Impulse

The video lesson for this topic is coming soon.

Essential Question

How does a force change an object’s momentum?

Momentum only changes when a force acts, and it matters how long that force acts. By the end of this lesson, you should be able to use J = FΔt = Δp to find impulse, force, time, or a change in velocity, and explain why airbags and helmets keep people safe.

What Is Impulse?

A force acting over time
Impulse = force × time
Unit: newton-seconds (N·s)

A red force arrow labeled F pushes a cart to the right on the ground. A stopwatch next to it is labeled delta t. Caption: impulse = force × time.

An impulse is a force acting over a period of time. Impulse equals the force times the time it acts, so a big force for a short time can give the same impulse as a small force for a long time. Because it is a force in newtons times a time in seconds, impulse is measured in newton-seconds, N·s.

Impulse Changes Momentum

J = FΔt = Δp

J = impulse, in newton-seconds (N·s)
F = average force, in newtons (N)
Δt = time the force acts, in seconds (s)
Δp = change in momentum, in kg·m/s, where Δp = mΔv = m(v_f − v_i)

The equation J = F delta t = delta p in a yellow box. Below it, four cards: J is impulse in newton-seconds, F is force in newtons, delta t is time in seconds, and delta p is the change in momentum in kilogram meters per second.

This is the impulse-momentum theorem. The impulse on an object equals its change in momentum. A force that acts in the direction of motion speeds the object up, and a force against the motion slows it down. Since mass usually stays the same, Δp = mΔv.

Where It Comes From

Start with Newton’s 2nd law
Acceleration: a = Δv/Δt
Multiply both sides by Δt
So N·s = kg·m/s

A cyan card showing three steps: F = ma, then F = m delta v over delta t, then F delta t = m delta v.

The impulse-momentum theorem is really Newton’s second law in disguise. Start with F = ma. Acceleration is the change in velocity divided by time, so F = mΔv/Δt. Multiply both sides by Δt and you get FΔt = mΔv. This also shows that a newton-second is the same unit as a kilogram meter per second.

Solving for Any Variable

Impulse: J = FΔt
Force: F = Δp/Δt
Time: Δt = Δp/F
Δp = m(v_f − v_i)

A triangle split into three parts with delta p on top and F and delta t on the bottom. Caption: cover the one you want.

Rearrange FΔt = Δp for whatever the problem asks for. Cover the quantity you want in the triangle: covering Δp leaves F next to Δt, so multiply. Covering F leaves Δp over Δt, so divide. To find Δp from velocities, use Δp = m(v_f − v_i).

More Time, Less Force

Stopping needs the same Δp
Longer time: smaller force
Shorter time: bigger force

A force versus time graph. A tall, narrow red spike is labeled dashboard: big force, short time. A low, wide blue hump is labeled airbag: small force, long time. Caption: same area, same impulse.

To stop a moving person, you have to take away all of their momentum, so Δp is fixed. Since FΔt = Δp, stretching the stopping time makes the force smaller. On a force-time graph, a dashboard gives a tall, narrow spike and an airbag gives a low, wide hump. The areas are the same, because the impulse is the same, but the airbag’s force is much smaller.

Stretching the Time

Airbags and crumple zones
Helmet foam
Bend your knees when you land
Catch an egg gently

Two eggs dropped from the same height. The left egg lands on a pillow, labeled longer time, smaller force, egg survives. The right egg lands on a hard floor, labeled shorter time, bigger force, egg breaks.

Safety gear works by stretching out the time it takes to stop. An airbag, the crumple zone of a car, and the foam in a helmet all make a collision last longer, so the force on your body is smaller. Bending your knees when you land from a jump does the same thing. An egg dropped on a pillow survives for the same reason, while the same egg dropped on a hard floor stops almost instantly and breaks.

Force-Time Graphs

Area under the curve = impulse
Rectangle: F × Δt
Triangle: ½ × base × height

A force versus time graph with a constant force of 40 N from 0 to 0.50 s. The rectangle under the line is shaded and labeled area = 20 N·s.

When force changes during a collision, use a force-time graph. The area under the curve equals the impulse. For a constant force, the area is a rectangle: here 40 N for 0.50 s gives 20 N·s. For a force that rises and falls in a straight line, the area is a triangle, ½ × base × height.

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: Finding Impulse

A hockey stick pushes a puck with an average force of 150 N for 0.080 s. What impulse does the stick give the puck?

A black hockey puck on ice with a red arrow labeled F = 150 N pushing it and the label delta t = 0.080 s.

Knowns: F = 150 N, Δt = 0.080 s
Equation: J = FΔt
Substitute: J = (150 N)(0.080 s)
Answer: J = 12 N·s in the direction of the push. Two significant figures, because 0.080 s has two.

Example 2: Catching a Baseball

A 0.150 kg baseball moving right at 30.0 m/s is caught and stopped by a glove in 0.0500 s. What average force does the glove exert on the ball?

A plus arrow points right. A baseball labeled 0.150 kg moves right at +30.0 m/s toward a brown glove.

Knowns: m = 0.150 kg, v_i = +30.0 m/s, v_f = 0 m/s, Δt = 0.0500 s
Change in momentum: Δp = m(v_f − v_i) = (0.150 kg)(0 − 30.0 m/s) = −4.50 kg·m/s
Equation: FΔt = Δp, so F = Δp/Δt
Substitute: F = (−4.50 kg·m/s) / (0.0500 s)
Answer: F = −90.0 N, or 90.0 N to the left. The glove pushes opposite the ball’s motion.

Example 3: Airbag vs. Dashboard

A 70.0 kg driver moving at 15.0 m/s is brought to a stop. Find the size of the average force on the driver if an airbag stops him in 0.300 s, and if the dashboard stops him in 0.0150 s.

Two cards side by side. The blue airbag card reads delta t = 0.300 s. The pink dashboard card reads delta t = 0.0150 s. Both say delta p is the same.

Change in momentum: Δp = m(v_f − v_i) = (70.0 kg)(0 − 15.0 m/s) = −1,050 kg·m/s, so its size is 1,050 kg·m/s
Equation: F = Δp/Δt
Airbag: F = (1,050 kg·m/s) / (0.300 s) = 3.50 × 10³ N
Dashboard: F = (1,050 kg·m/s) / (0.0150 s) = 7.00 × 10⁴ N
Compare: The dashboard force is 20 times bigger, because the stop takes 1/20 of the time.

Example 4: Force-Time Graph

A 0.40 kg soccer ball at rest is kicked. The graph shows the force on the ball: it rises in a straight line from 0 N to 600 N at 0.010 s, then falls back to 0 N at 0.020 s. Find the impulse and the ball’s final speed.

A force versus time graph shaped like a triangle. The force rises from 0 N at 0 s to 600 N at 0.010 s, then falls back to 0 N at 0.020 s.

Impulse is the area: The graph is a triangle with base 0.020 s and height 600 N.
Area: J = ½ × base × height = ½ × (0.020 s)(600 N) = 6.0 N·s
Change in momentum: J = Δp = m(v_f − v_i), and v_i = 0, so v_f = J/m
Answer: v_f = (6.0 N·s) / (0.40 kg) = 15 m/s

Big Ideas

Impulse = force × time, J = FΔt
Impulse = change in momentum, Δp
More time means less force
Area under F–t graph = impulse

Back to the essential question: a force changes an object’s momentum by giving it an impulse, the force times the time it acts, J = FΔt = Δp. Spreading a change in momentum over more time means a smaller force. Next up, we’ll see what happens to momentum when objects push on each other in conservation of momentum.