Impulse: F∆t = ∆p and Newton's second law

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This unit is A Level (IA2) content, and it assumes the AS work on forces and momentum from Unit 1: $p = mv$ , Newton's laws and conservation of momentum in one dimension.

It is examined in the Unit 4 paper (WPH14), where every recent series has set a two-dimensional collision and the impulse idea has appeared both as a multiple-choice item and as a written explanation.

Its two core practicals, CP9 and CP10, can also be examined in the Unit 6 practical paper (WPH16).

How the marks work

Each marking point is worth one mark. We write it as "(1)"; the physics mark schemes use no M1, A1 or B1 codes.

A calculation earns a "use of" mark for the equation with every value substituted, then a mark for the answer with its unit. A missing or wrong unit loses that last mark (not on a show that).

Error carried forward (ecf): a wrong value from an earlier part is followed through, so always carry on.

Show that: give your final value to at least one more significant figure than the printed value.

Deduce: the last mark is for comparing two values and stating a conclusion in words.

The one asterisked 6-mark question counts indicative points (up to 4 marks) plus up to 2 marks for a linked, logical line of reasoning.

Learn it: not on the list

The newton second and the momentum unit are the same: $1\ \text{N s} = 1\ \text{kg m s}^{-1}$ .

The area under a force-time graph is the impulse (the change of momentum).

Components of a momentum $p$ at angle $\theta$ to a direction: $p\cos\theta$ along it and $p\sin\theta$ at right angles to it; the resultant of two perpendicular components from Pythagoras, its direction from $\tan\theta$ .

Elastic collision: the total kinetic energy is conserved.

$1\ \text{MeV} = 1.60 \times 10^{-13}$ J (from the electronvolt on the list).

On the list (Unit 4 block): impulse $F\Delta t = \Delta p$ ; kinetic energy of a non-relativistic particle $E_k = p^2/2m$ .

From the Unit 1 block: $p = mv$ , $\Sigma F = ma$ , $E_k = \tfrac{1}{2}mv^2$ and the equations of uniformly accelerated motion.

Synoptic hooks

$p = mv$ and the principle of conservation of linear momentum in one dimension (Unit 1): the starting point of every collision question here.

Resolving a vector into two perpendicular components, and adding two perpendicular vectors (Unit 1): every two-dimensional momentum problem.

$\Sigma F = ma$ and Newton's third law (Unit 1): impulse is Newton's second law in another form, and conservation of momentum comes from the third law.

$E_k = \tfrac{1}{2}mv^2$ (Unit 1, kinetic energy): the elastic or inelastic test.

The equations of uniformly accelerated motion (Unit 1): the speed before an impact, and stopping distances.

The de Broglie wavelength $\lambda = h/p$ (Unit 2): uses the momentum you find from $E_k = p^2/2m$ .

The idea: a force acting for a time changes momentum

Push a stationary supermarket trolley and it starts to move.

Push harder, or push for longer, and it ends up moving faster.

Both the size of the force and the time it acts for matter.

The quantity that combines them, force multiplied by the time it acts, is called the impulse, and what it changes is the body's momentum.

You met momentum in Unit 1: $p = mv$ , a vector in the direction of the velocity.

In Unit 1 you also wrote Newton's second law as $\Sigma F = ma$ .

Newton himself stated the law in terms of momentum, and that form is more general.

The resultant force on a body equals its rate of change of momentum:

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

Multiply both sides by $\Delta t$ and you have the equation you need to use:

\text{impulse} = F\Delta t = \Delta p

Here $F$ is the resultant force (or its average value, if it varies), $\Delta t$ is the time for which it acts and $\Delta p$ is the change in momentum it produces.

The equation is on the list, under Unit 4.

Why it is Newton's second law

For a body of constant mass, $\Delta p = m\Delta v$ , so

F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma

which is the $\Sigma F = ma$ you already know.

So the momentum form contains the Unit 1 form.

It is more general because it also works when the mass changes or when mass flows: a rocket throwing out exhaust gas, a jet of water hitting a wall, rain falling into a moving truck.

In those cases you count the momentum carried by the moving mass each second, and that rate is the force.

Units, and impulse as a vector

Impulse has the unit N s. Momentum has the unit kg m s⁻¹. They are the same unit, because $1\ \text{N} = 1\ \text{kg m s}^{-2}$ , so $1\ \text{N s} = 1\ \text{kg m s}^{-1}$ .

Impulse is a vector, in the direction of the force. The change in momentum is in that direction too, and it need not be in the direction the body was moving.

That second point is where most errors start.

A ball that hits a wall and bounces back has its velocity reversed, so the change in momentum is larger than either momentum on its own.

