Impulse & Momentum

This lesson covers impulse and momentum at A-Level. Momentum is a fundamental concept in mechanics that is especially useful for analysing collisions and impacts. The principle of conservation of momentum is one of the most important laws in physics.

Momentum

The momentum of an object is the product of its mass and velocity:

$p = mv$

Property	Detail
Unit	kg m/s (or Ns)
Type	Vector (has direction)
Direction	Same as the velocity

Exam Tip: Momentum is a vector quantity. Always define a positive direction and use signs consistently. A common error is to forget that velocity (and hence momentum) can be negative.

Impulse

Impulse is the change in momentum of an object:

$I = \Delta p = mv - mu$

Impulse is also equal to force multiplied by time:

$I = Ft \quad \text{(for a constant force)}$

More generally (for a variable force):

$I = \int F \, dt$

The unit of impulse is Ns (newton-seconds) or equivalently kg m/s.

Impulse-Momentum Theorem

$Ft = mv - mu$

This is Newton's Second Law in its impulse form.

Example: A 0.5 kg ball moving at 8 m/s is hit by a bat and returns at 12 m/s. Find the impulse.

Taking the initial direction as positive:

$I = mv - mu = 0.5 \times (-12) - 0.5 \times 8 = -6 - 4 = -10 \text{ Ns}$

The magnitude of the impulse is 10 Ns (in the direction opposite to the initial motion).

Conservation of Momentum

The principle of conservation of momentum states:

In any collision or interaction between objects, the total momentum before the event equals the total momentum after the event, provided no external forces act on the system.

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

This principle applies to all types of collisions and explosions.

Types of Collisions

Type	Momentum conserved?	Kinetic energy conserved?
Elastic	Yes	Yes
Inelastic	Yes	No (some KE is lost)
Perfectly inelastic	Yes	No (maximum KE lost — objects stick together)

Elastic Collision

In an elastic collision between two particles:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{(momentum)}$ $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \text{(kinetic energy)}$

Perfectly Inelastic Collision

Objects coalesce (stick together) after the collision:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$

$v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Newton's Experimental Law (Coefficient of Restitution)

The coefficient of restitution $e$ relates the relative speeds before and after a collision:

$e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$

Value of $e$	Type of collision
$e = 1$	Perfectly elastic
$0 < e < 1$	Inelastic
$e = 0$	Perfectly inelastic (objects coalesce)

Explosions

In an explosion (e.g., two objects pushed apart by a spring), the total momentum before and after is the same. If the system starts at rest:

$0 = m_1 v_1 + m_2 v_2$

$m_1 v_1 = -m_2 v_2$

The objects move in opposite directions with momenta equal in magnitude.

Exam Tip: In explosion problems, the total momentum is conserved (usually zero if the system starts at rest). The objects move in opposite directions. Use kinetic energy to find the energy released by the explosion.

Impulse on a Force-Time Graph

The impulse is the area under a force-time graph.

For a constant force: impulse = $F \times t$ (rectangle). For a linearly increasing force: impulse = $\frac{1}{2}F_{\max} \times t$ (triangle). For a variable force: integrate or find the area under the curve.

Summary

Momentum: $p = mv$ (vector, unit: kg m/s or Ns).
Impulse: $I = Ft = mv - mu$ (change in momentum).
Conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ (no external forces).
Elastic collisions conserve both momentum and kinetic energy.
Inelastic collisions conserve momentum but not kinetic energy.
Perfectly inelastic: objects coalesce; maximum KE is lost.
Coefficient of restitution: $e = \frac{\text{separation speed}}{\text{approach speed}}$ .
Impulse = area under a force-time graph.

Exam Tip: When solving collision problems, always start by defining a positive direction and writing the conservation of momentum equation. If the collision is elastic, you have a second equation (conservation of KE). If $e$ is given, use Newton's experimental law as the second equation.

A-Level Deep Dive: Impulse and Momentum

Spec mapping

Beyond Mathematics 7357 — Further Maths / Physics overlap. Impulse and momentum do not appear in the AQA A-Level Mathematics (7357) core specification. Their natural home is AQA Further Mathematics (7367), section M (Mechanics options), which lists "momentum and impulse, conservation of linear momentum, direct impact of elastic spheres, Newton's experimental law, coefficient of restitution $e$ ", and AQA A-Level Physics (7408), section 3.4 (Mechanics and materials), which examines momentum $p = mv$ , impulse $J = F\Delta t = \Delta p$ , and the principle of conservation of linear momentum in 1D and 2D collisions. Students taking 7357 alone will not be examined on collisions, but the mathematical machinery — vectors, integration, Newton's laws — is shared with 7357 sections G (Vectors), N (Numerical methods) and O (Mechanics: kinematics and forces). Treat this Deep Dive as enrichment for prospective Further Maths, Physics, or Engineering applicants.

