Mechanics: momentum and connected bodies (Former VCE Specialist Mathematics study design)

Syllabus dot point

“Momentum $p = mv$ and impulse as the change in momentum, the impulse-momentum relationship, and the analysis of connected particles such as bodies linked by a string over a pulley or in contact, which share a common acceleration”

7 min readUpdated 23 Sep 2026

Quick answer

Momentum is $p = mv$ , a vector pointing along the velocity. Impulse is the change in momentum, equal to force times time for a constant force: $\text{impulse} = F\,t = \Delta p = m v - m u$ . For connected bodies linked by an inextensible string over a smooth pulley, the two masses share the same acceleration magnitude and the string tension is common. Apply Newton's second law to each body separately, taking the direction of motion as positive, then solve the simultaneous equations for the acceleration and tension. The heavier side drives the motion.

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What this dot point is asking
Momentum and impulse
Connected bodies
Strategy
Examples in context
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Exam-style questions

What this dot point is asking

Former study design. Dynamics (Newton's laws, momentum and connected particles) was part of the previous VCE Specialist Mathematics study design and last appeared on the 2022 examinations. The current study design (from 2023) has no dynamics in Units 3 and 4: forces appear only in Units 1 and 2 (vectors applied to statics and motion under a constant force), and the Unit 4 calculus is kinematics (rectilinear motion and vector calculus). Keep this page for Units 1 and 2 vectors, for Physics, or for older resources.

VCAA wants you to use momentum and impulse to analyse changes in motion, and to handle connected bodies, such as two masses joined by a string over a pulley, that move together with a common acceleration. You apply Newton's second law to each body and combine the equations.

Momentum and impulse

The momentum of a particle is

p = mv,

a vector in the direction of the velocity, measured in kilogram metres per second. Impulse is the change in momentum a force produces. For a constant force $F$ acting for time $t$ ,

\text{impulse} = Ft = \Delta p = mv - mu,

where $u$ and $v$ are the initial and final velocities. This follows from Newton's second law: $F = ma = m\frac{v - u}{t}$ , so $Ft = m(v - u)$ . Impulse measured as force times time equals the change in momentum it causes.

Connected bodies

When two particles are joined by a light inextensible string passing over a smooth pulley, two facts simplify the analysis:

the string is inextensible, so both bodies have the same acceleration magnitude $a$ ;
the pulley is smooth and the string light, so the tension $T$ is the same throughout.

To solve, apply Newton's second law to each body separately, choosing the direction of motion as positive for each. This gives two equations in the two unknowns $a$ and $T$ . Add or substitute to solve. Bodies in direct contact (one pushing another) likewise share a common acceleration, and you analyse the system as a whole and then each body to find the contact force.

Strategy

Decide the direction of motion (the heavier mass falls, dragging the lighter one).
Draw the forces on each body: weight $mg$ and tension $T$ .
Write $F = ma$ for each, with consistent positive directions.
Solve the pair of equations for $a$ and $T$ .

Worked example

Two masses over a pulley

Masses of $3\ \text{kg}$ and $5\ \text{kg}$ hang from a light string over a smooth pulley. Find the acceleration of the system and the tension in the string. Take $g = 9.8\ \text{m/s}^2$ .

The heavier $5\ \text{kg}$ mass falls and the $3\ \text{kg}$ mass rises, with common acceleration $a$ and common tension $T$ .

Step 1: Write the equation of motion for the $5\ \text{kg}$ mass

Taking downward as positive for this mass, net force equals mass times acceleration. The weight $5g$ drives the motion downward while tension $T$ opposes it.

5g - T = 5a.

Step 2: Write the equation of motion for the $3\ \text{kg}$ mass

Taking upward as positive for this mass, tension $T$ pulls it up while its weight $3g$ pulls it down. Choosing the direction of motion as positive for each body keeps the algebra consistent.

T - 3g = 3a.

Step 3: Add the equations to find the acceleration

Adding the two equations eliminates the unknown tension $T$ , leaving a single equation in $a$ . This is the standard strategy for connected-body problems.

5g - 3g = 5a + 3a \quad\Rightarrow\quad 2g = 8a \quad\Rightarrow\quad a = \frac{2g}{8} = \frac{g}{4}.

With $g = 9.8$ , $a = \frac{9.8}{4} = 2.45\ \text{m/s}^2$ .

Step 4: Find the tension and verify

Substitute the acceleration back into either motion equation to recover $T$ , then check the result against the other equation to confirm no arithmetic errors.

T = 3g + 3a = 3(9.8) + 3(2.45) = 29.4 + 7.35 = 36.75\ \text{N}.

