text{m s}^{-1}\) in B's direction keeps A's final velocity as \(+2\), ignoring that A rebounded. 7.67 treats both initial momenta as positive, and the last option has the right size but the wrong direction.
First: \(3 \times 5 = 5v_1\), so \(v_1 = 3\ \text{m s}^{-1}\). Second: \(5 \times 3 + 4 \times 1 = 9v_2\), so \(v_2 = \dfrac{19}{9} = 2.11\ \text{m s}^{-1}\). The tempting 1.22 gives C a negative velocity, but C moves away from B, the same way as A. 1.67 leaves out C's momentum, and 3 is the velocity after the first collision only.
First collision: \(12 = 2 \times 0.6 + 3v_B\), so \(v_B = 3.6\ \text{m s}^{-1}\). Second: \(3 \times 3.6 = 8v\), so \(v = 1.35\ \text{m s}^{-1}\). A moves at only \(0.6\ \text{m s}^{-1}\), slower than 1.35, so it never catches up. The tempting "collides again" compares the wrong speeds. 1.5 uses A's full 12 N s for the second collision, and 2.16 divides by C's mass only.
\(6m + 2 \times (-1.5) = 0 + 2 \times 3\), so \(6m - 3 = 6\) and \(m = 1.5\). The tempting 0.5 gives Q's initial momentum a positive sign (\(6m + 3 = 6\)), ignoring that it moved towards P. 1 leaves out Q's initial momentum, and 0.75 forgets to multiply Q's velocities by its mass (\(6m - 1.5 = 3\)).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.