Chapter 1 · Mechanics

1-7 Momentum and Impulse

❓ Lesson question

Why is a phone less likely to break when it lands on a bed than when it lands on a tiled floor, even if it is dropped from the same height?

You will learn to

1Describe momentum as the product of mass and velocity, p = mv.
2Determine the impulse of a force and relate it to the change in momentum, J = FΔt = Δp.
3Explain why increasing the time over which a momentum change occurs reduces the average resultant force.

A phone is dropped from the same height onto two different surfaces. It lands on a bed and comes to rest undamaged. It then lands on a tiled floor from the same height, and the screen cracks.

In both cases the phone reaches the surface with the same mass and the same speed, so the change in momentum is the same. The difference lies entirely in the interaction during the impact — not in the fall.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) What does the bed change: the change in momentum, the stopping time, or both?

(b) Does a ball bouncing off a wall experience a greater, smaller, or the same force as a ball that sticks to the wall, for the same initial speed and interaction time?

Write a line for each before this opens.

🔬 Explore — a cart, a force sensor and two bumpers

What you’ll do: measure the impulse acting during a collision and the change in momentum it produces, then compare them.

⏱️ 40 min🧰 a dynamics cart and level track · a force sensor with a stiff and a soft bumper · a motion sensor · a data logger · a balance

1Measure the cart’s total mass m. Fix the force sensor and its spring bumper at one end of the track, with the motion sensor at the other.
2Zero the force sensor with no force on it, and set the software to plot force and velocity against the same time axis.
3Push the cart so it moves at approximately constant speed before striking the bumper and rebounding — keeping your hand clear during the collision.
4From the velocity–time graph read the velocity just before (vᵢ) and just after (v_f) the collision. Taking the initial direction as positive makes v_f negative, so Δp = m(v_f − vᵢ).
5From the force–time graph, find the impulse J as the area under the pulse, and record the peak force and the contact time Δt.
6Repeat twice more: at a higher incident speed, then with the softer bumper at about the original speed.

🎛️ Simulation — same impulse, different force

The area under each pulse is the impulse, and both areas are equal — because the momentum change is the same however the ball is stopped. Only the shape differs.

Mass0.2 kg

Speed on impact6 m/s

Rebound speed0 m/s

Contact time — soft surface0.4 s

Contact time — hard surface0.02 s

Momentum change Δp

1.20 kg m/s

Drag either contact time — this does not move. The surface cannot change it.

Force — soft

3 N

Force — hard

60 N

Hard ÷ soft

20.0×

exactly the inverse of the time ratio

Press sticks, then rebounds at full speed. The momentum change doubles, from mv to 2mv — reversing the motion is a bigger change than merely stopping it. So a ball that bounces off a wall pushes on it harder than one that sticks, for the same contact time. That is prediction (b), settled.

Analyse what you collected

(a) For each trial, compare the impulse $J$ (the area under the force–time graph) with the change in momentum $Δ p = m (v_{f} - v_{i})$ . Are they equal within experimental uncertainty?

(b) The softer bumper increases the contact time. For a similar change in momentum, explain how this affects the force during the collision.

✍️ Write, in two lines, why the area under your force–time graph should equal $m (v_{f} - v_{i})$ , before reading on.

Measurement note

Read

v_{i}

and

v_{f}

from the flat (constant-velocity) parts of the velocity–time graph, just before and just after the collision — never during it, where the value is changing every instant. Take the impulse from the force pulse only, and re-zero the force sensor before each run.

In the Explore, comparing Runs 1 and 3, the cart had the same mass and about the same incident speed. The softer bumper exerted a smaller average force over a longer time; the firmer bumper a larger force over a shorter one. You may have predicted that the softer bumper reduces the cart’s change in momentum. It does not — for the same initial and rebound speeds the momentum change is fixed. Only the interaction time, and therefore the force, differed.

Momentum and impulse

A body of mass $m$ moving with velocity $v$ has momentum:

p = m v

It is a vector in the same direction as the velocity, measured in $kg m/s$ . A change in momentum is produced by a resultant force acting over a time interval, and for a constant resultant force:

J = F Δ t = Δ p

Rearranging gives Newton’s second law in momentum form:

F = \frac{Δ p}{Δ t}

Read that as the sentence it is: for the same change in momentum, increasing the interaction time reduces the average force, and shortening it increases the force. That single sentence explains the bed, the helmet, the crumple zone and the airbag.

Point

The product of a body’s mass and velocity is its momentum, $p = m v$ , representing the quantity of motion. Its unit is $kg m/s$ .
The product $F Δ t$ is the impulse, which equals the change in momentum. Its unit is the N s. If the momentum before is $m v$ and after is $m v^{'}$ , then $m v^{'} - m v = J = F Δ t$ .
A longer interaction time reduces the average force, for the same change in momentum. The average force is the impulse divided by the time.
Both are vectors. When the direction reverses, the change is the difference of two opposite vectors — which is a sum of magnitudes, not a difference.

📐 Equation box

Momentum

p = m v

Impulse

J = F Δ t = Δ p

Newton's second law (momentum form)

F = \frac{Δ p}{Δ t}

Reversal at the same speed

∣Δ p ∣ = 2 m v

Key terms

Momentum: — Mass × velocity — a vector, in kg m/s, representing quantity of motion.
Impulse: — Force × time — a vector, in N s, equal to the change in momentum it produces.
Impulse–momentum theorem: — The area under a force–time graph equals the change in momentum.
Average force: — The impulse divided by the contact time.

🛠️ Worked example

Situation

A ball of mass $6.0 \times 1 0^{- 2} kg$ moves eastward at 20 m/s. Determine the magnitude and direction of the impulse applied to the ball when (a) it is caught, (b) it is hit back westward at the same speed, and (c) it is hit back northward at the same speed.