A tennis ball moves towards a wall at 22 m s⁻¹ and rebounds along the same line at 17 m s⁻¹; the direction towards the wall is taken as positive

Take the direction towards the wall as positive.

Before the bounce the tennis ball (mass 0.058 kg) has momentum $0.058 \times 22 = +1.276$ kg m s⁻¹.

Afterwards it has $0.058 \times (-17) = -0.986$ kg m s⁻¹.

The change is final minus initial:

\Delta p = (-0.986) - (+1.276) = -2.262\ \text{kg m s}^{-1}

The minus sign says the change is directed away from the wall, which is the direction of the force the wall exerts.

Its size, 2.262 kg m s⁻¹, is $m(u + v)$ for a rebound along the same line: the speeds add.

Key terms

Impulse: impulse = $F\Delta t = \Delta p$ (Newton's second law of motion)

Newton's second law of motion (momentum form): the resultant force on a body is equal to its rate of change of momentum, $F = \Delta p/\Delta t$ .

Unit of impulse: the newton second, N s, equal to the unit of momentum, kg m s⁻¹.

Learn the impulse line exactly as written.

When asked for Newton's second law in words, the key words are rate of change of momentum; "force equals mass times acceleration" is only the constant-mass case.

What the answer means

An impulse tells you how much momentum a force delivers or removes.

The same change in momentum can come from a large force for a short time or a small force for a long time.

That trade-off is the physics of crumple zones, airbags, cushioned shoes and catching a ball with "soft hands", which the next section develops with force-time graphs.

When you calculate a force from $F = \Delta p/\Delta t$ , remember that you get the average resultant force over the time; the peak force during a real impact is larger.

Worked example 1: a direct substitution

A hockey ball of mass 0.16 kg is at rest.

A stick hits it and it leaves at 30 m s⁻¹.

The stick is in contact with the ball for 2.5 ms. Calculate the average force on the ball.

(2)

1. $F\Delta t = \Delta p$ , so $F = \dfrac{0.16 \times 30}{2.5 \times 10^{-3}}$ (1) use of $F\Delta t = \Delta p$ with $p = mv$

2. $F = 1920$ N

Answer: $1.9 \times 10^{3}$ N (1) answer with unit.

Write the millisecond conversion in the substitution line, where the examiner can see it.

Worked example 2: a show that with a rebound

The tennis ball in the diagram above is in contact with the wall for 4.5 ms. Show that the average force exerted by the wall on the ball is about 500 N. (3)

1. Towards the wall positive: $\Delta p = 0.058 \times (-17) - 0.058 \times 22 = -2.262$ kg m s⁻¹ (1) change in momentum with the speeds added

2. $F = \dfrac{\Delta p}{\Delta t} = \dfrac{-2.262}{4.5 \times 10^{-3}}$ (1) use of $F\Delta t = \Delta p$

3. $F = -503$ N (1) value to 3 s.f., one more than the printed "500"

Answer: 503 N, directed away from the wall.

Using $0.058 \times (22 - 17)$ gives 64 N and loses two marks: the rebound reverses the velocity, so the speeds add.

Worked example 3: a flowing mass

A hose sprays water horizontally at a wall.

The water leaves the hose at 12 m s⁻¹ and 0.35 kg of water hits the wall each second.

The water runs down the wall and does not bounce back.

Calculate the force the water exerts on the wall.

(3)

1. In 1.0 s, a mass of 0.35 kg loses a horizontal velocity of 12 m s⁻¹: $\Delta p = 0.35 \times 12 = 4.2$ kg m s⁻¹ (1) use of $p = mv$ for the mass arriving in a time $\Delta t$

2. $F = \dfrac{\Delta p}{\Delta t} = \dfrac{4.2}{1.0}$ (1) use of $F\Delta t = \Delta p$

3. $F = 4.2$ N

Answer: 4.2 N (1) answer with unit.

By Newton's third law the wall pushes the water with 4.2 N in the opposite direction, which is what stops it.

Here $\Sigma F = ma$ does not apply directly, because no single body is accelerating: the momentum form handles a stream of mass.

Worked example 4: an exam-length problem with a synoptic step

A train of mass $3.8 \times 10^{5}$ kg is travelling at 24 m s⁻¹ along a level track.

Its brakes apply a constant resultant force of $1.9 \times 10^{5}$ N.

(a) Calculate the time taken for the train to stop. (3)

(b) Calculate the stopping distance. (2)

(a)

1. $\Delta p = mv = 3.8 \times 10^{5} \times 24 = 9.12 \times 10^{6}$ kg m s⁻¹ (1) use of $p = mv$

2. $\Delta t = \dfrac{\Delta p}{F} = \dfrac{9.12 \times 10^{6}}{1.9 \times 10^{5}}$ (1) use of $F\Delta t = \Delta p$

3. $\Delta t = 48$ s (1) answer with unit

(b)

1. Constant force, so uniform deceleration: $s = \dfrac{(u + v)t}{2} = \dfrac{(24 + 0) \times 48}{2}$ (1) use of a suvat equation (ecf from (a))

2. $s = 576$ m

Answer: (a) 48 s; (b) $5.8 \times 10^{2}$ m (1) answer with unit.

