Impulse — Force, Time and the Area Under an F–t Graph

Spec mapping: OCR H556 Module 3.3 — Work, energy and power; momentum (impulse as the product of force and time; impulse-momentum theorem $J = F\Delta t = \Delta p$ ; area under a force-time graph as impulse; applications including safety in vehicles). PAG 6 (Newtonian mechanics) anchors the experimental side. Refer to the official OCR H556 specification document for exact wording.

Impulse is the formal way of measuring how a force changes an object's momentum over time. It turns Newton's Second Law from a rate equation ( $\vec{F} = d\vec{p}/dt$ ) into a product equation ( $\vec{J} = \vec{F}\,\Delta t = \Delta\vec{p}$ ), gives us a powerful graphical interpretation (area under the force-time graph), and explains a huge range of real phenomena — from crumple zones in cars and airbags in cockpits to why cricketers "follow through" when catching a ball and why packaging is filled with foam.

For OCR A-Level Physics A, impulse is assessed in Module 3.3 (the momentum chapter) and tested in PAG 6 (the Newtonian mechanics practical, where force-time graphs from collision force-sensors give direct experimental impulse measurements). It appears in both numerical calculations and extended written answers on real-world safety applications. This lesson develops the impulse-momentum theorem from Newton 2, works through the graphical interpretation, and applies the framework to a range of A-Level scenarios.

Definition

Impulse is the product of the resultant force on a body and the time for which the force acts:

$\vec{J} = \vec{F}\,\Delta t \quad \text{(constant force)}$

$\vec{J}$ is the impulse — a vector quantity (same direction as $\vec{F}$ ).
$\vec{F}$ is the resultant force in newtons (N).
$\Delta t$ is the time interval over which the force acts, in seconds (s).
SI units: N s (equivalently kg m s $^{-1}$ — see units below).

From Newton 2 ( $\vec{F} = d\vec{p}/dt$ ), multiplying both sides by $dt$ and integrating over the contact time:

$\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \int_{p_1}^{p_2} d\vec{p} = \Delta\vec{p}$

So impulse equals change in momentum. A 5 N s impulse delivered to a body changes its momentum by 5 kg m s $^{-1}$ , regardless of the body's mass, initial velocity, or direction of motion. This is the impulse-momentum theorem, one of the cleanest applications of Newton 2 in a form mark schemes ask candidates to state and apply.

The Impulse-Momentum Theorem

In words:

The change in momentum of a body equals the impulse applied to it.

In symbols (most general form, variable force):

$\Delta\vec{p} = \int_{t_1}^{t_2} \vec{F}(t)\,dt$

For a constant force, the integral collapses to a product:

$\Delta\vec{p} = \vec{F}\,\Delta t$

The integral form is the key insight: if the force varies with time, the total impulse is the area under the force-time graph from $t_1$ to $t_2$ . A-Level candidates are not expected to evaluate integrals algebraically, but they are expected to interpret graphs geometrically — counting squares, computing triangle/rectangle/trapezium areas, or estimating areas under curves.

Diagram — Impulse as the Area Under F-t

flowchart TD
  A[Force varies with time F t] --> B[Plot F on y-axis vs t on x-axis]
  B --> C[Compute area under curve = integral of F dt]
  C --> D[Area = impulse J = change in p]
  D --> E[For constant F: area = rectangle F times dt]
  D --> F[For triangular pulse: area = half times base times peak]
  D --> G[For arbitrary F t: count squares or estimate]

Worked Example 1 — Constant Force on a Ball

A $3.0$ kg ball at rest experiences a constant force of $12$ N for $0.50$ s. Find the impulse, change in momentum, and final velocity.

$J = F\,\Delta t = 12 \times 0.50 = 6.0$ N s.
$\Delta p = J = 6.0$ kg m s $^{-1}$ .
Starting from rest ( $p_0 = 0$ ): final momentum $p = 6.0$ kg m s $^{-1}$ .
Final velocity: $v = p/m = 6.0/3.0 = 2.0$ m s $^{-1}$ .

Cross-check via Newton 2 + SUVAT: $a = F/m = 12/3.0 = 4.0$ m s $^{-2}$ , $v = u + at = 0 + 4.0 \times 0.50 = 2.0$ m s $^{-1}$ . ✓ Both routes agree, as they must.

