C. Energy and Momentum · Momentum and Collisions

Force–Time Graphs: Impulse as an Area

C2.5 A1.8 A1.12About 65 minutes

Learning goals

I can find the impulse of a changing force as the area under a force–time graph.
I can find areas of rectangles, triangles and trapezoids, and add positive and negative areas.
I can estimate impulse from sensor data using trapezoids or by counting squares.
I can use the impulse from a graph to find a change in velocity and an average force.

Words to know

force–time graph
A graph of force (vertical axis) against time (horizontal axis) during an interaction.

area under the graph
The region between the graph and the time axis; for a net-force–time graph it equals the impulse.

force sensor (force plate)
A device that measures force many times per second and records it as data.

trapezoid method
Estimating an area by joining data points with straight lines and adding the areas of the trapezoids formed.

peak force
The largest force reached during a collision.

Real forces are not constant

In the last lesson we used an average force. Real collision forces change from moment to moment. When a hockey stick hits a puck, the force starts at zero, rises to a peak as the blade and puck squash together, then falls back to zero as they separate. A force sensor or force plate in a physics lab records this as a force–time graph.

How do you find the impulse when the force keeps changing? The answer is beautiful and simple: the impulse is the area under the force–time graph.

Why area?

For a constant force, J = FΔt. On a force–time graph, that is a rectangle with height F and width Δt, so the impulse is the rectangle's area.

For a changing force, imagine cutting the graph into many thin vertical strips. Over each very short strip the force is almost constant, so each strip's impulse is almost F × (width). Adding all the strips gives the total area. (In calculus, MCV4U and beyond, this area is called an integral: J = ∫F dt.)

Key facts:

Impulse = area under the net-force–time graph = Δp.
Area above the time axis is positive impulse; area below the axis is negative impulse. Add them with signs.
Average force = area ÷ total time. It is the height of the rectangle with the same area and the same width.
Units: (N)(s) = N·s. Watch out for milliseconds: 1 ms = 0.001 s.

A triangular force pulse lasting 10 ms. The impulse is the shaded triangle's area, 6.0 N·s. The average force is 6.0 N·s ÷ 0.010 s = 600 N, half the peak force.

Worked example

Example 1: A triangular pulse

A 0.40 kg ball at rest is struck. The net force rises steadily from 0 to 1200 N in 5.0 ms, then falls steadily back to 0 at 10.0 ms (the graph above). Find the impulse, the ball's final speed and the average force.

Area of the triangle: J = ½(base)(height) = ½(0.010 s)(1200 N) = 6.0 N·s.
J = Δp = m(v_f − 0), so v_f = 6.0 ÷ 0.40 = 15 m/s.
F_avg = J ÷ Δt = 6.0 ÷ 0.010 = 600 N.

J = 6.0 N·s, v_f = 15 m/s, F_avg = 600 N. For a symmetric triangle, the average force is half the peak.

Worked example

Example 2: A trapezoid

The force of a stick on a 0.16 kg puck (at rest) rises from 0 to 800 N in 2.0 ms, stays at 800 N for 6.0 ms, then falls to 0 in 2.0 ms. Find the impulse and the puck's speed.

Split the shape: two triangles and a rectangle.
Rising triangle: ½(0.0020)(800) = 0.80 N·s. Falling triangle: also 0.80 N·s.
Rectangle: (0.0060)(800) = 4.8 N·s.
Total J = 0.80 + 4.8 + 0.80 = 6.4 N·s. (Or use the trapezoid formula: ½(a + b)h = ½(0.0060 + 0.010)(800) = 6.4 N·s.)
v_f = J ÷ m = 6.4 ÷ 0.16 = 40 m/s.

J = 6.4 N·s and v_f = 40 m/s (about 144 km/h). The average force is 6.4 ÷ 0.010 = 640 N.

A force that pushes forward and then backward. The area above the axis (+3.0 N·s) and the area below it (−1.0 N·s) add with signs to give the net impulse.

Worked example

Example 3: Positive and negative areas

A 0.50 kg cart is moving at 2.0 m/s [W]. Take east as positive. A net force acts on it as shown in the graph above: +6.0 N for 0.50 s, then −2.0 N for the next 0.50 s. Find the cart's final velocity.

Area above the axis: (6.0)(0.50) = +3.0 N·s. Area below: (−2.0)(0.50) = −1.0 N·s.
Net impulse J = 3.0 + (−1.0) = +2.0 N·s.
v_i = −2.0 m/s. J = m(v_f − v_i), so v_f = v_i + J ÷ m = −2.0 + 2.0 ÷ 0.50 = −2.0 + 4.0 = +2.0 m/s.

v_f = 2.0 m/s [E]. The cart turned around.

Areas from data: the trapezoid method

A force sensor does not give you a neat triangle. It gives a table of readings taken at equal time steps (for example, every millisecond). To find the area:

Join neighbouring points with straight lines. Each strip is then a trapezoid with area ½(F₁ + F₂)Δt.
Add the areas of all the strips.

When the readings start and end at zero and are equally spaced by Δt, this simplifies to J ≈ Δt × (sum of all the readings between the first and last).

Another method, useful with a printed graph, is counting squares: find the impulse represented by one grid square, count the squares under the curve (estimating part-squares), and multiply.

Force readings every 1 ms while a racket hits a tennis ball. Joining the points makes eight trapezoids. Their total area, 3.34 N·s, is the impulse.

Worked example

Example 4: Impulse from sensor data

A tennis racket serves a 0.058 kg ball that is momentarily at rest. A force sensor records the net force on the ball every 1.0 ms: 0, 150, 420, 700, 800, 700, 420, 150, 0 N. Estimate the impulse and the speed of the ball.

