Impulse on a Force-Time Graph (OCR AS Physics): Revision Note

A ball of mass 3.0 kg, initially at rest, is acted on by a force F which varies with t as shown by the graph.

Calculate the velocity of the ball after 16 s.

Answer:

Step 1: List the known quantities

• Mass, m = 3.0 kg

• Initial velocity, u = 0 m s^-1 (since it is initially at rest)

Step 2: Calculate the impulse

• The impulse is the area under the graph

• The graph can be split up into two right-angled triangles with a base of 8 s and a height of 4 kN

Area = Impulse = 32 × 10³ N s

Step 3: Write the equation for impulse

Impulse, I = Δp = m(v – u)

Step 4: Substitute in the values

I = mv

32 × 10³ = 3.0 × v

v = (32 × 10³) ÷ 3.0

v = 10666 m s^–1 = 11 km s^-1

Some maths tips for this section: Rate of Change

• ‘Rate of change’ describes how one variable changes with respect to another

• In maths, how fast something changes with time is represented as dividing by Δt (e.g. acceleration is the rate of change in velocity)

• More specifically, Δt is used for finite and quantifiable changes such as the difference in time between two events

Areas

• The area under a graph may be split up into different shapes, so make sure you’re comfortable with calculating the area of squares, rectangles, right-angled triangles and trapeziums!