momentum

The concept of momentum developed slowly. Aristotle (384-322 BCE) described motion as something that needed a continuing cause, which seems reasonable when you watch carts, rocks, and people eventually stop.

In 1020 CE, Ibn Sina said that a thrown object carries an impressed tendency of motion, with outside effects like air resistance gradually reducing that motion. In the early 1600s, Galileo argued that objects can keep moving without a constant push, especially when friction is reduced.

In 1687, Newton put these ideas into a mathematical theory in the Principia. Newton called momentum the quantity of motion: how much matter an object has multiplied by how fast it is moving.


This is a simulation of a "Newton's cradle". A real Newton's cradle is made of metal balls suspended by two strings. Click and drag a ball to fling it.




When two bodies collide, momentum is transferred. Momentum is the velocity of a body multiplied by its mass. A small force can quickly stop an object with low momentum, but a large or prolonged force is required to stop an object with high momentum.

m

$$p = mv$$

\(p\) = momentum [kg m/s] vector
\(m\) = mass [kg]
\(v\) = velocity [m/s] vector
Example: What is the momentum of a bowling ball (in kg m/s)?
A typical mass is 5 kg and a typical and velocity is around 18 mph.
solution $$ 18 \left( \mathrm{ \frac{\color{red}{mile}}{\color{Teal}{hour}}} \right)\left(\frac{1609\,\mathrm{m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{Teal}{ \mathrm{hour} }}{3600\,\mathrm{s}}\right) = 8.0 \mathrm{\tfrac{m}{s}} $$
$$p=mv$$ $$p=(5.0 \, \mathrm{kg})(8.0 \, \mathrm{\tfrac{m}{s}})$$ $$p=40 \,\mathrm{kg\tfrac{m}{s}}$$
Example: Which has more momentum? A 10 kg bicycle moving at 8 m/s or a 15 kg bicycle moving at 6 m/s?
solution
$$p_1=mv$$ $$p_1=(10)(8)$$ $$p_1=80 \,\mathrm{kg\tfrac{m}{s}}$$
$$p_2=mv$$ $$p_2=(15)(6)$$ $$p_2=90 \,\mathrm{kg\tfrac{m}{s}}$$

The 15 kg bike moving at 6 m/s has more momentum.

Newton's Law and Momentum

Newton's second law can be written as a change in momentum with time. Force changes an object's momentum.

derivation

If the mass stays constant, the change in momentum comes from the change in velocity.

$$\sum F = \frac{\Delta p}{\Delta t}$$ $$\Delta p = m\Delta v$$ $$\sum F = \frac{m\Delta v}{\Delta t}$$ $$a = \frac{\Delta v}{\Delta t}$$ $$\sum F = ma$$

So force equals mass times acceleration is a special case of the momentum version of Newton's second law.

$$\sum F = \frac{\Delta p}{\Delta t}$$

\(\sum F\) = net force [N]
\(\Delta p\) = change in momentum [kg m/s]
\(\Delta t\) = time interval [s]
Example: A 1200 kg car speeds up from 5 m/s to 25 m/s in 8.0 s. What is the car's change in momentum, and what average net force acted on it?
solution $$\Delta p=m\Delta v$$ $$\Delta p=m(v_f-v_i)$$ $$\Delta p=(1200)(25-5)$$ $$\Delta p=24000\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\sum F=\frac{24000}{8.0}$$ $$\sum F=3000\,\mathrm{N}$$

The car's momentum increases by 24000 kg m/s, so the average net force is 3000 N forward.

Example: A 0.145 kg baseball is moving 40 m/s toward a bat. After the hit, it moves 30 m/s in the opposite direction. If the bat touches the ball for 0.010 s, what is the average net force on the ball?
solution

Let the ball's original direction be positive. The final velocity is negative because the ball reverses direction.

$$\Delta p=m(v_f-v_i)$$ $$\Delta p=(0.145)(-30-40)$$ $$\Delta p=-10.15\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\sum F=\frac{-10.15}{0.010}$$ $$\sum F=-1015\,\mathrm{N}$$

The negative sign means the force is opposite the ball's original motion.

Impulse

An impulse is defined as a force applied over a period of time. Applying a larger force or longer lasting force produces a larger impulse. A large impulse produces a large change in momentum.

