# 48 6.13 The Impulse-Momentum Theorem

When thinking about how to reduce forces during collisions we intuitively know that increasing the duration of the collision is helpful. The combination of the force and collision duration is known as the impulse. The impulse can be calculated by multiplying the average net force (Fave) by the duration of the collision (Δt). (Alternatively, the impulse is equal to the area underneath the force vs. time curve for the collision such as those in the previous example). The impulse-momentum theorem states that  the impulse applied to an object will be equal to the change in its momentum.

$\boldsymbol{\Delta{\vec{\textbf{t}}}\textbf{F}}=\boldsymbol{m({v_f}) - m( {v_i})}$

Notice that we have calculated the change in momentum as the  initial momentum (mivi) subtracted from the final momentum (mfvf). If the mass of the object doesn’t change during the collision, then the initial and final mass are the same. In this case we call it m and factor it out on the right side of the equation:

$\boldsymbol{\Delta{\vec{\textbf{t}}}\textbf{F}}=\boldsymbol{m({v_f} - {v_i})}$

Now we see that the impulse-momentum theorem shows us how a small net force applied over a long time can be used to produce the same velocity change as a large net force applied over a short time.

### Everyday Example: Landing

A person jumping from a height of 5 m, or about 20 ft, hits the ground with a speed of nearly 10 m/s, or about 22 mph (we’ll learn how to figure that out later). Let’s calculate the average force applied to a 100 kg person during such a landing if the collision with the ground lasts 1/10 of a second. We start with the impulse-momentum theorem.

$\boldsymbol{\Delta{\vec{\textbf{t}}}\textbf{F}}=\boldsymbol{m({v_f} - {v_i})}$

We want force, so let’s divide over the collision duration:

$\boldsymbol{{F}}=\boldsymbol{(m({v_f} - {v_i}))/\Delta{\vec{\textbf{t}}}}$

Remembering that direction is important when working with forces and velocities, we need to define some directions. Let’s make downward negative so the initial velocity is -10 m/s. The final velocity is 0 m/s because the person comes to rest on the ground during landing. The stated collision duration was 0.1 s, so we are ready to calculate the average net force:

$\boldsymbol{{F}}=\boldsymbol{(100kg({0 m/s} - {-10 m/s}))/{0.1s} = 10, 000N}$

We see that the net force is positive, meaning that it points upward because we chose downward as the negative direction. This makes sense because the ground pushes up on the person to provide the impulse to stop the persons downward motion.

Finally, we need to remember that we have calculated the average net force, which how much the forces are out of balance. This person has a weight of about 1,000 N  (100 kg x 9.8 m/s/s = 1000 N). Weight acts downward, so to get the required 10,000 of net force upward there must actually be a 11,000 applied upward on their feet, with 1000 N of that being cancelled out by their weight.

Spreading the force out over a longer time would reduce the average force (and peak force) applied to the person. For example, the the collision were made to last 5/10 of a second instead of 1/10 of a second, the net force would be five times smaller:

$\boldsymbol{{F}}=\boldsymbol{(100kg({0 m/s} - {-10 m/s}))/{0.5s} = 5, 000N}$

And adding the 1000 N body weight to get the total force on the feet we get 6,000 N.

The people in this video are well practiced at techniques for reducing forces by extending impact time.

### Reinforcement Exercises: Fall time

Apply the impulse-momentum theorem to calculate the fall time for the person who fell from the 5 m height in the previous example. [Hint: If we ignore air resistance, then the only force on them during the fall is their weight, so that is the net force. You already know the initial velocity at the start of the fall is zero, and the final velocity was given to be 10 m/s.]

It’s important to recognize that we have been applying the impulse-momentum theorem to only one object involved in the collision. We know from the Principle of Momentum Conservation that the total combined momentum change of all objects involved in a collision is zero, so applying the impulse-momentum theorem to all of the objects would just tell us that the total net force on ALL objects during the collision is zero.