The impulse route is shorter (one multiplication) and avoids needing $a$ as an intermediate step. For constant-mass problems either route works; for variable-force or variable-mass problems, the impulse route is the natural choice.

Worked Example 2 — Variable Force From a Triangular Pulse

A $0.5$ kg ball experiences a force that varies with time as a triangular pulse:

$t = 0$ to $t = 0.1$ s: force rises linearly from $0$ to $50$ N.
$t = 0.1$ s to $t = 0.2$ s: force falls linearly from $50$ N back to $0$ .

Find the total impulse and the ball's change in velocity.

The F-t graph is a triangle with base $\Delta t = 0.20$ s and peak height $F_{\text{peak}} = 50$ N. The area:

$J = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 0.20 \times 50 = 5.0 \text{ N s}$

So $\Delta p = 5.0$ kg m s $^{-1}$ and $\Delta v = \Delta p/m = 5.0/0.5 = 10$ m s $^{-1}$ .

Insight on average force: the average force over the pulse is $\bar{F} = J/\Delta t = 5.0/0.20 = 25$ N — exactly half the peak for a symmetric triangular pulse. Geometrically, the average force is the height of a rectangle of the same width as the actual pulse and the same area. For more complex pulses (sinusoidal, exponential decay), the average force can be calculated by dividing total impulse by total duration: $\bar{F} = J/\Delta t$ .

This is the operational principle of crash-test dummies and impact sensors: a force-time trace from an accelerometer gives the integral (impulse), and dividing by the contact time gives the average force the passenger experiences. Peak force is what causes injury at any moment; average force is what determines the total momentum change.

Worked Example 3 — Cricket Ball: Hard Catch vs Soft Catch

A $0.16$ kg cricket ball travels at $30$ m s $^{-1}$ before being caught. Compare the average force on the catcher's hand if the ball is stopped in (a) $0.010$ s (a rigid catch — hands rigid against chest) and (b) $0.10$ s (a "soft" catch with the hands recoiling backward to extend the contact time).

In both cases, $\Delta p = m\,\Delta v = 0.16 \times 30 = 4.8$ kg m s $^{-1}$ (magnitude — the ball must be brought to rest).

(a) $\bar{F} = \Delta p/\Delta t = 4.8/0.010 = 480$ N.
(b) $\bar{F} = \Delta p/\Delta t = 4.8/0.10 = 48$ N.

By extending the catch over a ten-times-longer time, the average force drops by a factor of 10. The total impulse is unchanged (the momentum has to be removed regardless of how slowly you do it), but the peak force is dramatically reduced. This is the universal principle of every sports catching technique, every crumple zone, every airbag, and every shock-absorbing material:

Same $\Delta p$ + larger $\Delta t$ ⇒ smaller $\bar{F}$ .

Conversely, anything that shortens the impact time raises the force — which is why bare-knuckle boxing without gloves causes more facial fractures than gloved boxing (the gloves extend the contact time, reducing peak force at the same overall impulse delivered).

Real-World Applications

Seat Belts and Crumple Zones

In a crash, a passenger's momentum must be reduced from $mv$ to zero. The impulse $\Delta p$ is determined entirely by the mass and initial velocity — it cannot be reduced. What can be controlled is the time $\Delta t$ over which the impulse is delivered. A seat belt extends $\Delta t$ from ~ $10$ ms (chest hitting the windscreen) to ~ $100$ ms (gradual deceleration against the belt and crumpling car structure). Same $\Delta p$ , ten times the $\Delta t$ , ten times smaller average force on the passenger:

$\bar{F} = \frac{\Delta p}{\Delta t}$

For a $70$ kg passenger at $20$ m s $^{-1}$ ( $45$ mph), $\Delta p = 1400$ kg m s $^{-1}$ . Without restraints, $\Delta t \approx 10$ ms gives $\bar{F} = 140\,000$ N — roughly $200g$ of deceleration, fatal. With seatbelt + airbag, $\Delta t \approx 100$ ms gives $\bar{F} = 14\,000$ N — about $20g$ , painful but survivable.

