What is momentum — mass times velocity — Physics

Before this

This leads to

Fast and light, or slow and heavy?

A tennis ball flies towards you at 50 m/s. A truck rolls towards you at 1 m/s, a slow walking pace. Which would you rather stop with your hands?

Almost everyone chooses the tennis ball, even though it is fifty times faster. Your instinct knows something that speed alone does not capture. The truck is hard to stop because it has a huge mass. The ball is easy to stop because it has a tiny one. "How hard is this to stop?" depends on mass and velocity together. Physics has a name for that combination: momentum.

This lecture defines momentum, shows why it is the right combination, and explains why its direction matters as much as its size.

Momentum is mass times velocity

Momentum

The momentum $p$ of an object is its mass times its velocity: $p = m v$ .

Mass in kg and velocity in m/s give momentum in kg m/s. Momentum is a vector: it has the same direction as the velocity.

Compare the two objects. A tennis ball has a mass of about 0.057 kg (the rules allow 56.0 to 59.4 g). An invented delivery truck has a mass of 8000 kg.

Tennis ball: $p = 0.057 \times 50 \approx 2.85$ kg m/s.
Truck: $p = 8000 \times 1 = 8000$ kg m/s.

The truck has about 2800 times the momentum of the ball, although the ball is 50 times faster. Mass and velocity count equally in the product. Double either one, and the momentum doubles.

Some books write the unit as N s (newton seconds) instead of kg m/s. They are the same unit, and the lecture on impulse shows why. In this lecture we use kg m/s.

Why mass times velocity?

Why should "hard to stop" be measured by $m v$ , and not by $m + v$ , or $m v^{2}$ , or anything else? The answer comes from Newton's second law. This is the key step of the lecture, so take it slowly.

The second law says $F = ma$ . Acceleration is the change in velocity divided by the time taken: $a = Δ v \div Δ t$ . So:

F = m \times \frac{Δ v}{Δ t} = \frac{m Δ v}{Δ t} = \frac{Δ ( m v )}{Δ t}

The last step works because the mass does not change, so the change in $m v$ is just $m$ times the change in $v$ . Read the result in words: the resultant force on an object equals the rate of change of its momentum. To stop an object, you must remove all its momentum. A bigger momentum needs either a bigger force or a longer time.

Now the truck and the ball can be compared exactly. Suppose you must stop each one in half a second. The truck needs a force of $8000 \div 0.5 = 16000$ N. The ball needs only $2.85 \div 0.5 = 5.7$ N. That is why your instinct was right. That is it. That is the whole idea. Momentum is the quantity a force changes: the force times the time it acts equals the change in momentum.

The reasoning also shows why the answer is not $m v^{2}$ or $m + v$ . A force acting for a time changes velocity in proportion to that time, and the effect on a heavier object is smaller in proportion to its mass. Only the product $m v$ changes by exactly "force times time". (The quantity with $v^{2}$ in it is also important, but it measures something else, kinetic energy, which a later topic studies.)

This is also closer to how Newton himself wrote the second law. He described motion by its "quantity of motion", which is mass times velocity, and said that a force changes it.

Direction matters

Velocity is a vector, and mass is a positive number, so momentum is a vector with the same direction as the velocity. In one dimension, choose a positive direction and use signs, as the Notation sheet asks for every mechanics problem.

Car A (1000 kg) moves right at 15 m/s; car B (1500 kg) moves left at 10 m/s. Taking right as positive, their momenta are +15 000 and −15 000 kg m/s. They are equal in size and opposite in direction, so the total momentum of the two cars is zero, even though both are moving.

Taking right as positive, car A has momentum $1000 \times (+ 15) = + 15000$ kg m/s, and car B has $1500 \times (- 10) = - 15000$ kg m/s. The total momentum of the two cars is zero. That may feel strange, because both cars are clearly moving. But momentum adds like a vector, and these two cancel. This total will matter a great deal in the next lecture, on collisions.

Before you go on, make a prediction about a change in momentum.

Predict, then reveal — a ball that bounces

A 0.06 kg ball hits a wall at 20 m/s and bounces straight back at 20 m/s. What is the size of its change in momentum?

The change in momentum of a 0.06 kg ball that hits a wall at 20 m/s and bounces back at 20 m/s. Many people answer zero, because the speed is the same before and after.

Take the direction towards the wall as positive. Before: 0.06 × 20 = +1.2 kg m/s. After: 0.06 × (−20) = −1.2 kg m/s. Change = after − before = −1.2 − 1.2 = −2.4 kg m/s. The size of the change is 2.4 kg m/s, away from the wall.

