Visighta

Momentum, Impulse, and Conservation Laws

8.2 Momentum Change from Force over Time

The Egg, the Concrete, and the Pillow

Hold an egg at shoulder height and drop it onto a concrete floor. It shatters. Now drop an identical egg from the same height onto a thick pillow. It survives.

In both cases, the egg starts with the same speed just before impact (same height, same $g$). In both cases, the egg ends at rest. The momentum change is identical --- the egg goes from $m\vec{v}$ to zero. The impulse delivered to the egg is exactly the same.

And yet one egg breaks and the other does not.

The difference is not how much force is delivered. It is how the force is delivered --- spread across time. The concrete stops the egg in a fraction of a millisecond. The pillow stops it over tens of milliseconds. The total impulse is the same, but the force profile is completely different. The concrete delivers a single enormous spike. The pillow delivers a gentle, drawn-out push.

This distinction --- between the total impulse and the shape of the force that delivers it --- is what Section 8.2 is about. It is the reason airbags exist, the reason boxers roll with punches, and the reason you bend your knees when you land from a jump.

Before you read on: A car airbag increases the stopping time of an occupant's head from 0.02 s (hitting the dashboard) to 0.2 s (inflating into the bag). The momentum change is the same in both cases.

By what factor does the average force on the occupant decrease?

Commit to a number before continuing.

Quick Recall: Impulse and Momentum

Before we go further, make sure the key results from Section 8.1 are fresh.

Recall prompt: What is the definition of momentum? What is the definition of impulse? How are they related by the impulse-momentum theorem?

If these feel fuzzy, revisit Section 8.1. Everything here builds on the equation $\vec{J} = \Delta\vec{p}$.

Exploration: Shaping the Force Profile

[Interactive: Force Profile Explorer. Two force-versus-time graphs are displayed side by side. On the left, a tall, narrow triangular pulse represents the "concrete" scenario --- a large peak force over a very short time. On the right, a shorter, wider triangular pulse represents the "pillow" scenario --- a smaller peak force spread over a longer time. Both pulses have the same area (same impulse).

A slider labeled "Interaction time" controls the width of the right-hand pulse. As the student widens the pulse, the peak force automatically decreases to keep the area constant. A numerical readout displays the peak force, the interaction time, and the impulse (area) for each profile. A bar chart below compares the peak forces of the two profiles.

Guided prompts: - "Set the interaction time to twice the original. What happened to the peak force?" - "Now set it to five times the original. What is the peak force now compared to the narrow spike?" - "Keep the impulse fixed. What happens to the peak force as you make the pulse wider and wider?" - "Can you find a pulse width that makes the peak force less than 100 N? What about less than 50 N?"]

Here is the key observation. When you widen the force pulse while keeping the area constant, the peak force drops. Double the time, and the peak force halves. Triple the time, and the peak force drops to one-third. The product of peak force and time stays roughly constant (exactly constant for rectangular pulses, approximately so for triangular or smooth ones).

This is not a coincidence. It is a direct consequence of the impulse-momentum theorem.

Concept Reveal: The Impulse-Time Tradeoff

From Section 8.1, you know the impulse-momentum theorem:

$$\vec{J} = \int_{t_i}^{t_f} \vec{F}(t) \, dt = \Delta\vec{p}$$

The left side is the impulse --- the integral of force over time. Graphically, it is the area under the $F(t)$ curve. The right side is the change in momentum. The theorem says these are equal.

Now here is the crucial point. The momentum change $\Delta\vec{p}$ depends only on the initial and final states of the object --- its mass and how its velocity changes. For the egg, $\Delta\vec{p}$ is fixed: the egg goes from moving to stopped, regardless of whether it hits concrete or a pillow.

Since $\Delta\vec{p}$ is fixed, the impulse $\vec{J}$ must also be fixed. That means the area under the $F(t)$ curve is the same in both scenarios. But the shape of the curve can be anything --- as long as the area is right.

A tall, narrow curve (large force, short time) and a short, wide curve (small force, long time) can have the same area. They deliver the same impulse. They produce the same momentum change. But they subject the object to very different peak forces.

For a constant force, the integral simplifies:

$$J = F_{\text{avg}} \cdot \Delta t = \Delta p$$

This gives us the average-force form:

$$F_{\text{avg}} = \frac{\Delta p}{\Delta t}$$

This equation captures the tradeoff directly. For a fixed $\Delta p$, increasing $\Delta t$ decreases $F_{\text{avg}}$ in exact proportion. Ten times the stopping time means one-tenth the average force.