Airbags

Airbags work on the same impulse-time-stretching principle. Instead of the chest hitting the hard steering wheel in $\sim 5$ ms (peak force ~ $300\,000$ N for a $70$ kg torso at $20$ m s $^{-1}$ ), the airbag extends the interaction to $\sim 100$ ms, delivering the same impulse but at $\sim 20$ times smaller force. The bag deflates as it absorbs energy, extending the interaction further.

Packaging

Egg cartons, polystyrene, bubble wrap, foam-in-place packaging — all of these are impulse-time stretchers. Drop an egg onto a hard floor: $\Delta t \approx 1$ ms, peak force enormous, shell shatters. Drop the same egg onto $5$ cm of foam: $\Delta t \approx 50$ ms, peak force $50\times$ smaller, shell intact. The mass and impact velocity are the same; the bubble wrap stretches the impulse over much more time.

Follow-Through in Sport

A golfer who "follows through" after hitting the ball keeps the clubhead in contact with the ball for slightly longer. The peak force from the muscles is limited by the swing's strength, but extending $\Delta t$ increases the total impulse $J = \bar{F}\,\Delta t$ , giving the ball more $\Delta p$ and a longer drive. The same logic explains why cricket bowlers and tennis players follow through their swings.

PAG 6 — Newtonian Mechanics Practical

The OCR PAG 6 practical anchors the impulse-momentum theorem experimentally. A typical setup: a force sensor attached to a wall, with a trolley (mass and velocity measured by light gates) colliding into it. The force-time trace from the sensor gives $J = \int F\,dt$ (area under the curve, computed by data-logger software or by hand counting squares); the trolley's measured $m\,\Delta v$ gives the predicted $\Delta p$ . The two should match — and they do, within ~ $5\%$ , validating the theorem.

Worked Example 4 — Step-Function Force on a Trolley

A force on a $2.0$ kg trolley varies with time as follows:

$0 \leq t \leq 2$ s: $F = +5$ N
$2 \leq t \leq 4$ s: $F = -3$ N
$4 \leq t \leq 5$ s: $F = 0$ N

Find the total impulse from $t = 0$ to $t = 5$ s and the final velocity if the trolley starts from rest.

Area under the (signed) F-t graph:

Rectangle 1 ( $0$ to $2$ s, $F = +5$ N): $A_1 = 5 \times 2 = +10$ N s.
Rectangle 2 ( $2$ to $4$ s, $F = -3$ N): $A_2 = -3 \times 2 = -6$ N s.
Interval 3 ( $4$ to $5$ s, $F = 0$ ): $A_3 = 0$ .

Total impulse: $J = A_1 + A_2 + A_3 = 10 - 6 + 0 = +4$ N s.

$\Delta p = +4$ kg m s $^{-1}$ , so final velocity $v = \Delta p/m = 4/2.0 = +2.0$ m s $^{-1}$ in the positive direction.

Intermediate state check: after the first 2 s the trolley has $p = +10$ kg m s $^{-1}$ , $v = +5.0$ m s $^{-1}$ . The next 2 s (under $-3$ N) decelerates it: $\Delta p = -6$ kg m s $^{-1}$ , so $p_{4\text{s}} = +4$ kg m s $^{-1}$ , $v_{4\text{s}} = +2.0$ m s $^{-1}$ . The final 1 s with $F = 0$ leaves $v$ unchanged at $+2.0$ m s $^{-1}$ . ✓ Consistent with the total-impulse calculation.

The trolley's $v$ - $t$ profile would be a piecewise-linear curve: rising linearly at gradient $+2.5$ m s $^{-2}$ for 2 s (reaching $+5.0$ m s $^{-1}$ ), then falling linearly at gradient $-1.5$ m s $^{-2}$ for 2 s (reaching $+2.0$ m s $^{-1}$ ), then flat for 1 s.

Rebounds and Sign Conventions

When a ball rebounds from a wall, both initial and final velocities matter — and they have different signs in the chosen convention. This is the most common source of impulse-calculation errors at A-Level.

Worked Example 5 — Ball Rebound from a Wall

A $0.25$ kg ball strikes a wall at $8.0$ m s $^{-1}$ and rebounds at $6.0$ m s $^{-1}$ . The contact time is $0.020$ s. Find the average force on the ball, and by Newton 3 the force the ball exerts on the wall.

Take toward the wall as positive.