Examples of impulse:
cars speeding up
cars crashing
rockets accelerating
kicking a soccer ball
punching
jumping
derivation of impulse

Newton's 2nd law (F=ma) was originally written in terms of momentum, not mass and acceleration. We can rearrange F=ma to see it in terms of momentum.

$$a = \color{green}\frac{\Delta v}{\Delta t}$$
$$F = ma$$ $$F = m \color{green}\frac{\Delta v}{\Delta t}$$ $$F \Delta t = m \Delta v$$
$$F \Delta t = m(v_f - v_i)$$ $$F \Delta t = mv_f - mv_i$$ $$F \Delta t = p_f - p_i$$ $$F \Delta t = \Delta p$$

$$J = \Delta p$$ $$F \Delta t = \Delta p $$

\(J\) = impulse [Ns, kg m/s] vector
\(F\) = force [N, kg m/s²] vector
\(\Delta t\) = time period [s]
\(\Delta p\) = change in momentum [kg m/s] vector

When using impulse in solving problems you might want to unpack momentum into mass and velocity.

$$F \Delta t = mv - mu$$
Example: Which will produce a greater change in velocity: doubling the force or doubling the time the force is applied?
solution

Doubling force or time applied will produce the same increase in velocity. This is because F and Δt are in the same position in the equation.

$$\Delta v = \frac{\color{blue}F \Delta t}{m} $$
Example: A medicine ball is hurtling towards you at 15 m/s. You weakly try to stop it by applying a 100 N force for 0.1 s. You don't stop the ball, but it slows to 14 m/s. Calculate the mass of the ball.
solution $$F \Delta t = m\Delta v$$ $$\frac{F \Delta t}{\Delta v} = m$$ $$\frac{(-100)(0.1)}{14-15} = m$$ $$\frac{(-100)(0.1)}{-1} = m$$ $$10 \, \mathrm{kg} = m$$

Electric cars are capable of high accelerations because electric motors provide instant torque without the need to shift gears.

Example: The Porsche Taycan Turbo GT (2025) has a mass of around 2250 kg. It has one of the fastest 0 to 60 miles/hour accelerations at 1.898 seconds. Convert the miles/hour into m/s, and then find the force produced by the car.
solution $$ 60 \left( \mathrm{ \frac{\color{red}{mile}}{\color{blue}{hour}}} \right)\left(\frac{1609\,\mathrm{ m}}{1 \,\color{red}{\mathrm{mile} }}\right)\left(\frac{1\, \color{blue}{ \mathrm{hour} }}{3600\, \mathrm{s} }\right) = 26.8 \mathrm{\tfrac{m}{s}} $$
$$F \Delta t = m\Delta v$$ $$F = \frac{m\Delta v}{\Delta t}$$ $$F = \frac{(2250)(26.8)}{1.898}$$ $$F = \frac{60300}{1.898}$$ $$F = 31770 \, \mathrm{N}$$

How does the 0 to 60 miles/hour acceleration compare to the acceleration of gravity?
solution $$F = 21536 N$$ $$F = ma$$ $$\frac{F}{m}=a$$ $$\frac{31770}{2250}=a$$ $$a = 14.12 \,\mathrm{ \tfrac{m}{s^2} }$$

It's almost one and half times the acceleration of gravity. How would that feel?

$$ g = 9.81 \, \mathrm{\tfrac{m}{s^2}} $$

We can rearrange our impulse equation to show the relationship between velocity and time.

$$F \Delta t = m\Delta v$$ $$\Delta v = \frac{F}{m} \Delta t$$
force = N
mass = kg
Example: The slope of this graph is acceleration, and a high acceleration can be dangerous. Use the graph to identify what types of vehicles will be safer in collisions?
solution

A vehicle with more mass will experience less acceleration. A large SUV is safer in a collision then a motorcycle. On the other hand, a larger vehicle is safer for you, but more dangerous for whatever you crash into.

Another way to improve safety is to reduce the force. This can be done by increasing the total time for the collision. Like hitting an airbag instead of a steering wheel. Modern cars also lengthen the time of a collision with crumple zones.

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Example: A 6000 g backpack slides across the floor at 3 m/s after being nudged in a hallway. What is its momentum?
solution $$6000\,\mathrm{g}=6\,\mathrm{kg}$$ $$p=mv$$ $$p=(6\,\mathrm{kg})(3\,\mathrm{m/s})$$ $$p=18\,\mathrm{kg\,m/s}$$
Example: A 0.060 kg tennis ball is moving left at 25 m/s just after leaving a racket. The court is 24 m long, but the momentum only depends on mass and velocity. What is the ball's momentum?
solution

Left is the negative direction, so the velocity is -25 m/s.

$$p=mv$$ $$p=(0.060\,\mathrm{kg})(-25\,\mathrm{m/s})$$ $$p=-1.5\,\mathrm{kg\,m/s}$$

The negative sign tells us the momentum points left.

Example: Which has more momentum: a 45 kg skateboarder moving at 4 m/s or a 12 kg scooter moving at 14 m/s?
solution $$p_1=mv$$ $$p_1=(45)(4)$$ $$p_1=180\,\mathrm{kg\,m/s}$$ $$p_2=mv$$ $$p_2=(12)(14)$$ $$p_2=168\,\mathrm{kg\,m/s}$$

The skateboarder has slightly more momentum.

