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Change in Momentum and Impulse

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, impulse is a force applied over a period of time, and it's exactly what causes an object's momentum to change.

Why this matters

Picture a major league batter at the plate. The pitcher throws a 95 mph fastball. The batter can do two very different things. They could lay down a bunt, where the bat just barely touches the ball, gently nudging it into the infield. The bat and ball are in contact for a tiny fraction of a second.

Or, they could swing for the fences. In a home run swing, the batter follows through, keeping the bat in contact with the ball for as long as possible to drive it over the wall.

In both cases, the bat applies a force to the ball. But the outcome—a gentle bunt versus a 400-foot home run—is dramatically different. Why? It's not just about the force, but also about how long that force is applied. This combination of force and time is what we call impulse, and it's the key to understanding any change in motion, from a car crash to a rocket launch. Today, we'll unpack this powerful idea.

Diagram

Force vs. Time Graph: Calculating Impulse A Force versus Time graph shows a triangular force pulse. The force starts at zero, rises linearly to 100 Newtons at 0.02 seconds, and returns to zero at 0.04 seconds. The shaded area under the triangle is labeled as Impulse and is calculated to be 2.0 N·s, which equals the change in momentum. Time, t (s) Force, F (N) 0 0.01 0.02 0.03 0.04 0.05 0 50 100 150 Calculating Impulse (J) Impulse = Area under F-t curve J = Area of Triangle = ½ × base × height J = ½ × (0.04 s) × (100 N) = 2.0 N·s By theorem J = Δp, so Δp = 2.0 kg·m/s Area = Impulse, J
A Force versus Time graph shows a triangular force pulse. The force starts at zero, rises to 100 Newtons at 0.02 seconds, and returns to zero at 0.04 seconds. The shaded area under the triangle is labeled as Impulse and calculated to be 2.0 Newton-seconds.

Concept map

flowchart TD
    A[Net Force, F_net] -->|Applied over time, Δt| B(Impulse, J);
    B -->|Causes| C(Change in Momentum, Δp);
    C -->|Is defined as| D(p_final - p_initial);
    C -->|Results in| E(Change in Velocity, Δv);
    subgraph "Impulse-Momentum Theorem"
        B -.->|J = Δp| C;
    end
    subgraph "Graphical Analysis"
        F[F-t Graph] -- Area --> B;
        G[p-t Graph] -- Slope --> A;
    end

Core explanation

Hello everyone! Let's dive into one of the most fundamental relationships in physics. It's a concept that elegantly connects forces with the changes they produce.

From Newton's Second Law to Momentum

You're already good friends with Newton's Second Law: F⃗_net = ma⃗. It's been your go-to for solving problems involving forces and acceleration. But we can write it in a way that gives us an even deeper insight.

Remember the definition of acceleration: it's the rate of change of velocity, or a⃗ = Δv⃗ / Δt.

Let's substitute that into Newton's Second Law:

F⃗_net = m(Δv⃗ / Δt)

Now, let's do a little algebra and multiply both sides by Δt:

F⃗_net * Δt = mΔv⃗

Look at the right side of that equation, mΔv⃗. Since momentum (p⃗) is mv⃗, then mΔv⃗ is just the change in momentum, Δp⃗.

So, we can rewrite our equation as:

F⃗_net * Δt = Δp⃗

This equation is a powerhouse. It tells us that a net force applied for a certain amount of time causes a change in an object's momentum.

Defining Impulse

Physicists give a special name to that left side of the equation, F⃗_net * Δt. We call it impulse, and we give it the symbol J⃗.

J⃗ = F⃗_avg * Δt

Impulse is the product of the average force exerted on an object and the time interval over which that force acts.

This is where some students get a little confused. They ask, "Why 'average' force?" Think back to the baseball bat. The force the bat exerts on the ball isn't constant. It ramps up from zero to some huge peak value and then drops back to zero, all in a few milliseconds. It's messy. Using the average force lets us simplify the situation and still get the correct result.

Since impulse equals F⃗_avg * Δt, its units are Newton-seconds (N·s).

And because force is a vector, impulse is also a vector. Its direction is the same as the direction of the average net force that causes it.

The Impulse-Momentum Theorem

Now we can put it all together. We established that F⃗_net * Δt = Δp⃗, and we just defined J⃗ = F⃗_avg * Δt. Therefore:

J⃗ = Δp⃗

This simple, beautiful equation is called the Impulse-Momentum Theorem. It states that the impulse delivered to an object is exactly equal to the change in that object's momentum.

Think of it like this: Momentum is a state of being, like the amount of money in your bank account. Impulse is the transaction—the deposit or withdrawal—that changes that state. You can't change your account balance without a transaction; you can't change an object's momentum without an impulse.

This theorem is actually a more fundamental statement than F=ma. Why? Because it works even when mass changes, like in a rocket burning fuel. For AP Physics 1, we'll stick to constant mass, but it's cool to know! In fact, we can get F=ma right back from the theorem:

F⃗_net = Δp⃗ / Δt = m(Δv⃗ / Δt) = ma⃗

Finding Impulse from a Graph

What if the force isn't constant or we don't know the average force? This is where graphs become our best friend.