$p_{\text{initial}} = 0.25 \times (+8.0) = +2.0$ kg m s $^{-1}$ .
$p_{\text{final}} = 0.25 \times (-6.0) = -1.5$ kg m s $^{-1}$ (negative because rebound is in the opposite direction).
$\Delta p = p_{\text{final}} - p_{\text{initial}} = -1.5 - 2.0 = -3.5$ kg m s $^{-1}$ .
$\bar{F}_{\text{on ball}} = \Delta p/\Delta t = -3.5/0.020 = -175$ N.

Magnitude $175$ N, directed away from the wall (the wall pushes the ball back toward where it came from). By Newton 3, the ball pushes the wall with $+175$ N (toward and into the wall) over the same $0.020$ s.

Crucially, $|\Delta p| = 3.5$ kg m s $^{-1}$ is larger than either initial or final $|p|$ alone — because the direction reversed. A "stop-dead" scenario (no rebound) would give $|\Delta p| = 2.0$ kg m s $^{-1}$ , and so a force of only $100$ N. The rebound adds another $1.5$ kg m s $^{-1}$ of momentum change, raising the force by $75\%$ . Rebound impacts deliver more impulse than stop-dead impacts at the same initial speed.

Common mid-band error: computing $\Delta p = 1.5 - 2.0 = -0.5$ kg m s $^{-1}$ by forgetting to sign-flip the rebound velocity. Always make the sign convention explicit at the start, and substitute signed values for every velocity.

Graphical Interpretation — Curved F(t)

If the force varies continuously with time — a sinusoidal half-period, an exponential decay, a sharp asymmetric collision pulse — the impulse is still defined as "the area under the F-t graph". You can estimate this by:

Counting squares on a printed grid (each square has area $F_{\text{grid}} \times t_{\text{grid}}$ ).
Approximating as a triangle or trapezium if the actual shape is close.
Splitting into multiple rectangles or triangles for piecewise-defined pulses.
Symbolic integration if you know calculus (post-A-Level).

OCR mark schemes frequently ask candidates to estimate the area under a non-rectangular F-t curve and use it to find $\Delta p$ . State your method ("counted squares: 18 squares × 0.1 N × 0.05 s/square = 0.090 N s") and quote units N s (or equivalently kg m s $^{-1}$ ).

Synoptic Links

Newton's second law (Module 3.2, lesson 2) — impulse is just Newton 2 integrated over time: $\int F\,dt = \Delta p$ . Same physics, different formulation (product vs rate).
Linear momentum (Module 3.3, lesson 4) — impulse is the integral version of the definition $p = mv$ in differential form; both reduce to "momentum change" in the integrated picture.
Conservation of momentum (Module 3.3, lesson 6) — when external impulse on a system is zero (no external forces), total momentum is conserved; this is the workhorse for collision and explosion problems.
Work and energy (Module 3.3) — work is $\int F\,ds$ (area under $F$ - $s$ graph); impulse is $\int F\,dt$ (area under $F$ - $t$ graph). The two are different quantities with different units and different physical meanings. Confusing them is a high-cost A-Level error.
Circular motion (Module 5.4) — at constant speed in a circle the centripetal force delivers continuous impulse, changing the direction of momentum without changing its magnitude.
PAG 6 — Newtonian mechanics practical — force-sensor + trolley collisions give direct experimental verification of $J = \Delta p$ , with the F-t trace and the $m\,\Delta v$ value computed independently and compared.

Impulse — Force, Time and the Area Under an F–t Graph

Impulse — Force, Time and the Area Under an F–t Graph

Definition

The Impulse-Momentum Theorem

Diagram — Impulse as the Area Under F-t

Worked Example 1 — Constant Force on a Ball

Worked Example 2 — Variable Force From a Triangular Pulse

Worked Example 3 — Cricket Ball: Hard Catch vs Soft Catch

Real-World Applications

Seat Belts and Crumple Zones

Airbags

Packaging

Follow-Through in Sport

PAG 6 — Newtonian Mechanics Practical

Worked Example 4 — Step-Function Force on a Trolley

Rebounds and Sign Conventions

Worked Example 5 — Ball Rebound from a Wall

Graphical Interpretation — Curved F(t)

Synoptic Links

Specimen question modelled on the OCR H556 paper format

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