4.1 Impulse and momentum in two dimensions
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A2 content for Unit 4 (WPH14), specification points 81–86, including Core Practicals 9 and 10. Momentum in one dimension is on the Unit 1 momentum page.
What this page covers
8 points
- Newton's second law in momentum form, \(F = \Delta p/\Delta t\), and impulse \(F\Delta t = \Delta p\)
- Impulse as the area under a force–time graph; why longer contact times mean smaller forces
- Core Practical 9: investigating the relationship between force and change of momentum
- Conservation of momentum in two dimensions: glancing collisions
- Explosions into several directions
- Elastic and inelastic collisions, and how to decide which one you have
- Deriving and using \(E_k = p^2/2m\)
- Core Practical 10: using video analysis to study collisions between small spheres
In one minute
6 points
- Newton's second law says the resultant force equals the rate of change of momentum, \(F = \Delta p/\Delta t\), so a force acting for a time gives an impulse \(F\Delta t = \Delta p\).
- Impulse has unit N s (= kg m s⁻¹). For a varying force it is the area under the force–time graph.
- For the same change in momentum, a longer contact time means a smaller average force.
- In two dimensions, momentum is conserved separately in each of two perpendicular directions (if no resultant external force acts).
- Elastic: total kinetic energy is unchanged. Inelastic: some kinetic energy is transferred to other stores. Momentum is conserved in both.
- \(E_k = p^2/2m\): for the same momentum, the lighter body has more kinetic energy.
Contents
14 sections
- What is impulse, and how is it linked to Newton's second law?
- How do you find an impulse from a force–time graph?
- Core Practical 9: how do you investigate force and change of momentum?
- How do you apply conservation of momentum in two dimensions?
- How do you handle an explosion into several directions?
- What makes a collision elastic or inelastic?
- Where does \(E_k = p^2/2m\) come from, and when is it useful?
- Core Practical 10: how do you analyse collisions between spheres using video?
- Exam tips for 4.1
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is impulse, and how is it linked to Newton's second law?
In Unit 1 you met momentum, p = mv, and its conservation along a line. At A2 force itself is defined through momentum.
For a body of constant mass, \(\Delta p/\Delta t = m\Delta v/\Delta t = ma\), so F = ma is a special case. The momentum form also works when the mass changes, for example a jet of water hitting a wall, where the force is (mass arriving per second) × (change in velocity). Multiplying both sides by Δt gives the impulse of the force:
\[F\Delta t = \Delta p\]Here F is the resultant force (N), Δt the time for which it acts (s) and Δp the change of momentum (kg m s⁻¹). This equation is in the formulae list. If the force varies, F is the average resultant force over Δt. Impulse is a vector in the direction of the resultant force, and its unit N s is the same as kg m s⁻¹.
A volleyball of mass 0.27 kg travels horizontally towards a player at 9.0 m s⁻¹. She hits it straight back along the same line at 21 m s⁻¹. Her hand is in contact with the ball for 0.015 s. Calculate the average force her hand exerts on the ball.
Take the return direction as positive, so u = −9.0 m s⁻¹ and v = +21 m s⁻¹. \(\Delta p = 0.27 \times (21 - (-9.0)) = 0.27 \times 30 = 8.1\ \text{N s}\).
\(F = \dfrac{\Delta p}{\Delta t} = \dfrac{8.1}{0.015} =\) 540 N in the return direction
Using 21 − 9.0 instead gives 220 N: the reversal of direction has been ignored.
How do you find an impulse from a force–time graph?
In a real impact the force is not constant: it rises, peaks and falls in a few milliseconds. Each thin strip under a force–time graph has area FΔt, so the area under the graph is the impulse, and therefore the change of momentum. For straight-line graphs, split the area into triangles and rectangles. For a curve, count squares and multiply by the impulse that one square represents. The average force is the area divided by the total contact time.
