The Second Law and Momentum: Quick Revision

A brief note on The Second Law and Momentum. Read it before a test. Try each question at the end before you open its answer.

The big ideas

• Momentum $p = mv$, in kg m/s, with a direction.
• Second law: $F = ma$, where $F$ is the net force. $1\ \text{N} = 1\ \text{kg m/s}^2$.
• The same law as force $=$ change of momentum $\div$ time: a longer stop needs a smaller force.
• Four steps: draw the forces, find the net force, $a = F/m$, then the equations of motion.

The facts at a glance

A worked example

A $120\ \text{kg}$ scooter with its rider has a net force of $360\ \text{N}$ forward for $4\ \text{s}$, then a braking force of $600\ \text{N}$ for $2.4\ \text{s}$. Describe its motion.

1. Speeding up. $a = \dfrac{360\ \text{N}}{120\ \text{kg}} = 3\ \text{m/s}^2$, so it gains $12\ \text{m/s}$.
2. Braking. $a = -\dfrac{600\ \text{N}}{120\ \text{kg}} = -5\ \text{m/s}^2$, so it loses $12\ \text{m/s}$.
3. So it ends at rest again.

Practice with new numbers

A program to try

The program works out the scooter.

Mistakes to avoid

• Using one force instead of the net force in $F = ma$.
• Forgetting direction in momentum. A bounce changes it by $m(u + v)$.
• Mixing grams and kilograms. Change to kg first.
• Thinking airbags reduce the momentum change. They only make it take longer.

Check yourself

1. What is the momentum of a $0.2\ \text{kg}$ ball at $15\ \text{m/s}$?

2. What net force gives a $4\ \text{kg}$ trolley an acceleration of $2.5\ \text{m/s}^2$?

3. A $70\ \text{kg}$ runner speeds up from $2\ \text{m/s}$ to $8\ \text{m/s}$ in $3\ \text{s}$. What average force acts?

4. Why do high jumpers land on thick mats?

5. The same force acts on a $1\ \text{kg}$ and a $5\ \text{kg}$ block. Which accelerates more, and by how many times?