Worked example with full mark scheme

Question (8 marks, Further-Maths style):

A particle $A$ of mass $3\,\text{kg}$ moves on a smooth horizontal table with velocity $4\,\text{m s}^{-1}$ in the positive $x$ -direction. It collides directly with a stationary particle $B$ of mass $2\,\text{kg}$ . After the collision, $A$ continues to move in the positive $x$ -direction with velocity $1.2\,\text{m s}^{-1}$ .

(a) Find the velocity of $B$ after the collision. (3)

(b) Find the magnitude of the impulse exerted on $B$ by $A$ during the collision. (2)

(d) State, with reasoning, whether the collision is elastic. (1)

Solution with mark scheme:

(a) Step 1 — apply conservation of linear momentum.

Take the positive $x$ -direction as positive. Before: total momentum $= 3 \cdot 4 + 2 \cdot 0 = 12\,\text{kg m s}^{-1}$ . After: $3 \cdot 1.2 + 2 v_B = 3.6 + 2 v_B$ .

M1 — applying conservation of momentum with correct signs and masses on both sides.

Step 2 — equate and solve.

$12 = 3.6 + 2 v_B \implies 2 v_B = 8.4 \implies v_B = 4.2\,\text{m s}^{-1}$

A1 — correct value. A1 — direction stated (positive $x$ -direction, i.e. same direction as $A$ 's original motion).

(b) Step 1 — impulse on $B$ equals change in momentum of $B$ .

$J_B = m_B v_B - m_B u_B = 2 \cdot 4.2 - 2 \cdot 0 = 8.4\,\text{N s}$

M1 — using $J = \Delta p$ on particle $B$ specifically (not on $A$ , not the system).

A1 — $|J_B| = 8.4\,\text{N s}$ , with units. (Equivalently: impulse on $A$ from $B$ is $3 \cdot 1.2 - 3 \cdot 4 = -8.4\,\text{N s}$ , equal in magnitude and opposite in sign — Newton's third law.)

$F_{\text{avg}} = \dfrac{J}{\Delta t} = \dfrac{8.4}{0.05} = 168\,\text{N}$

M1 — correct rearrangement $F = J / \Delta t$ .

A1 — $168\,\text{N}$ with units, in the direction of $B$ 's motion.

(d) Step 1 — compare KE before and after.

KE before $= \tfrac{1}{2} \cdot 3 \cdot 4^2 + 0 = 24\,\text{J}$ . KE after $= \tfrac{1}{2} \cdot 3 \cdot 1.2^2 + \tfrac{1}{2} \cdot 2 \cdot 4.2^2 = 2.16 + 17.64 = 19.8\,\text{J}$ .

KE is not conserved ( $24 \neq 19.8$ ), so the collision is inelastic.

B1 — correct conclusion supported by KE comparison.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on AQA Further Maths 7367 mechanics format

Question (6 marks): Two smooth spheres $P$ and $Q$ of equal mass $m$ move along the same straight line. $P$ has velocity $5u$ and $Q$ has velocity $-u$ (i.e. moving toward $P$ ). They collide directly. The coefficient of restitution between the spheres is $e = \tfrac{1}{2}$ .

(a) Find the velocities of $P$ and $Q$ after the collision. (5)

(b) State the impulse exerted on $Q$ by $P$ . (1)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — conservation of momentum: $m(5u) + m(-u) = m v_P + m v_Q \implies 4u = v_P + v_Q$ .
M1 (AO1.1a) — Newton's experimental law: $v_Q - v_P = -e(u_Q - u_P) = -\tfrac{1}{2}(-u - 5u) = 3u$ .
A1 (AO1.1b) — solving simultaneously: adding the equations gives $2 v_Q = 7u \implies v_Q = \tfrac{7u}{2}$ .
A1 (AO1.1b) — $v_P = 4u - v_Q = 4u - \tfrac{7u}{2} = \tfrac{u}{2}$ .
A1 (AO2.5) — direction commentary: both $v_P, v_Q > 0$ , so both spheres now move in $P$ 's original direction; $Q$ has reversed.