The bounce changes the momentum twice as much as stopping would. This is why a bouncing ball pushes harder on a wall than a ball of clay that sticks: the wall must stop the ball and then send it back.

Now try your own comparisons. Choose two objects and their velocities, and compare their momenta and the total.

Explore — comparing momenta

Choose objects A and B and their velocities (negative means moving left). The arrows show each momentum and the total, to the same scale. Can you make the total zero?

A's velocity: 30 m/s

B's velocity: 1 m/s

Momenta of some everyday objects (masses: tennis ball 0.057 kg, cricket ball 0.16 kg, person 60 kg, car 1200 kg, truck 8000 kg; the person, car and truck are typical invented values):

Object	Velocity	Momentum (kg m/s)
tennis ball	30 m/s	≈ 1.71
cricket ball	30 m/s	4.8
person	2 m/s	120
car	−20 m/s	−24 000
truck	1 m/s	8000

Momenta in opposite directions have opposite signs. A car at 20 m/s to the left and three trucks at 1 m/s to the right would have a total momentum of exactly zero.

To make the total zero, the two momenta must be equal in size and opposite in direction. With a heavy object and a light one, the light one must move very much faster.

A system's total momentum

The third-law lecture had two skaters, 50 kg and 70 kg, pushing each other apart with 140 N for 0.5 s. Look at them again with momentum. Starting at rest, their total momentum is zero.

Skater A reaches $2.8 \times 0.5 = 1.4$ m/s, so A's momentum is $50 \times 1.4 = 70$ kg m/s one way. Skater B reaches $2.0 \times 0.5 = 1.0$ m/s, so B's momentum is $70 \times 1.0 = 70$ kg m/s the other way. Equal and opposite. The total is still zero.

That is not a coincidence. Each skater received the same force for the same time, in opposite directions (the third law). Force times time is the change in momentum. So their changes were equal and opposite, and the total could not change. The last lecture found that when internal forces cancel, the motion of the whole system depends only on outside forces. Momentum is the quantity that makes this exact. The next lecture turns it into one of the most useful laws in physics.

Stop and think

A heavy truck parked on a hill has a huge mass. Does it have a huge momentum?

Show the answer

No. It is not moving, so its velocity is zero, and its momentum is $8000 \times 0 = 0$ . Momentum is not a store of "heaviness". It belongs to motion. A truck at rest is hard to start moving because of its large mass (its inertia, from the first law), but it has no momentum to remove. Do not confuse momentum with mass, or with force: a truck at rest has no momentum, and a ball in flight has momentum but no force pushing it forwards.

Worked examples

Worked Example 1 — Momentum from mass and velocity

(a) A 0.16 kg cricket ball is bowled at 35 m/s. (b) A 75 kg sprinter runs at 10 m/s. Find each momentum.

(a) $p = 0.16 \times 35 = 5.6$ kg m/s, in the direction of the ball's motion. A men's cricket ball must weigh between 155.9 g and 163 g, so 0.16 kg is a fair value.

(b) $p = 75 \times 10 = 750$ kg m/s. The sprinter is far slower than the ball but has over 100 times its momentum.

Worked Example 2 — Velocity from momentum

A 0.5 kg ball has a momentum of 12.5 kg m/s. Find its speed.

Rearrange $p = m v$ : $v = p \div m = 12.5 \div 0.5 = 25$ m/s.

Check: $0.5 \times 25 = 12.5$ .

Worked Example 3 — Units first

A 1200 kg car travels at 72 km/h. Find its momentum.

Convert first: $72 km/h = 72 \times 1000 \div 3600 = 20$ m/s.

Then $p = 1200 \times 20 = 24000$ kg m/s. Using 72 directly would give an answer 3.6 times too large.

Worked Example 4 — Matching momentum

The 1200 kg car of Worked Example 3 has a momentum of 24 000 kg m/s. How fast must the 8000 kg truck move to have the same momentum?

$v = p \div m = 24000 \div 8000 = 3$ m/s.

The truck has about 6.7 times the car's mass, so it needs about 6.7 times less speed for the same momentum: 20 m/s becomes 3 m/s. Momentum lets you compare a heavy, slow object with a light, fast one on one scale.