A hockey puck of mass 0.17 kg is at rest on the ice. A stick strikes it. The force on the puck rises steadily from zero to 450 N in 2.0 ms, stays at 450 N for 4.0 ms, then falls steadily to zero over 2.0 ms (figure above). Friction from the ice is negligible during the strike. (a) Calculate the speed of the puck as it leaves the stick. (b) Calculate the average force on the puck during the contact.
(a) Area = two triangles + one rectangle, with times in seconds: \(2 \times \tfrac12 \times 450 \times 2.0\times10^{-3} + 450 \times 4.0\times10^{-3} = 0.90 + 1.80 = 2.7\ \text{N s}\).
Impulse = change of momentum, from rest: \(v = \dfrac{2.7}{0.17} = 15.9\ \text{m s}^{-1}\) ≈ 16 m s⁻¹
(b) \(F_{av} = \dfrac{2.7}{8.0\times10^{-3}} = 337.5\ \text{N}\) ≈ 340 N
450 N × 4.0 ms is 1.8 N s, not 1800 N s. Convert ms to s before you multiply, or the impulse comes out a thousand times too big.
Why longer contact times protect you
When a moving body is brought to rest, its change of momentum is fixed by its mass and speed. Only the contact time can be changed. Since F = Δp/Δt, increasing Δt reduces the average force. Crumple zones, airbags, helmet foam, gym mats and phone cases all work this way. The same idea works in reverse: to give a ball a large change of momentum, "follow through" so the force acts for longer.
Core Practical 9: how do you investigate force and change of momentum?
The aim is to show that the impulse of a force (the area under its force–time graph) equals the change of momentum it produces. One way to do it:
- Set-up: a trolley carrying a card runs down a runway towards a force sensor fixed at the end. A pad of modelling clay on the sensor makes the trolley stop dead rather than bounce. A light gate just before the sensor times the card. Both connect to a data logger.
- Friction compensation: tilt the runway until a trolley given a small push moves at constant speed (the light-gate times at two points are equal). Then there is no resultant external force on the trolley before the impact.
- Measurements: mass of trolley and card on a top-pan balance; card length with a ruler or vernier calipers; the time for the card to pass through the gate; the force–time graph of the impact, sampled at a high rate (at least 1000 readings per second, because the impact lasts only tens of milliseconds).
- Calculations: u = card length ÷ time, so the change of momentum is mu (final momentum zero). The logger's software finds the area under the force–time graph.
- Range: release the trolley from different points to change u, and repeat each run. Plot impulse (area) against mΔv: a straight line through the origin with gradient 1 supports FΔt = Δp.
- Safety: stop the trolley at the end of the bench; keep the runway stable when it is tilted.
In one run, the trolley and card have mass (0.640 ± 0.005) kg. The card is (0.0500 ± 0.0005) m long and passes through the light gate in (0.0625 ± 0.0010) s. The trolley stops against the pad. The data logger gives the area under the force–time graph as 0.501 N s. (a) Calculate the change of momentum of the trolley and its percentage uncertainty. (b) Deduce whether the result supports impulse = change of momentum.
(a) \(u = \dfrac{0.0500}{0.0625} = 0.800\ \text{m s}^{-1}\), so \(\Delta p = 0.640 \times 0.800 = 0.512\ \text{N s}\).
Δp = mL/t, so add the percentage uncertainties: 0.78% + 1.0% + 1.6% = 3.4%. Δp = 0.512 N s ± 3.4% (± 0.017 N s)
(b) Δp lies between 0.495 and 0.529 N s, and the impulse, 0.501 N s, is inside this range. The two values differ by 2.1%, less than the 3.4% uncertainty, so the result supports FΔt = Δp.
The timing is the largest source of uncertainty; a longer card would reduce it.
In Unit 4 you are expected to combine uncertainties. For a quantity found by multiplying or dividing (such as mL/t), add the percentage uncertainties. For a quantity found by adding or subtracting, add the absolute uncertainties.
How do you apply conservation of momentum in two dimensions?