Returning to the Prediction

You were asked: if an airbag increases the stopping time from 0.02 s to 0.2 s, by what factor does the average force decrease?

The stopping time increases by a factor of $0.2 / 0.02 = 10$. Since $F_{\text{avg}} = \Delta p / \Delta t$ and $\Delta p$ is the same in both cases:

$$\frac{F_{\text{with airbag}}}{F_{\text{without}}} = \frac{\Delta p / 0.2}{\Delta p / 0.02} = \frac{0.02}{0.2} = \frac{1}{10}$$

The average force decreases by a factor of 10. The airbag does not reduce the impulse --- it cannot, because the occupant must go from moving to stopped regardless. What the airbag does is spread that impulse over ten times as much time, reducing the force to one-tenth of what the dashboard would deliver.

This is a factor-of-ten reduction in force from a simple engineering choice: make the collision last longer. No new physics. No exotic materials. Just more time.

[Video: Side-by-side slow-motion footage of a crash test dummy hitting a dashboard versus an airbag. Force-versus-time graphs overlay each clip. The dashboard collision produces a sharp, enormous spike lasting about 20 ms. The airbag collision produces a broad, rounded curve lasting about 200 ms. Both curves have the same shaded area. A narrator says: "Same impulse. Same momentum change. But the peak force with the airbag is an order of magnitude smaller --- and that's the difference between a fatal injury and a survivable one."]

Graphical Interpretation: Area Under the $F(t)$ Curve

This section uses the same area-under-the-curve reasoning you developed in Section 7.2, where work was the area under the $F(x)$ graph. The mathematical structure is identical --- only the axes have changed.

Section 7.2 Section 8.2
Force $F(x)$ versus position Force $F(t)$ versus time
Area under curve $= \int F \, dx =$ work Area under curve $= \int F \, dt =$ impulse
Determines energy transfer Determines momentum change
Shape of $F(x)$ affects how energy is delivered Shape of $F(t)$ affects how momentum is delivered

In Section 7.2, a constant force gave a rectangular area, and a varying force gave a more complex shape that required integration. The same applies here. If $F(t)$ is constant during the interaction, the impulse is simply $F \cdot \Delta t$ (a rectangle). If $F(t)$ varies --- rising sharply, peaking, then falling --- you need the full integral.

But here is a physical difference worth noting. In Section 7.2, the shape of $F(x)$ told you how efficiently energy was transferred at each position. Here, the shape of $F(t)$ tells you something more visceral: it tells you the peak force experienced during the interaction. And peak force is what breaks bones, cracks eggshells, and damages structures. Engineers do not just care about the total impulse. They care intensely about the shape of the force profile, because the shape determines whether something survives.

Worked Example: Reading Impulse from a Force-Time Graph

Problem: During a collision, the force on a 0.5 kg ball varies with time as follows:

  • From $t = 0$ to $t = 0.01$ s: $F$ increases linearly from 0 to 600 N.
  • From $t = 0.01$ to $t = 0.03$ s: $F$ decreases linearly from 600 N to 0.

Find the impulse delivered to the ball and the ball's change in velocity.

Solution:

The $F(t)$ graph is a triangle with base $\Delta t = 0.03$ s and height $F_{\max} = 600$ N.

$$J = \text{area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}(0.03)(600) = 9.0 \text{ N} \cdot \text{s}$$

By the impulse-momentum theorem:

$$J = \Delta p = m \Delta v$$

$$\Delta v = \frac{J}{m} = \frac{9.0}{0.5} = 18 \text{ m/s}$$

Sanity check: The average force over the 0.03 s interval is $F_{\text{avg}} = J / \Delta t = 9.0 / 0.03 = 300$ N. This is half the peak force, which makes sense for a triangular pulse --- just as a triangle has half the area of its enclosing rectangle. The same geometric reasoning from Section 7.2 appears again.

Faded Example: Comparing Two Force Profiles

Two different force profiles act on the same 2 kg object, initially at rest.

  • Profile A: A constant force of 400 N for 0.005 s.
  • Profile B: A constant force of $F_B$ for 0.020 s.

Both profiles deliver the same impulse.

Step 1: Find the impulse from Profile A.

$$J_A = F_A \cdot \Delta t_A = 400 \times 0.005 = 2.0 \text{ N} \cdot \text{s}$$

Step 2: Find the force $F_B$ in Profile B.

Your turn: If Profile B delivers the same impulse over 0.020 s, what is $F_B$?