Momentum and safety

The second-law form $F = Δ p \div Δ t$ explains much of modern safety design. In a crash, a car and its passengers must lose all their momentum. That change is fixed by the mass and the speed. What engineers can change is the time it takes. A seat belt that stretches a little, an airbag that slows the head gently, and a front end that crumples all make the stop last longer. A longer time means a smaller force for the same change in momentum, and a smaller force means less injury. Sports equipment works the same way: padded gloves, soft landing mats and bending your knees when you land all spread the change in momentum over more time. The impulse lecture puts numbers on this.

Three classic mistakes

Trap 1 — Ignoring direction

Momentum is a vector. A ball that bounces back at the same speed has a change in momentum of $2 m v$ , not zero. Always choose a positive direction and give each momentum its sign.

Trap 2 — Wrong units

Mass must be in kg and velocity in m/s. A 57 g ball is 0.057 kg, and 72 km/h is 20 m/s. Mixing grams or km/h into $p = m v$ gives answers that are 1000 or 3.6 times wrong.

Trap 3 — Momentum is not force

Momentum is what an object has because it moves. Force is what changes momentum. A ball flying through the air has momentum, but no forward force acts on it, as the free-body lecture showed.

Try it yourself

Take the direction stated in each question as positive. Show units.

A 0.010 kg bullet travels at 400 m/s. Find its momentum. How fast would a 70 kg person have to walk to have the same momentum?

Hint 1
$p = m v$ for the bullet first.

Hint 2
Then $v = p \div 70$ .

Full answer
Bullet: $0.010 \times 400 = 4.0$ kg m/s. Person: $v = 4.0 \div 70 \approx 0.057$ m/s, less than 6 cm per second. A very slow shuffle has the momentum of a bullet. What makes a bullet dangerous is that its momentum is delivered to a very small area in a very short time.
Two trolleys move towards each other. Trolley P (2 kg) moves right at 3 m/s; trolley Q (3 kg) moves left at 2 m/s. Find the total momentum, taking right as positive.

Hint 1
Q's velocity is −2 m/s.

Hint 2
Add $2 \times 3$ and $3 \times (- 2)$ .

Full answer
$p_{P} = + 6$ kg m/s and $p_{Q} = - 6$ kg m/s, so the total is 0. Both trolleys move, but the system's total momentum is zero.
A 0.4 kg ball falls onto the floor at 8 m/s and rebounds upwards at 6 m/s. Taking upwards as positive, find its change in momentum.

Hint 1
Before: velocity −8 m/s. After: +6 m/s.

Hint 2
Change = after − before.

Full answer
Before: $0.4 \times (- 8) = - 3.2$ kg m/s. After: $0.4 \times 6 = + 2.4$ kg m/s. Change $= 2.4 - (- 3.2) = 5.6$ kg m/s, upwards. The change is larger than either momentum alone, because the direction reversed.
A goalkeeper catches a 0.43 kg football moving at 25 m/s and stops it in 0.1 s. Use $F = Δ p \div Δ t$ to find the average force on the ball. Then explain why goalkeepers pull their hands back as they catch.

Hint 1
The change in momentum is from $0.43 \times 25$ to 0.

Hint 2
A longer stopping time means a smaller force for the same change in momentum.

Full answer
Take the ball's direction of motion as positive. $Δ p = 0 - 0.43 \times 25 = - 10.75$ kg m/s. Average force $= 10.75 \div 0.1 = 107.5$ N, against the ball's motion. Pulling the hands back makes the stop take longer, say 0.2 s, which halves the force to about 54 N. The change in momentum is fixed; only the time can change. The impulse lecture develops this idea.

The bigger idea

Momentum was named before Newton. René Descartes, in 1644, argued that the "quantity of motion" in the world is conserved, but he measured it by speed without direction. In 1668, Christiaan Huygens and John Wallis showed that collisions only make sense when the direction is included, with opposite directions given opposite signs. Newton then built "quantity of motion" into his laws in 1687. The idea grew from a hunch into a vector law in about forty years.

Momentum matters because it is the quantity forces change, and because, inside a system, forces come in equal and opposite pairs. The next lecture shows what that means for collisions: whatever happens inside a system, its total momentum stays the same.

Key concepts in this lesson

Momentum is mass times velocity, p = mv, measured in kg m/s.
Momentum is a vector with the direction of the velocity, so a positive direction must be chosen and signs used.
The resultant force on an object equals its rate of change of momentum, so a larger momentum needs a larger force or a longer time to stop.
The total momentum of a system is the sum of the momenta, with signs, and it can be zero even when every part is moving.
An object at rest has no momentum, however large its mass.