(b)

B1 (AO1.1b) — impulse on $Q$ $= m(v_Q - u_Q) = m(\tfrac{7u}{2} - (-u)) = \tfrac{9mu}{2}$ in $P$ 's original direction.

Total: 6 marks split AO1 = 5, AO2 = 1.

Synoptic links

Connects to:

A-Level Physics 7408, section 3.4 — Newton's laws of motion: Newton's second law in its general form is $F = \dfrac{dp}{dt}$ , not $F = ma$ . The latter is a special case for constant mass. Variable-mass problems (rockets, raindrops accreting moisture) require the momentum form. Impulse $J = \int F\,dt$ is the time-integral of this differential law.
7357 section G — Vectors: momentum is a vector ( $\mathbf{p} = m\mathbf{v}$ ). 2D collision problems require vector addition: $\sum \mathbf{p}_{\text{before}} = \sum \mathbf{p}_{\text{after}}$ in component form. Resolving along and perpendicular to the line of centres of two colliding spheres is identical in technique to resolving forces on an inclined plane.
7357 section M — Integration: for a time-varying force $F(t)$ , the impulse over $[t_1, t_2]$ is $J = \int_{t_1}^{t_2} F(t)\,dt$ . This is the area under the force–time graph. The cubic-impulse problem (force $F(t) = kt(T - t)$ over a contact duration $T$ ) is a clean integration exercise dressed in mechanics language.
Further Maths 7367, section M — Newton's experimental law: the coefficient of restitution $e$ relates relative velocities: $e = -\dfrac{v_2 - v_1}{u_2 - u_1}$ (after / before, along the line of impact). $e = 1$ is perfectly elastic (KE conserved), $e = 0$ is perfectly inelastic (objects stick together). The 7357-only candidate will not see this, but it is the central tool for any 7367 collision question.
A-Level Physics 7408, section 3.4 — energy considerations: in any collision, momentum is conserved (provided no external impulse), but kinetic energy is generally not. The energy "lost" appears as heat, sound, deformation. Only in perfectly elastic collisions (between hard, frictionless bodies, idealised) is KE also conserved. Newton's cradle is the canonical demonstration.

Mark-scheme literacy

Impulse / momentum questions on 7367 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Stating conservation of momentum, applying $J = \Delta p$ , computing post-collision velocities
AO2 (reasoning / interpretation)	20–30%	Justifying choice of positive direction, reasoning about elastic vs inelastic from KE comparison, identifying when external impulse invalidates conservation
AO3 (problem-solving)	10–20%	Multi-stage collision problems, oblique impact with line-of-centres resolution, variable-force impulse via integration

Examiner-rewarded phrasing: "taking the positive direction as ..."; "by conservation of linear momentum (no external impulse acts)"; "since KE before $\neq$ KE after, the collision is inelastic"; "by Newton's third law, the impulse on $A$ from $B$ is equal and opposite". Phrases that lose marks: omitting the units ( $\text{N s}$ for impulse, $\text{kg m s}^{-1}$ for momentum); failing to state direction on a vector quantity; using $F = ma$ where $F = dp/dt$ is needed (variable-mass problems).

A specific 7367 pattern: questions phrased "find the impulse exerted on $X$ by $Y$ " demand the impulse on the named body — not the system, not the other body. Computing $\Delta p$ for the wrong particle is the single most common mark-loser.

Grade-band model answers

3-mark question

Question: A ball of mass $0.4\,\text{kg}$ travelling at $8\,\text{m s}^{-1}$ strikes a wall and rebounds at $5\,\text{m s}^{-1}$ in the opposite direction. Find the magnitude of the impulse exerted on the ball by the wall.

Grade C response (~140 words):

Take the initial direction as positive. Before: $p_1 = 0.4 \cdot 8 = 3.2\,\text{kg m s}^{-1}$ . After: $p_2 = 0.4 \cdot (-5) = -2\,\text{kg m s}^{-1}$ .

Impulse $= p_2 - p_1 = -2 - 3.2 = -5.2\,\text{N s}$ .

So the magnitude is $5.2\,\text{N s}$ .

Impulse & Momentum

Impulse & Momentum

Momentum

Impulse

Impulse-Momentum Theorem

Conservation of Momentum

Types of Collisions

Elastic Collision

Perfectly Inelastic Collision

Newton's Experimental Law (Coefficient of Restitution)

Explosions

Impulse on a Force-Time Graph

Summary

A-Level Deep Dive: Impulse and Momentum

Spec mapping

Worked example with full mark scheme

Specimen question modelled on AQA Further Maths 7367 mechanics format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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