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

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

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 $t$,

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

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

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 $T - 2g = 2a$.