Momentum is a vector. When bodies move in a plane, the total momentum is the vector sum of the individual momenta. If no resultant external force acts, the total momentum vector is constant, so its component in every direction is constant. That turns one vector equation into two ordinary ones.
- Choose two perpendicular directions (usually the original line of motion, x, and at right angles to it, y) and write one conservation equation for each.
- Resolve each momentum: \(p_x = p\cos\theta\), \(p_y = p\sin\theta\), with θ measured from the x direction. Signs show which side of the line a body moves.
- Recombine at the end: magnitude \(\sqrt{p_x^2 + p_y^2}\), direction \(\tan^{-1}(p_y/p_x)\).
A glancing (oblique) collision is one where the bodies do not move along the line joining their centres, so they move off at angles to the original line.
A curling stone A of mass 19.0 kg slides at 1.50 m s⁻¹ and strikes a stationary stone B of the same mass a glancing blow. Afterwards A moves at 0.75 m s⁻¹ at 55.0° to its original direction. Friction is negligible during the collision. Determine the velocity of B just after the collision.
Take x along A's original direction and y towards the side A moves to. Momentum before: x: 19.0 × 1.50 = 28.5 kg m s⁻¹; y: 0.
A after: x: 19.0 × 0.75 cos 55.0° = 8.17 kg m s⁻¹; y: 19.0 × 0.75 sin 55.0° = 11.67 kg m s⁻¹.
Conserve x: \(28.5 = 8.17 + 19.0v_x\), so \(v_x = 1.070\ \text{m s}^{-1}\). Conserve y: \(0 = 11.67 + 19.0v_y\), so \(v_y = -0.614\ \text{m s}^{-1}\).
Recombine: \(v = \sqrt{1.070^2 + 0.614^2} = 1.23\ \text{m s}^{-1}\) at \(\tan^{-1}(0.614/1.070) = 29.9^\circ\). 1.23 m s⁻¹ at 29.9° to A's original direction, on the opposite side of the line from A
You can also solve a two-body problem with a scale drawing or the cosine rule on the vector triangle, but components are safer and are expected when angles and speeds are both unknown. Always state the direction of a final velocity as an angle and a reference line.
How do you handle an explosion into several directions?
An object at rest has zero momentum, so after it breaks apart the vector sum of the fragments' momenta is still zero. With two fragments, they fly apart in exactly opposite directions. With three, their momenta drawn head to tail form a closed triangle, so the three velocities lie in one plane. If the object was moving before the explosion, the total momentum after equals its momentum before, in both components. Kinetic energy is not conserved: chemical or elastic potential energy is transferred to kinetic energy of the fragments.
A piece of space debris of mass 0.200 kg is at rest relative to an observer. It breaks into three fragments. A (0.050 kg) moves at 3.0 m s⁻¹ in the +x direction; B (0.080 kg) moves at 2.0 m s⁻¹ in the +y direction. Determine the velocity of the third fragment, C.
\(m_C = 0.200 - 0.050 - 0.080 = 0.070\ \text{kg}\); \(p_A = 0.15\ \text{kg m s}^{-1}\) (+x); \(p_B = 0.16\ \text{kg m s}^{-1}\) (+y).
Total momentum is zero in each direction: \(p_{C,x} = -0.15\), \(p_{C,y} = -0.16\), so \(p_C = \sqrt{0.15^2 + 0.16^2} = 0.219\ \text{kg m s}^{-1}\).
\(v_C = \dfrac{0.219}{0.070} = 3.13\ \text{m s}^{-1}\), at \(\tan^{-1}(0.16/0.15) = 46.8^\circ\). 3.1 m s⁻¹ at 46.8° below the −x direction
C moves into the quadrant opposite to both A and B, as it must for the total to be zero.