Check your answer $$J_B = J_A = 2.0 \text{ N} \cdot \text{s}$$ $$F_B = \frac{J_B}{\Delta t_B} = \frac{2.0}{0.020} = 100 \text{ N}$$ Profile B has one-quarter the force of Profile A, because the interaction time is four times longer.

Step 3: Find the final velocity of the object.

Your turn: Both profiles deliver the same impulse to the same 2 kg object starting from rest. What is the final velocity in each case?

Check your answer $$\Delta v = \frac{J}{m} = \frac{2.0}{2} = 1.0 \text{ m/s}$$ The final velocity is 1.0 m/s in both cases. The momentum change is identical, so the outcome is identical --- even though one profile used four times the force. The object does not "know" or "care" about the force profile. It only responds to the total impulse.

Why the Shape Matters: Peak Force and Damage

If the momentum change is the same regardless of the force profile, why should anyone care about the shape?

Because the object's structural integrity depends on the peak force, not the total impulse. The egg does not break because it received too much impulse. It breaks because the force at some instant exceeded the strength of the shell. A human skull does not fracture because the momentum change was too large. It fractures because the force at some instant exceeded the bone's tolerance.

This is why engineers care about the shape of $F(t)$:

  • Same impulse, different peak forces. A 10 N$\cdot$s impulse delivered in 0.001 s requires an average force of 10,000 N. The same impulse delivered in 1 s requires only 10 N. The momentum change is identical. The damage potential is vastly different.

  • The design principle. To protect something fragile, extend the interaction time. Every millisecond you add to the collision reduces the peak force. This is not about reducing the impulse (you cannot --- the momentum must change by a fixed amount). It is about reshaping the force profile from a tall, dangerous spike into a low, survivable spread.

This principle underlies some of the most important safety engineering of the past century.

Practice

Layer 1: Concrete

A 0.4 kg ball moving at 15 m/s strikes a wall and bounces back at 10 m/s. The contact time with the wall is 0.008 s.

(a) Find the impulse delivered to the ball.

(b) Find the average force exerted by the wall on the ball.

Check your answer **(a)** Take the initial direction of motion as positive. The initial momentum is $p_i = 0.4 \times 15 = 6.0$ kg$\cdot$m/s. After bouncing back, $p_f = 0.4 \times (-10) = -4.0$ kg$\cdot$m/s. $$J = \Delta p = p_f - p_i = -4.0 - 6.0 = -10.0 \text{ N} \cdot \text{s}$$ The impulse is 10.0 N$\cdot$s directed opposite to the initial motion (toward the wall's push on the ball). Note that the ball *reverses direction*, so the momentum change is larger than it would be if the ball simply stopped. Stopping would give $|\Delta p| = 6.0$ N$\cdot$s. Reversing adds the outgoing momentum: $|\Delta p| = 6.0 + 4.0 = 10.0$ N$\cdot$s. **(b)** $$F_{\text{avg}} = \frac{|J|}{\Delta t} = \frac{10.0}{0.008} = 1250 \text{ N}$$ The wall exerts an average force of 1250 N on the ball during the 8 ms contact. This is far larger than the ball's weight ($mg = 3.9$ N), which is why we can safely ignore gravity during the collision.

Layer 2: Pattern

Design two different $F(t)$ profiles (you choose the shape) that deliver an impulse of exactly 6.0 N$\cdot$s, but with different peak forces.

(a) Design a profile with a peak force of 1200 N. What is the minimum interaction time?

(b) Design a profile with a peak force of no more than 200 N. What is the minimum interaction time?

(c) What is the ratio of the peak forces? What is the ratio of the interaction times?

Check your answer The simplest shape is a rectangle (constant force), which minimizes the interaction time for a given peak force. **(a)** Rectangular pulse: $F = 1200$ N, $\Delta t = J / F = 6.0 / 1200 = 0.005$ s $= 5$ ms. **(b)** Rectangular pulse: $F = 200$ N, $\Delta t = 6.0 / 200 = 0.03$ s $= 30$ ms. **(c)** Peak force ratio: $1200 / 200 = 6$. Interaction time ratio: $5 / 30 = 1/6$. The ratios are reciprocals. When the impulse is fixed and the profile is rectangular, reducing the peak force by a factor of $n$ requires increasing the time by a factor of $n$. This is the impulse-time tradeoff in its simplest form. If you chose a triangular profile instead, the interaction time would be longer (by a factor of 2 for the same peak force, because a triangle has half the area of a rectangle). The tradeoff still holds, but the specific numbers differ with the shape.

Layer 3: Structure

Why do engineers care about the shape of the force profile, not just the total impulse?

