What are different tensions?

With a light string over a smooth pulley the tension is the same throughout.

What are inconsistent positive directions?

Set up each equation with the actual direction of motion as positive for that body, or signs will conflict.

What is impulse as force alone?

Impulse is force times time, equal to the change in momentum, not just the force.

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

A constant 6\ N force acts for 5\ s. State the impulse. [1 mark]

VCAA VCAA 2022 Exam 1-style practice: A 2\ kg ball travelling at 5\ m/s strikes a wall and rebounds along the same line at 3\ m/s. Taking the initial direction as positive, calculate the impulse exerted on the ball by the wall.

Impulse equals the change in momentum, p = mv - mu, taking the initial direction as positive. Initial momentum: mu = 2 × 5 = 10\ kg m/s. After rebounding, the velocity is -3\ m/s, so final momentum mv = 2 × (-3) = -6\ kg m/s. Impulse = mv - mu = -6 - 10 = -16\ kg m/s, i.e. 16\ kg m/s directed away from the wall. Markers reward the sign convention, both momenta, and the impulse magnitude and direction.

VCAA VCAA 2023 Exam 2-style practice: Masses of 4\ kg and 6\ kg are joined by a light inextensible string over a smooth pulley and released from rest. Take g = 9.8\ 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\ seconds.

**(a)** The 6\ kg mass falls, the 4\ kg mass rises, with common acceleration a and tension T. For the 6\ kg mass (down positive): 6g - T = 6a. For the 4\ kg mass (up positive): T - 4g = 4a. Add: 6g - 4g = 10a, so 2g = 10a and a = 2 × 9.810 = 1.96\ m/s^2. Substitute into T - 4g = 4a: T = 4(9.8) + 4(1.96) = 39.2 + 7.84 = 47.04\ N. **(b)** From rest with a = 1.96: v = u + at = 0 + 1.96 × 2 = 3.92\ m/s. Markers reward both motion equations, solving for a and T, and the constant-acceleration speed.

§-Syllabus dot point

VICSpecialist MathematicsSyllabus dot point

How do momentum and impulse describe changes in motion, and how do we analyse connected bodies that share a common acceleration?

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

A focused answer to the VCE Specialist Mathematics Unit 4 key-knowledge point on momentum and connected bodies. Momentum, impulse as change in momentum, and analysing connected particles with a shared acceleration, with a verified worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Momentum and impulse
Connected bodies
Strategy
Examples in context
Try this

What this dot point is asking

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

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 practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 13 marksA

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.

VCAA 2023 Exam 25 marksMasses 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.