The readings start and end at 0, and Δt = 0.0010 s.
Sum of the middle readings: 150 + 420 + 700 + 800 + 700 + 420 + 150 = 3340 N.
J ≈ (0.0010 s)(3340 N) = 3.34 N·s.
v_f = J ÷ m = 3.34 ÷ 0.058 ≈ 57.6 m/s.

J ≈ 3.3 N·s and v ≈ 58 m/s (about 210 km/h).

Same area, different shape

Two collisions with the same area under their force–time graphs give the same change in momentum, even if one is tall and narrow and the other low and wide. Safety engineers try to make the graph low and wide (a small peak force over a long time). Athletes hitting a ball try to make the area as large as possible.

Just for fun

Why did the physics student bring graph paper to the hockey game?

She wanted to find the area under every slapshot.

Hands-on activity

Graph your own push

You need: a bathroom scale (digital or dial), a phone that can record video, graph paper (or a spreadsheet) and a ruler.

Put the scale on the floor or on a sturdy table. Start recording a video of the display.
Press down on the scale with one hand. Over about 4 seconds, push gradually harder, hold, then gradually ease off to zero.
Play the video back and read the scale every 0.5 s. Scales show kilograms, so convert each reading to newtons: F = (reading in kg) × 9.8 N/kg.
Make a table and plot force against time. Find the area under your graph using trapezoids.
Think about it: your hand gave the scale an impulse of several newton-seconds, yet the scale did not start moving. Why? Which other force acted on the scale, and what was its impulse?

What to look for: a hump-shaped graph and an impulse of perhaps 50 to 200 N·s. The scale stayed still because the floor (or table) pushed up with an equal and opposite impulse, so the net impulse on the scale was zero, and so was its change in momentum. Digital scales update only a few times a second, so your graph will look a bit stepped.

Your turn

A constant net force of 50 N acts for 0.30 s. Sketch the F–t graph and find the impulse.

A triangular force pulse has a peak of 900 N and lasts 4.0 ms. Find the impulse.

For the pulse in the previous question, find the average force.

A force rises from 0 to 300 N in 0.10 s, stays at 300 N for 0.20 s, then falls to 0 in 0.10 s. It acts on a 45 kg crate at rest on frictionless ice. Find the impulse and the crate's final speed.

A 5.0 kg object at rest feels a net force of +20 N for 2.0 s, then −10 N for 1.0 s. Find its final velocity.

Which gives the larger impulse: a triangular pulse with peak 1000 N lasting 2.0 ms, or a constant 100 N lasting 15 ms?

More practice: warm-up

What does the area under a net-force–time graph represent?

What are the units of the area under a force–time graph?

Find the area of a rectangle on an F–t graph that is 10 N high and 2.0 s wide.

Find the area of a triangle on an F–t graph with base 0.20 s and height 40 N.

What does area below the time axis mean?

More practice: on your own

A 0.16 kg puck slides at 10 m/s. A stick hits it in its direction of motion with a triangular pulse: peak 1500 N, duration 4.0 ms. Find the puck's new speed.

A sensor records 0, 300, 500, 300 and 0 N at 2.0 ms intervals. Estimate the impulse and the average force.

In a crash test, the area under the force–time graph for a 1200 kg car is 3.0 × 10⁴ N·s, and the crash lasts 0.12 s. Find the average force and the car's speed before the crash, if it stopped completely.

Two force–time graphs have the same area, but graph B lasts 4 times as long as graph A. Compare the changes in momentum and the average forces.

The net force on a 4.0 kg object at rest increases steadily from 0 N at t = 0 to 60 N at t = 3.0 s. Find its speed at t = 3.0 s.

Challenge

(For students taking calculus.) The net force on a 0.050 kg ball at rest is F(t) = (3.0 × 10⁶)t(0.010 − t), in newtons, for 0 ≤ t ≤ 0.010 s. Find the impulse and the ball's final speed. Compare with a triangle that has the same peak and duration.

A 60 kg student stands on a force plate, which reads 588 N. She crouches and jumps: the plate reads a steady 1200 N for 0.30 s, then she leaves the plate. Find her take-off speed and how high her centre of mass rises. (Hint: use the net force.)

A symmetric trapezoid force pulse lasts 10 ms in total, with 2.0 ms ramps at each end. It changes the velocity of a 0.20 kg ball by 30 m/s. Find the peak force.

Check your understanding

Quiz

On a graph of net force against time, the area under the graph represents:

A triangular force pulse has a peak of 600 N and lasts 0.020 s. What is the impulse?

How do you find the average force from a force–time graph?

Two force–time graphs have the same area. Graph A is low and wide; graph B is tall and narrow. Which statement is true?

Part of a net-force–time graph lies below the time axis. That part:

A force sensor reads 0, 200, 200 and 0 N at 1.0 ms intervals. Using trapezoids, the impulse is closest to:

For parents and teachers

Big idea (C2.5): the impulse of a changing force is the area under its force–time graph, and it equals the change in momentum. Average force = area ÷ time.

How to help: Sketch a hump-shaped force–time graph on grid paper and ask your teen to estimate its area by counting squares, then by trapezoids. Ask: "If the hump were twice as wide but half as tall, what would change?" (Not the impulse; the average force would halve.)

Common difficulties:

Milliseconds. The time axis is often in ms. Convert to seconds before multiplying.
Signs. Area below the axis is negative impulse.
Using the peak force as the average. The average is area ÷ duration.
Using a single force instead of the net force, especially on a force plate, where the person's weight must be subtracted (Challenge 2).
The calculus challenge is optional; it is for students taking MCV4U at the same time.