1. Force vs. Time Graph

Imagine you have a graph that plots the net force on an object versus time.

Force vs. Time Graph

The equation for impulse is J = F_avg * Δt. If the force were constant, this would just be the area of a rectangle on the graph (height F times width Δt). It turns out this is true even for a changing force!

The impulse J is the area under the curve of a Force vs. Time graph.

For the triangular force pulse shown in the visual, we can find the impulse by calculating the area of the triangle:

Impulse = Area = ½ × base × height J = ½ × (0.04 s) × (100 N) = 2.0 N·s

So, the total impulse delivered to the object is 2.0 N·s. By the Impulse-Momentum Theorem, this means the object's change in momentum, Δp, is 2.0 kg·m/s.

2. Momentum vs. Time Graph

We can also flip our perspective. Remember our starting point:

F⃗_net = Δp⃗ / Δt

Look at that right side. Δp / Δt is "rise over run" for a momentum vs. time graph.

The net force F_net is the slope of a Momentum vs. Time graph.

This is a powerful pair of ideas to remember for the exam:

  • F-t graph
    Area = Impulse (which equals Δp)
  • p-t graph
    Slope = Net Force

Mastering this graphical relationship will serve you well.

Worked examples

Let's make this concrete with a couple of examples.

Example 1

Kicking a Soccer Ball

Aaliyah kicks a 0.45 kg soccer ball that is initially at rest. Her foot is in contact with the ball for 0.050 seconds, and the ball leaves her foot with a velocity of 25 m/s. What was the average force she exerted on the ball?

1. Identify your goal. We need to find the average force, F_avg. The problem gives us mass, a time interval, and initial/final velocities. This smells like an impulse-momentum problem.

2. Find the change in momentum (Δp). The formula for change in momentum is Δp = p_f - p_i.

  • Initial momentum: p_i = mv_i = (0.45 kg)(0 m/s) = 0 kg·m/s.
  • Final momentum: p_f = mv_f = (0.45 kg)(25 m/s) = 11.25 kg·m/s.
  • Change in momentum: Δp = 11.25 - 0 = 11.25 kg·m/s.

3. Use the Impulse-Momentum Theorem. The theorem states F_avg * Δt = Δp. We know Δt and we just found Δp. Now we can solve for the force.

F_avg * (0.050 s) = 11.25 kg·m/s

F_avg = (11.25 kg·m/s) / (0.050 s)

F_avg = 225 N

So, the average force Aaliyah exerted on the ball was 225 Newtons. That's about 50 pounds of force!

Example 2

Using a Force-Time Graph

A 0.50 kg cart is initially at rest on a frictionless track. It is then pushed with a force that varies with time, as shown in the graph below. What is the cart's final speed at t = 0.04 s?

(Imagine the triangular Force vs. Time graph from the visual plan is shown here, peaking at 100 N at 0.02 s and ending at 0.04 s).

1. Realize what the graph gives you. This is a Force vs. Time graph. The key piece of information we can get from this graph is the impulse, which is the area under the curve.

2. Calculate the impulse (J). The shape is a triangle. The area of a triangle is ½ * base * height.

  • Base: The force is applied from t=0 to t=0.04 s, so the base is 0.04 s.
  • Height: The peak force is 100 N.

J = Area = ½ * (0.04 s) * (100 N) = 2.0 N·s

3. Apply the Impulse-Momentum Theorem. The theorem tells us J = Δp. So, the change in the cart's momentum is 2.0 N·s. Since 1 N = 1 kg·m/s², a N·s is the same as a kg·m/s.

Δp = 2.0 kg·m/s

4. Solve for the final velocity. We know Δp = p_f - p_i = mv_f - mv_i. The cart starts from rest, so v_i = 0 and p_i = 0. This simplifies the equation to Δp = mv_f.

2.0 kg·m/s = (0.50 kg) * v_f

v_f = (2.0 kg·m/s) / (0.50 kg)

v_f = 4.0 m/s

The final speed of the cart is 4.0 m/s.

This is a classic AP problem. They love asking you to find impulse from the area of a graph and then use it to find a final velocity. Don't let the graph intimidate you; just remember that "area under the F-t curve" means impulse.

Try it yourself

Ready to try one on your own?

Problem 1: A 150 g (that's 0.150 kg!) baseball is thrown with a speed of 40 m/s. It is hit straight back at the pitcher with a speed of 50 m/s. The bat was in contact with the ball for 5.0 milliseconds (0.005 s). What was the average force exerted by the bat on the ball?

Hint 1: Be careful with your signs! Define a direction as positive (e.g., the direction of the pitched ball). The final velocity will be negative. Hint 2: Remember that Δp = p_final - p_initial.

Problem 2: A box is pulled across the floor. The graph of the net force on the box vs. time is a rectangle, with a constant force of 10 N applied from t=2s to t=5s. If the box started from rest, what is its change in momentum?

Hint: What's the area of a rectangle?

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