What makes a collision elastic or inelastic?
| Type of collision | Momentum | Total kinetic energy | Example |
|---|---|---|---|
| Elastic | conserved | conserved (same before and after) | collisions between gas molecules; very nearly, two steel spheres |
| Inelastic | conserved | decreases: some is transferred to internal (thermal) energy, sound and deformation | most real collisions: cars, balls, curling stones |
| Bodies stick together (totally inelastic) | conserved | the largest decrease that momentum conservation allows | a dart sticking in a target |
| Explosion | conserved | increases: energy comes from a chemical or elastic store | a spring-loaded separation; a firework |
Momentum is conserved in every one of these, provided no resultant external force acts, because the internal forces are equal and opposite. Kinetic energy is what tells them apart. To test a collision, calculate the total kinetic energy before and after and compare. Kinetic energy is a scalar, so just add the values; no components are needed.
Applied to worked example 4: before, \(\tfrac12 \times 19.0 \times 1.50^2 = 21.4\ \text{J}\); after, 5.34 J + 14.5 J = 19.8 J. About 1.6 J (7.4%) has been transferred, so the collision is inelastic.
If a body hits an identical stationary body elastically, they move off at 90° to each other. Momentum gives \(\vec u = \vec v_A + \vec v_B\), and kinetic energy gives \(u^2 = v_A^2 + v_B^2\); together they are Pythagoras' theorem. In worked example 4 the angle between the paths is 55.0° + 29.9° = 84.9°, less than 90°, which agrees with the collision being inelastic.
Where does \(E_k = p^2/2m\) come from, and when is it useful?
Start from \(E_k = \tfrac12 mv^2\) and p = mv, so v = p/m. Substitute:
\[E_k = \tfrac12 m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}\]Here \(E_k\) is in J, p in kg m s⁻¹ and m in kg. It is in the formulae list, but you may be asked to derive it. It is valid for speeds much less than the speed of light. It is useful when you know the momentum rather than the velocity, especially after an explosion or decay from rest, where the fragments have momenta of equal size. Then \(E_k \propto 1/m\): the light fragment takes almost all the kinetic energy. Rearranged, \(p = \sqrt{2mE_k}\) gives the momentum of a particle of known kinetic energy.
A nucleus at rest emits an alpha particle (mass 6.64 × 10⁻²⁷ kg) with kinetic energy 5.30 MeV. The nucleus left behind has mass 3.42 × 10⁻²⁵ kg. 1 eV = 1.60 × 10⁻¹⁹ J. (a) Calculate the momentum of the alpha particle. (b) Calculate the kinetic energy of the recoiling nucleus.
(a) \(E_k = 5.30\times10^{6} \times 1.60\times10^{-19} = 8.48\times10^{-13}\ \text{J}\). \(p = \sqrt{2 \times 6.64\times10^{-27} \times 8.48\times10^{-13}} =\) \(1.06 \times 10^{-19}\ \text{kg m s}^{-1}\)
(b) Momentum is conserved from rest, so the nucleus has a momentum of the same size: \(E_k = \dfrac{(1.06\times10^{-19})^2}{2 \times 3.42\times10^{-25}} =\) \(1.65 \times 10^{-14}\ \text{J}\) (≈ 0.103 MeV)
Quick route: with equal momenta, \(E_k \propto 1/m\), so the recoil energy is 5.30 MeV × (6.64 × 10⁻²⁷ / 3.42 × 10⁻²⁵). The alpha particle carries 98.1% of the kinetic energy released.
Core Practical 10: how do you analyse collisions between spheres using video?
- Set-up: two small spheres (for example a steel ball bearing and a glass marble) on a smooth, level surface. Sphere B is at rest; sphere A is rolled at it slightly off-centre so that they collide at a glancing angle.
- Camera: fixed directly above, pointing straight down, so the plane of motion is perpendicular to the line of sight. A ruler or grid of known spacing in the plane of the spheres calibrates distances and avoids parallax errors.
- Recording: use a high frame rate and good lighting to reduce motion blur. The frame rate gives the time between frames.
- Analysis (ICT): video-analysis software tracks the centre of each sphere frame by frame and outputs x and y coordinates against time. Velocity components = change in coordinate ÷ time interval (or the gradient of the position–time graphs).