Check your answer Because damage, injury, and structural failure depend on the **peak force** (or more precisely, the peak stress), not on the total impulse. The impulse is fixed by the momentum change: $J = \Delta p$. An engineer cannot change this. If a car occupant must go from 30 m/s to 0 m/s, the impulse is $m \times 30$ regardless of how the collision unfolds. What the engineer *can* control is how that impulse is distributed in time. A tall, narrow $F(t)$ profile delivers the required impulse with a dangerously high peak force. A broad, flat profile delivers the same impulse with a much lower peak force. The total impulse is the same --- the area under the curve is the same --- but the peak height is different. This is why engineers design crumple zones (the car's front end collapses progressively, extending the collision time), airbags (the occupant decelerates into a soft, inflating cushion instead of a rigid dashboard), and helmets (foam liner crushes gradually, spreading the impact over more time). In every case, the strategy is the same: *extend the interaction time to reduce the peak force*, without changing the impulse. The shape of the force profile is the difference between a survivable collision and a fatal one.

Layer 4: Transfer

Consider these real-world scenarios. For each one, explain how the impulse-time tradeoff is being used, and identify what is being protected.

(a) A baseball outfielder catches a fly ball by letting the glove move backward with the ball ("soft hands") rather than holding the glove rigidly.

(b) A car's crumple zone collapses in a front-end collision, absorbing about 0.5 meters of crush distance over roughly 0.1 seconds.

(c) A parkour athlete rolls upon landing from a high jump instead of landing stiff-legged.

Check your answer **(a)** The ball must go from its incoming speed to rest --- the impulse is fixed by $\Delta p$. By pulling the glove backward, the outfielder extends the stopping time from a few milliseconds (rigid glove) to perhaps 50--100 milliseconds. The peak force on the hand and ball drops accordingly. Without soft hands, the sudden stop can sting painfully or even injure the hand. With soft hands, the same momentum change happens with a gentler force spread over a longer time. **(b)** The crumple zone extends the collision time from about 5 ms (a perfectly rigid car hitting a wall) to about 100 ms. The momentum change is the same --- the car goes from its initial speed to near-zero. But the extended time reduces the average force on the passenger compartment by roughly a factor of 20. The crumple zone sacrifices the car's structure (which can be replaced) to protect the occupants (who cannot). The crush distance also allows the use of the work-energy theorem (Chapter 7): the crumple zone absorbs kinetic energy through deformation work. **(c)** Landing stiff-legged concentrates all the deceleration into the brief instant when the feet hit the ground --- a very short $\Delta t$ and a very high force on the ankles, knees, and spine. Rolling converts the downward motion into rotational and horizontal motion over a much longer time (the duration of the roll). The impulse is the same, but the peak force on any single joint is dramatically reduced. Parkour athletes are using the impulse-time tradeoff with their bodies as the engineering medium. In every case, the strategy is identical: the momentum change is non-negotiable, so extend the interaction time to reduce the peak force.

Reflection

How does extending the interaction time protect you in a collision?

Think about what you learned in this section. The impulse --- the total "push" integrated over time --- is dictated by the momentum change. You cannot negotiate with $\Delta\vec{p}$. If you must go from 30 m/s to zero, the impulse is $m \times 30$ N$\cdot$s no matter what.

But you can choose how that impulse is distributed in time. A short interaction time forces the impulse into a tall, narrow spike --- a dangerously high force crammed into a tiny interval. A long interaction time lets the impulse spread into a low, wide curve --- a manageable force sustained over a longer period.

The area under both curves is the same. The height is not. And it is the height --- the peak force --- that breaks things.

Every safety device you encounter, from helmets to airbags to padded playground surfaces, is an engineering solution to a single physics principle: $F_{\text{avg}} = \Delta p / \Delta t$. Make $\Delta t$ bigger, and $F_{\text{avg}}$ gets smaller. That is the entire idea.

Looking Ahead

You now understand how the shape of a force profile determines the peak force during an interaction, even when the total impulse is fixed. This gives you a powerful way to reason about collisions, impacts, and safety engineering.

But so far, we have focused on a single object receiving an impulse. What happens when two objects interact with each other --- when the force on one is paired with an equal and opposite force on the other? In the next section, you will discover that Newton's third law, applied to the impulse-momentum theorem, leads to one of the most powerful results in all of physics: conservation of momentum for isolated systems. When no external forces act on a system, the total momentum does not change --- no matter how violent or complicated the internal interaction. This conservation law will let you predict the outcomes of collisions and explosions without knowing anything about the forces involved.