Step 1 of 6 — try it yourself before you open the next one.

A force of 5.0 N is applied to an object of mass 4.0 kg moving at 2.0 m/s for 2.0 s in the direction of its motion. Determine (i) the magnitude of the impulse, (ii) the increase in the object’s momentum, (iii) its final speed.

A ball of mass 0.15 kg moving horizontally at 30 m/s is hit back by a bat at the same speed, at an elevation angle of 60° above the horizontal. Determine (i) the magnitude of the impulse, (ii) the average force if the contact time is

2.0 \times 1 0^{- 2} s

In a New Context

When a motorcyclist falls and their head strikes the road, the change in momentum cannot be avoided — the head must slow to a stop. A helmet does not change that. What it changes is the time over which the change happens.

The helmet’s energy-absorbing foam compresses during the impact, increasing the stopping time and the stopping distance of the head. For the same change in momentum:

F = \frac{Δ p}{Δ t}

so a longer $Δ t$ means a smaller average impact force, and therefore a smaller average deceleration of the skull and its contents.

➡️ Your turn: Explain why a thicker, softer helmet liner gives better protection, even though the magnitude of the head’s change in momentum is unchanged.

The same idea, four times over

The bed under the phone, the foam in the helmet, the crumple zone in a car, the airbag: all four are the identical piece of physics. None of them reduces

Δ p

— that is fixed by the mass and the speed you arrive with. Every one of them increases

Δ t

. Once you have seen that, you have seen the whole of vehicle safety engineering in one equation.

🧪 Nature of Science — Newton's second law in momentum form

Newton (1687) did not express his second law as $F = ma$ . He stated that the rate of change of motion is proportional to the applied force and occurs in its direction. “Quantity of motion” corresponds to what we now call momentum. In modern notation: $F_{resultant} = \frac{d p}{d t}$ , and over an interval, $F_{avg} = Δ p /Δ t$ .

For constant mass this reduces to $F = ma$ — the familiar form is a special case of the momentum form, not the other way round. A scientific law describes a consistent relationship between physical quantities and allows predictions under stated conditions; part of understanding a law is knowing which of its forms is the more general.

Your turn: Suggest why the momentum form F = Δp/Δt is considered the more general statement of the law.

✅ Lesson question, answered

Why is a phone less likely to break when it lands on a bed than when it lands on a tiled floor, even if it is dropped from the same height?

If a phone hits a bed or tiles with the same speed and stops without bouncing, the change in momentum is the same in both cases. The difference is the stopping time.

A bed deforms, increasing the stopping time, which reduces the average force. A tiled floor stops the phone in a far shorter time, producing a much larger average force and a much greater chance of damage.

The surface cannot change $Δ p$ . It can only change $Δ t$ — and $F = Δ p /Δ t$ does the rest.

A force of 2.5 N was applied to an object of mass 2.0 kg moving at 1.0 m/s for 0.40 s in the direction of its motion. Determine (i) the impulse, (ii) the increase in momentum, (iii) the final speed.

A ball of mass 0.14 kg moving at 40 m/s is caught with a glove. Determine (i) the magnitude of the impulse received by the ball, (ii) the average force exerted by the ball on the glove, if the contact time until it stops is

2.0 \times 1 0^{- 2} s

A ball of mass

3.0 \times 1 0^{- 2} kg

moves eastward at 10 m/s. Determine the magnitude and direction of the impulse applied when (i) the ball is caught, (ii) it is hit back westward at the same speed, (iii) it is hit back northward at the same speed.

A ball of mass 0.40 kg moving horizontally at 10 m/s is hit back at the same speed in a direction making an angle of 120° with its initial direction. Determine the magnitude of the impulse received, and the angle between the impulse and the initial direction.

📝 Exam-style question

A 0.50 kg ball moving at 8.0 m/s is brought to rest by a glove in 0.040 s. Determine the impulse exerted on the ball.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Part 1 — multiple choice (50% of the paper)

The vector nature of momentum is where most of these go wrong. Read every direction.

A 3.0 kg object moves at 4.0 m/s. Its momentum is:

A phone lands on a bed rather than on tiles, from the same height. Compared with the tiles, the bed gives:

A 0.20 kg ball hits a wall at 5.0 m/s and rebounds at 5.0 m/s. The magnitude of its momentum change is:

The area under a force–time graph for a collision represents:

A constant force of 12 N acts for 0.50 s. The impulse is:

A car crash lasts 0.10 s. Fitting a crumple zone that extends it to 0.40 s changes the average force on the occupants to:

Part 2 — written answer (the other 50%)

State your positive direction before you compute anything — most lost marks in this topic are signs, not arithmetic. Each question has its own notebook page.

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

A 2.0 kg trolley moves at 3.0 m/s. Determine its momentum.

1.2

A force of 8.0 N acts for 0.25 s. Determine the impulse.

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

A 0.20 kg ball hits a wall at 15 m/s and rebounds at 12 m/s. Determine (i) the magnitude of the change in momentum and (ii) the average force, if contact lasts 0.050 s.

2.2

A 1200 kg car travelling at 20 m/s is brought to rest in a crash lasting 0.15 s. Determine the average force. Then determine it again if a crumple zone extends the collision to 0.60 s, and comment.

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

A 0.058 kg tennis ball arrives at 25 m/s and is returned at 30 m/s along the same line in the opposite direction. Contact lasts 5.0 ms. Determine (i) the change in momentum, (ii) the average force, (iii) how that force compares with the ball’s weight.

3.2

A 0.30 kg ball travelling east at 12 m/s is struck and leaves at 12 m/s due north. Determine the magnitude and direction of the impulse.

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