Part (b) is Unit 1 physics inside a Unit 4 question: this is how synoptic marks appear.

Exam tip

Before you substitute into $F\Delta t = \Delta p$ , choose a positive direction and write it down.

Then write $\Delta p$ as final momentum minus initial momentum with signs.

For a rebound the result is $m(u + v)$ , not $m(u - v)$ , and an examiner can award the method mark only if your substitution shows the signs.

Common mistakes

Subtracting speeds for a rebound. The velocity reverses, so the change in momentum is $m(u + v)$ . Students subtract because "change" sounds like "difference of the numbers"; with directions, final minus initial gives a sum.

Writing $\Delta p = F\Delta t$ with the force in the wrong place. $F$ is the resultant force on the body whose momentum changes. If a ball rests on the ground while it is struck, the force of the ground and the weight are part of the resultant.

Quoting $F = \Delta p/\Delta t$ as the peak force. It is the average resultant force over the contact time.

Mixing units. Contact times in ms must become s ( $10^{-3}$ ); masses in g must become kg.

Check your understanding

1. Which of the following is not a unit of momentum?

A kg m s⁻¹

B N s

C J s m⁻¹

D N s⁻¹

(1) Total [1]

2. A golf ball of mass 0.046 kg is at rest. A club strikes it and it leaves at 65 m s⁻¹. The club is in contact with the ball for 0.50 ms. Calculate the average force exerted on the ball. (2) Total [2]

3. A cricket ball of mass 0.16 kg falls vertically and hits the ground at 9.0 m s⁻¹. It rebounds vertically at 6.0 m s⁻¹. The ball is in contact with the ground for 0.020 s. Show that the average force exerted on the ball by the ground is about 120 N. (3) Total [3]

4. A student writes: "A ball that hits a wall at 8 m s⁻¹ and rebounds at 8 m s⁻¹ has the same momentum before and after, so its change in momentum is zero." Criticise this statement. (2) Total [2]

5. A space probe of mass 450 kg is driven by an ion engine. The engine ejects $3.0 \times 10^{-6}$ kg of xenon ions each second at a speed of $3.0 \times 10^{4}$ m s⁻¹ relative to the probe. (Invented values, of a realistic size.)

(a) Calculate the force the ions exert on the probe. (2)

(b) The engine runs for 30 days. Calculate the change in velocity of the probe. (3)

(c) The mass of the probe decreases slightly as it uses fuel. Explain whether your answer to (b) is too large or too small. (2)

Total [7]

Answers

1. D (1). N s⁻¹ is kg m s⁻³, the unit of rate of change of force. A is the definition unit; B is equal to it because 1 N = 1 kg m s⁻²; C is also equal, because 1 J = 1 N m, so J s m⁻¹ = N s.

2. $F = \Delta p/\Delta t = 0.046 \times 65/(0.50 \times 10^{-3})$ (1); $F = 6.0 \times 10^{3}$ N (1) (5980 N).

3. Upwards positive: $\Delta p = 0.16 \times 6.0 - 0.16 \times (-9.0) = 2.4$ kg m s⁻¹ (1); resultant force $= 2.4/0.020 = 120$ N (1); the resultant is the ground force minus the weight, so ground force $= 120 + 0.16 \times 9.81 = 122$ N (1) (to 3 s.f., one more than the printed value).

4. Momentum is a vector: the velocity reverses, so the momentum changes from $+8m$ to $-8m$ (1); the change is $16m$ (twice the original momentum), not zero, and a large force acts on the ball (1).

5. (a) In 1 s, $\Delta p = 3.0 \times 10^{-6} \times 3.0 \times 10^{4} = 0.090$ kg m s⁻¹ (1); $F = 0.090$ N (1). (b) $\Delta t = 30 \times 24 \times 3600 = 2.59 \times 10^{6}$ s (1); $\Delta p = F\Delta t = 0.090 \times 2.59 \times 10^{6} = 2.33 \times 10^{5}$ N s (1); $\Delta v = 2.33 \times 10^{5}/450 = 520$ m s⁻¹ (1) (518 m s⁻¹; ecf from (a)). (c) The force is constant but the mass decreases (by about 8 kg), so the acceleration increases (1); the real change in velocity is slightly larger, so the answer to (b) is slightly too small (1).

Impulse: F∆t = ∆p and Newton's second law

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