Force-time graphs and impulse
Newton's Laws & Momentum - OCR A-Level Physics
Key Definition
Area under a force-time graph
The area enclosed between a force-time curve and the time axis is equal to the impulseThe product of force and the time for which it acts. Equal to the change in momentum. delivered, which equals the change in momentum of the object.
The area enclosed between a force-time curve and the time axis is equal to the impulseThe product of force and the time for which it acts. Equal to the change in momentum. delivered, which equals the change in momentum of the object.
$$\text{Area} = F \, \Delta t = \Delta p$$
Case 1 : Constant force
The graph is a horizontal line. The area is a rectangle: $\text{impulse} = F \times \Delta t$. Numerical example: a 50 N force for 0.20 s gives $\Delta p = 50 \times 0.20 = 10 \; \text{N s}$.
Case 2 : Triangular pulse
The force rises to a peak and falls back to zero. Use $\text{area} = \frac{1}{2} \times \text{base} \times \text{height}$. Example: peak 200 N over 0.05 s gives $\Delta p = \frac{1}{2} \times 0.05 \times 200 = 5 \; \text{N s}$.
Case 3 : Arbitrary curve
Count squares on the grid, or split the area into trapeziums. The gradient at any point gives the (instantaneous) rate of change of momentum, which equals the net force at that moment.
- For the same change in momentum, a graph with a lower peak and a wider base delivers a smaller maximum force on the object. This is the principle behind every collision-protection device.
- If two force-time graphs enclose the same area, they represent the same change in momentum, even if their shapes are very different.
- Worked example: a 0.20 kg ball strikes a wall at 8.0 $\text{m s}^{-1}$ and rebounds at 6.0 $\text{m s}^{-1}$. $\Delta p = (0.20)(-6.0) - (0.20)(8.0) = -2.8 \; \text{kg m s}^{-1}$. The area under the F-t graph is 2.8 N s.
- The gradient of the force-time graph has units of $\text{N s}^{-1}$ and is rarely asked about directly.
Common Mistake
HIGH
Wrong: Reading the peak force from a force-time graph and multiplying by the total time to get the impulse.
Right: Find the AREA enclosed by the curve. For a triangular pulse this is $\frac{1}{2} \times \text{base} \times \text{height}$, not peak $\times$ total time. Peak force times total time over-estimates impulse for any non-rectangular shape.
Right: Find the AREA enclosed by the curve. For a triangular pulse this is $\frac{1}{2} \times \text{base} \times \text{height}$, not peak $\times$ total time. Peak force times total time over-estimates impulse for any non-rectangular shape.
Examiner Tips and Tricks
- Always check axis units before computing area. If the time axis is in milliseconds, convert to seconds first or your impulse will be 1000 times too large.
- If the question gives a curved graph and asks for impulse, count squares from the grid and multiply by the area of one square (in N s).
- If asked to compare two impacts with and without a safety feature, state that the AREA under both graphs is the same (same $\Delta p$), but the peak force is reduced and the duration is longer.
Two overlaid force-time graphs: tall narrow triangular pulse ("hard impact") vs short wide pulse ("with crumple zone"). Same area enclosed, different peak force. Label peak F and base $\Delta t$ on each.