# Is there an intuitive way to view the concept of momentum?

Ideas like distance, velocity and force are very intuitive to understand because you can "see" their real-world applications and so one can come to understand them without having any knowledge of their mathematical formulas.

Momentum as it is defined is the product of mass times the velocity. I can see how it came to be mathematically derived via combining Newton's second law and the equation for acceleration (change in velocity over change in time). However, besides just remembering that p=mv, is there an intuitive way to view what momentum is? Am I supposed to have an intuitive understanding of it? or is that just how every physics student thinks of it?

• Does this answer help? Commented Apr 27, 2021 at 12:30

Imagine two large iron rolling balls, one larger than the other. Imagine they are moving at the same speed.

Imagine trying and stopping them. Which would be easier? The lighter one, of course. And if the balls are the same size, their speed is what matters.

So basically momentum is the property of the object that makes it easier or harder to stop, i.e. the momentum of a body is what changes the force that must be applied to stop it.

This quantity is transferred during collisions, and of course we know that it is dependent on the mass and velocity ; $$p=mv$$ as you stated.

One way to think about momentum is to make an analogy with static equilibrium.

Let two weights, $$m_1$$ and $$m_2$$ be on the ends of a seesaw, the seesaw is such that you can move the position of the fulcrum.

The seesaw is in static equilibrium when the fulcrum is located at the Common Center of Mass (CCM) of $$m_1$$ and $$m_2$$. You have radial distance $$r_1$$ and $$r_2$$. Then we have for the static equilibrium:

$$m_1r_1 = m_2r_2$$

Now take two masses $$m_1$$ and $$m_2$$, and use the CCM of $$m_1$$ and $$m_2$$ as the zero point of the coordinate system that you use to represent the respective velocities of $$m_1$$ and $$m_2$$ .

Let the initial situation be that the two masses are moving towards each other, so there will be a collision. Let that be a perfectly elastic collision.

Among the first to examine that kind of collision was Huygens. He set up two pendulum bobs to collide with each other. By having the bobs hit each other dead center Huygens obtained fairly good elastic collisions. Huygens had no way of making accurate velocity measurements, but he could measure how high bobs swing back up after they have collided.

Huygens' experiments corroborated the following expectation:

If the bobs swing towards each other with their velocities in the following ratio:

$$m_1v_1 = m_2v_2$$

Then after the elastic collision the velocities are reversed, and still in the same ratio:

$$m_1v_1 = m_2v_2$$

The supposition then is that the quantity $$mv$$ expresses something that we can refer to as 'quantity of motion'.

The supposition is that two objects of different mass bring the same amount of quantity of motion to the CCM by virtue of the quantity $$mv$$ being equal.

The clincher is that change of momentum is galilean invariant; you can use any inertial coordinate system to describe the collision; the amount of change of momentum is independent of the choice of inertial coordinate system.

If an object collides with you plastically its momentum is what knocks you back.

• +1 Not necessarily plastically. Commented Apr 27, 2021 at 11:51
• Thanks @Vadim yes, for “intuitive” I just wanted to keep it simple and avoid comparisons of collisions with different degrees of elasticity. And a plastic collision is easier to identify than an elastic one.
– Dale
Commented Apr 27, 2021 at 12:19
• must mean elastically. Commented Apr 27, 2021 at 12:19
• @anna v no, I mean plastically
– Dale
Commented Apr 27, 2021 at 12:21
• @anna v a plastic collision is a perfectly inelastic collision. The two objects collide and afterwards move together with the same velocity as each other.
– Dale
Commented Apr 27, 2021 at 12:33

As stated in this answer,

Momentum is the quantity needed to completely remove all movement from a rigid body or a point mass.

You apply momentum in equal and opposite terms (Newton's 3rd law) in order to stop a moving body and the way momentum is applied does not matter. Being a force over time, or a short impulse yields the same result.

You can also view the situation in reverse time starting from resting body and applying momentum to bring it up to a certain state of motion.

• Translational Momentum (aka linear momentum) describes the motion of the center or mass only. $$\boldsymbol{p} = m \, \boldsymbol{v}_{\rm CM}$$
• Rotational Momentum (aka angular momentum) describes the motion about the center of mass. $$\boldsymbol{L}_{\rm CM} = \mathbf{I}_{\rm CM} \boldsymbol{\omega}$$

Rotational momentum describes where in space is translational momentum acting through. For an offset point particle $$\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$ is the product of momentum and the moment-arm. This is also known as moment-of-momentum. For an offset rigid body $$\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} + \boldsymbol{L}_{\rm CM}$$ which includes the intrinsic momentum due to rotation. This has the effect to move the axis of momentum (aka the percussion axis) a little bit away from the center of mass.

See the linked post above for some examples of how momentum describes the state of motion stuff.