- Masses on a top-pan balance. Then compare the total momentum in x and in y before and after, and the total kinetic energy.
- Limitations: rolling spheres also store rotational kinetic energy, and rolling friction slows them, so take velocities from frames close to the collision.
In the video above, sphere A (mass 28.0 g) moves 45.0 mm along x between images before the collision. Afterwards, between images, A moves 21.0 mm in x and +14.0 mm in y, while sphere B (mass 20.0 g) moves 33.0 mm in x and −19.2 mm in y. Images are 0.0500 s apart. Each position is read to ± 0.5 mm, and each mass to ± 0.1 g. (a) Calculate the total momentum in the x and y directions before and after. (b) Deduce whether momentum is conserved within the uncertainties. (c) State whether the collision is elastic.
(a) Velocities (displacement ÷ 0.0500 s): \(u_A = 0.900\ \text{m s}^{-1}\); \(v_A = (0.420, 0.280)\); \(v_B = (0.660, -0.384)\) m s⁻¹.
Before: x = 0.0280 × 0.900 = 0.0252 kg m s⁻¹, y = 0. After: x = 0.01176 + 0.0132 = 0.0250 kg m s⁻¹; y = 0.00784 − 0.00768 = 0.00016 kg m s⁻¹
(b) Each displacement is the difference of two readings, so ± 1.0 mm. Before: 1.0/45.0 + 0.1/28.0 = 2.6%, so ± 0.00065. After: ± 0.00060 (A) and ± 0.00047 (B), so ± 0.0011 in total.
The x totals differ by 0.0002 kg m s⁻¹ (0.95%), much less than the combined uncertainty, and the y total after is close to zero compared with the individual y momenta (about 0.0078). Momentum is conserved within experimental uncertainty.
(c) Kinetic energy before 11.3 mJ, after 9.4 mJ, about 17% less: inelastic. Part of the apparent loss is rotational kinetic energy of the rolling spheres, which the video analysis does not count.
Exam tips for 4.1
In a two-dimensional problem, label your axes and write two separate conservation equations, "in x: …" and "in y: …". Mixing a component with a whole momentum in one equation is the quickest way to lose the thread.
For a rebound, Δp = m(v − u) with opposite signs for u and v. State the direction of the force or impulse in words at the end.
Check the time unit on the axis. If you count squares, work out the impulse of one square first (its height in N × its width in s) and say how many squares you counted.
"Show that the collision is inelastic" needs numbers: total kinetic energy before, total after, and a sentence comparing them. "Energy is lost as heat" without a calculation is not enough.
Use the chain: the change of momentum is fixed → the padding increases the time over which it happens → F = Δp/Δt, so the average force is smaller. Mention all three links.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Padding reduces the change of momentum. | The body still goes from the same speed to rest, so Δp is the same. | Padding increases the contact time, which reduces the average force. |
| Impulse and force are the same thing. | Impulse includes how long the force acts; its unit is N s, not N. | Impulse = average force × time = change of momentum. |
| In two dimensions you can add the speeds of the bodies. | Velocities in different directions partly cancel; speed is a scalar. | Add momentum components, then recombine with Pythagoras. |
| Kinetic energy is conserved separately in x and y. | Kinetic energy is a scalar and has no components. | Conserve momentum in each direction; compare total kinetic energy once. |
| In an inelastic collision some momentum is lost. | Internal forces are equal and opposite, so the total momentum cannot change. | Only kinetic energy is transferred to other stores; momentum is still conserved. |
| Two fragments with equal momenta have equal kinetic energies. | \(E_k = p^2/2m\) depends on mass as well. | For equal momenta, the lighter fragment has more kinetic energy, in inverse proportion to mass. |
How this topic connects
Momentum, vectors and energy (Unit 1)
The definition p = mv, signs in one dimension and the conservation principle come from the Unit 1 momentum page. Resolving into components is the same skill used for projectiles and forces on slopes, and \(E_k = \tfrac12 mv^2\) is the starting point for \(E_k = p^2/2m\).