Example: A 70 kg runner has 420 kg m/s of momentum. How fast is the runner moving?
solution $$p=mv$$ $$v=\frac{p}{m}$$ $$v=\frac{420}{70}$$ $$v=6\,\mathrm{m/s}$$
Example: A 2 kg cart speeds up from 1 m/s to 5 m/s. What impulse acted on the cart?
solution $$J=\Delta p$$ $$\Delta p=m(v_f-v_i)$$ $$\Delta p=(2)(5-1)$$ $$\Delta p=8\,\mathrm{kg\,m/s}$$ $$J=8\,\mathrm{N\,s}$$
Example: A cart has 18 kg m/s of momentum and moves at 3.0 m/s. What force is acting on it?
answer This cannot be solved from the information given. Momentum and velocity can tell you the mass, but force needs information about how the momentum changes over time.
Example: A 75 N force acts on a box for 0.20 s. What impulse is delivered?
solution $$J=F\Delta t$$ $$J=(75\,\mathrm{N})(0.20\,\mathrm{s})$$ $$J=15\,\mathrm{N\,s}$$
Example: A 0.50 kg ball is moving right at 12 m/s. It bounces off a wall and leaves moving left at 8 m/s. If the contact time is 0.050 s, what average force acted on the ball? Let right be positive.
solution

The final velocity is negative because the ball leaves moving left.

$$\Delta p=m(v_f-v_i)$$ $$\Delta p=(0.50)(-8-12)$$ $$\Delta p=-10\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\sum F=\frac{-10}{0.050}$$ $$\sum F=-200\,\mathrm{N}$$

The average force is 200 N to the left.

Example: A 1200 kg car slows from 72 km/h to rest on a dry road. The average braking force is 6000 N opposite the motion, and the car has four passengers. How long does it take to stop?
solution

Let the original direction of the car be positive, so the braking force is negative.

$$72\,\mathrm{km/h}=20\,\mathrm{m/s}$$ $$\Delta p=m(v_f-v_i)$$ $$\Delta p=(1200)(0-20)$$ $$\Delta p=-24000\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\Delta t=\frac{\Delta p}{\sum F}$$ $$\Delta t=\frac{-24000}{-6000}$$ $$\Delta t=4.0\,\mathrm{s}$$
Example: A 0.145 kg baseball is thrown at 90 miles/hour. Convert its speed to m/s, then find its momentum.
solution $$90\left(\frac{1609\,\mathrm{m}}{1\,\mathrm{mile}}\right)\left(\frac{1\,\mathrm{hour}}{3600\,\mathrm{s}}\right)=40.2\,\mathrm{m/s}$$ $$p=mv$$ $$p=(0.145\,\mathrm{kg})(40.2\,\mathrm{m/s})$$ $$p=5.83\,\mathrm{kg\,m/s}$$
Example: A 5 kg cart starts from rest. A steady 10 N net force pushes it for 3.0 s. What is the cart's final velocity?
solution $$J=F\Delta t$$ $$J=(10)(3.0)$$ $$J=30\,\mathrm{N\,s}$$ $$J=\Delta p$$ $$\Delta p=m(v_f-v_i)$$ $$30=(5)(v_f-0)$$ $$v_f=6.0\,\mathrm{m/s}$$
Example: A 3 kg object starts from rest and accelerates at 2 m/s² for 4 s. What is its momentum after 4 s?
solution

First use motion to find the final velocity.

$$v_f=v_i+at$$ $$v_f=0+(2)(4)$$ $$v_f=8\,\mathrm{m/s}$$ $$p=mv$$ $$p=(3)(8)$$ $$p=24\,\mathrm{kg\,m/s}$$
Example: An 8 kg sled speeds up from 2 m/s to 11 m/s over 6 s. What average net force acted on it?
solution $$\Delta p=m(v_f-v_i)$$ $$\Delta p=(8)(11-2)$$ $$\Delta p=72\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\sum F=\frac{72}{6}$$ $$\sum F=12\,\mathrm{N}$$
Example: A 75 kg driver slows from 64.8 km/h to rest during a crash test. Compare the average force if the stopping time is 0.10 s with the average force if an airbag and crumple zone stretch the stopping time to 0.60 s.
solution $$64.8\,\mathrm{km/h}=18\,\mathrm{m/s}$$ $$\Delta p=m(v_f-v_i)$$ $$\Delta p=(75)(0-18)$$ $$\Delta p=-1350\,\mathrm{kg\,m/s}$$ $$\sum F=\frac{\Delta p}{\Delta t}$$ $$\sum F_1=\frac{-1350}{0.10}$$ $$\sum F_1=-13500\,\mathrm{N}$$ $$\sum F_2=\frac{-1350}{0.60}$$ $$\sum F_2=-2250\,\mathrm{N}$$

The longer stopping time produces a much smaller average force.