Check with the other equation: $5g - T = 49 - 36.75 = 12.25$ , and $5a = 5(2.45) = 12.25$ . They agree, confirming $a = 2.45\ \text{m/s}^2$ and $T = 36.75\ \text{N}$ .

Final answer: $a = 2.45\ \text{m/s}^2$ , $T = 36.75\ \text{N}$ .

Common exam traps

Different accelerations for connected bodies: An inextensible string forces the same acceleration magnitude on both. Do not give them separate values.
Different tensions: With a light string over a smooth pulley the tension is the same throughout.
Inconsistent positive directions: Set up each equation with the actual direction of motion as positive for that body, or signs will conflict.
Impulse as force alone: Impulse is force times time, equal to the change in momentum, not just the force.

Examples in context

Example 1. A $2\ \text{kg}$ ball moving at $3\ \text{m/s}$ has momentum $2\times 3 = 6\ \text{kg m/s}$ .

Example 2. A force of $10\ \text{N}$ acting for $4\ \text{s}$ delivers an impulse of $40\ \text{kg m/s}$ , changing the momentum by that amount.

Try this

Q1. Find the momentum of a $0.5\ \text{kg}$ object moving at $8\ \text{m/s}$ . [1 mark]

Cue. $0.5\times 8 = 4\ \text{kg m/s}$ .

Q2. A constant $6\ \text{N}$ force acts for $5\ \text{s}$ . State the impulse. [1 mark]

Cue. $6\times 5 = 30\ \text{kg m/s}$ .

Q3. For $2\ \text{kg}$ and $4\ \text{kg}$ masses over a smooth pulley, write the two motion equations. [2 marks]

Cue. $4g - T = 4a$ and $T - 2g = 2a$ .

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

2\ \text{kg}

ball travelling at

5\ \text{m/s}

strikes a wall and rebounds along the same line at

3\ \text{m/s}

. Taking the initial direction as positive, calculate the impulse exerted on the ball by the wall.

Show worked answer →

Impulse equals the change in momentum, $\Delta p = mv - mu$ , taking the initial direction as positive.

Initial momentum: $mu = 2 \times 5 = 10\ \text{kg m/s}$ .

After rebounding, the velocity is $-3\ \text{m/s}$ , so final momentum $mv = 2 \times (-3) = -6\ \text{kg m/s}$ .

Impulse $= mv - mu = -6 - 10 = -16\ \text{kg m/s}$ , i.e. $16\ \text{kg m/s}$ directed away from the wall.

Markers reward the sign convention, both momenta, and the impulse magnitude and direction.

Practice (former study design)5 marks

Masses of

4\ \text{kg}

and

6\ \text{kg}

are joined by a light inextensible string over a smooth pulley and released from rest. Take

g = 9.8\ \text{m/s}^2

. (a) Determine the acceleration of the system and the tension in the string. (b) Calculate the speed of the system after

2\ \text{seconds}

Show worked answer →

(a) The $6\ \text{kg}$ mass falls, the $4\ \text{kg}$ mass rises, with common acceleration $a$ and tension $T$ .

For the $6\ \text{kg}$ mass (down positive): $6g - T = 6a$ . For the $4\ \text{kg}$ mass (up positive): $T - 4g = 4a$ .

Add: $6g - 4g = 10a$ , so $2g = 10a$ and $a = \dfrac{2 \times 9.8}{10} = 1.96\ \text{m/s}^2$ .

Substitute into $T - 4g = 4a$ : $T = 4(9.8) + 4(1.96) = 39.2 + 7.84 = 47.04\ \text{N}$ .

(b) From rest with $a = 1.96$ : $v = u + at = 0 + 1.96 \times 2 = 3.92\ \text{m/s}$ .

Markers reward both motion equations, solving for $a$ and $T$ , and the constant-acceleration speed.

Mechanics: momentum and connected bodies (Former VCE Specialist Mathematics study design)

What this dot point is asking

Momentum and impulse

Connected bodies

Strategy

Two masses over a pulley

Step 1: Write the equation of motion for the $5\ \text{kg}$ mass

Step 2: Write the equation of motion for the $3\ \text{kg}$ mass

Step 3: Add the equations to find the acceleration

Step 4: Find the tension and verify

Examples in context

Try this

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

Momentum and impulse

Connected bodies

Strategy

Two masses over a pulley

Step 1: Write the equation of motion for the 5 kg5\ \text{kg}5 kg mass

Step 2: Write the equation of motion for the 3 kg3\ \text{kg}3 kg mass

Step 3: Add the equations to find the acceleration

Step 4: Find the tension and verify

Examples in context

Try this

Exam-style questions

Practise this

Sources & how we know this

Step 1: Write the equation of motion for the $5\ \text{kg}$ mass

Step 2: Write the equation of motion for the $3\ \text{kg}$ mass