4.5 Magnetic fields and 4.7–4.8 Particle physics
The radius of a charged particle's path in a magnetic field depends on its momentum, and momentum conservation in two dimensions is used to read particle tracks and interactions.
5.2 The ideal gas and 5.4 Radioactive decay
Gas pressure comes from the change of momentum of molecules hitting a wall (F = Δp/Δt again), and alpha-decay energies split between the fragments exactly as in worked example 6.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(F\Delta t = \Delta p\) | impulse; average force during an impact | Yes |
| \(F = \Delta p/\Delta t\) | Newton's second law in momentum form, including changing mass | No — rearrange FΔt = Δp |
| impulse = area under F–t graph | varying forces | No — learn it |
| p = mv | momentum of each body | Yes |
| \(p_x = p\cos\theta,\ p_y = p\sin\theta\) | resolving momentum in two dimensions | No — learn it |
| \(\Sigma p_x\) before = \(\Sigma p_x\) after; \(\Sigma p_y\) before = \(\Sigma p_y\) after | collisions and explosions in a plane | No — learn the principle |
| \(E_k = \tfrac12 mv^2\) | testing whether a collision is elastic | Yes |
| \(E_k = p^2/2m\) | kinetic energy from momentum; sharing of energy in a decay | Yes (but learn the derivation) |
Check your understanding
1. Show that the newton second and the kilogram metre per second are the same unit.
1 N = 1 kg m s⁻², so 1 N s = 1 kg m s⁻² × 1 s = 1 kg m s⁻¹.
2. A 0.050 kg ball hits a wall at 10 m s⁻¹ and rebounds along the same line at 8.0 m s⁻¹. Contact lasts 0.020 s. Find the average force on the ball.
\(\Delta p = 0.050 \times (8.0 - (-10)) = 0.90\ \text{N s}\), so \(F = 0.90 \div 0.020 = 45\ \text{N}\), directed away from the wall.
3. On an air table, puck P (0.30 kg) moving east at 1.2 m s⁻¹ and puck Q (0.20 kg) moving north at 2.4 m s⁻¹ collide and stick. Find their velocity afterwards.
East: 0.36 kg m s⁻¹; north: 0.48 kg m s⁻¹. Total \(= \sqrt{0.36^2 + 0.48^2} = 0.60\ \text{kg m s}^{-1}\), so v = 0.60 ÷ 0.50 = 1.2 m s⁻¹ at \(\tan^{-1}(0.48/0.36) = 53^\circ\) north of east.
4. An electron and a proton have the same kinetic energy. Which has the larger momentum, and by what factor?
\(p = \sqrt{2mE_k}\), so for equal \(E_k\), \(p \propto \sqrt m\). The proton has the larger momentum, by \(\sqrt{1.67\times10^{-27} / 9.11\times10^{-31}} \approx 43\) times.
5. Explain why the foam lining of a cycle helmet reduces the risk of head injury in a fall.
The head's change of momentum as it stops is fixed by its mass and speed. The foam crushes, so the head takes longer to stop. Since force = rate of change of momentum, a longer time means a smaller average force on the head.
Specification checklist
- (81) I can use Newton's second law as force = rate of change of momentum, and impulse FΔt = Δp.
- (81) I can find an impulse, and so a change of momentum, from the area under a force–time graph.
- (82) I can describe how to use a force sensor, light gates and a data logger to compare impulse with change of momentum, and analyse the results with uncertainties.
- (83) I can apply conservation of momentum in two dimensions by resolving into perpendicular components, for glancing collisions and for explosions.
- (84) I can describe how video recording and analysis software are used to measure the velocities of colliding spheres and test conservation of momentum.
- (85) I can say whether a collision is elastic or inelastic by comparing total kinetic energy before and after.
- (86) I can derive \(E_k = p^2/2m\) and use it, for example to share kinetic energy between fragments of equal momentum.