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3.8.4Forces and Motion

Impulse on a Force-Time Graph

5 min read 6 parts Lesson 65 of 66 Unit quiz Virtual lab

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Section 1 of 6Area Equals Impulse

3.8.4 Impulse on a Force-Time Graph

When force varies during a collision, the graph is the best way to find the total impulse.

Graph axes
  • A force-time graph plots force on the vertical axis against time on the horizontal axis.
Area
  • The area between the line and the time axis equals the impulse, because impulse is force multiplied by time.
Units
  • Because force is in newtons and time in seconds, the area is naturally in N s.
Change in momentum
  • The area equals the change in momentum of the object, which gives its final speed if the starting speed is known.
Direction
  • Areas below the time axis are negative and represent impulses in the opposite direction.

Section 2 of 6Finding the Area

Different graph shapes are handled by different methods.

1
For a constant force, the graph is a rectangle and the area is simply force multiplied by time.
2
For a force that rises or falls steadily, split the graph into triangles using ½ × base × height.
3
For a curved graph, count the full squares and part squares, then multiply by the impulse that each square represents.
4
Add up all the areas, treating any section below the axis as negative.
5
Set the total equal to Δp = mv − mu and solve for the unknown velocity or mass.

Section 3 of 6Average Force

Average Force
Definition
  • The average force is the total impulse divided by the total time of contact.
Rectangle equivalent
  • It is the height of a rectangle having the same area and the same width as the actual graph.
Peak force
  • The maximum force is usually much greater than the average, particularly in short collisions.
Comparison
  • Two collisions with equal area have the same impulse, even if one has a high short peak and the other a low long one.

Section 4 of 6Typical Graph Shapes

Bouncing ball
  • The force rises sharply from zero, peaks as the ball is most squashed and falls back to zero as it leaves.
Crash test
  • A rigid barrier gives a tall narrow spike while a crumple zone gives a lower, wider graph with the same area.
Rocket engine
  • A steady thrust gives a rectangle whose area is thrust multiplied by burn time.

Section 5 of 6Worked Example

Question

  • A ball of mass 0.25 kg at rest is struck.
  • The force rises linearly from 0 to 800 N in 0.010 s and then falls linearly to 0 in a further 0.020 s.
  • Find the impulse, the average force and the speed of the ball afterwards.

  1. The graph is a triangle of height 800 N and base 0.030 s.
  2. Impulse = ½ × 800 × 0.030 = 12 N s.
  3. Average force = 12 / 0.030 = 400 N.
  4. Δp = mv − 0 = 12, so v = 12 / 0.25 = 48 m s⁻¹.

Answer

  • Impulse = 12 N s;
    • average force = 400 N;
    • final speed = 48 m s⁻¹.

Section 6 of 6Exam Checklist & Tips

3.8.4 – Key Points for the Exam

  • Force-time graph A graph whose area gives the impulse delivered.
  • Impulse The change in momentum produced by a force acting over a time.
  • Average force Total impulse divided by contact time.
  • Peak force The greatest value of force during an interaction.

Common Mistakes

  • Reading the gradient of a force-time graph as impulse; the area, not the gradient, is the impulse.
  • Using the peak force as the average force.
  • Forgetting the factor ½ when the graph is a triangle.
  • Ignoring a negative area below the axis when the force reverses.
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