Mistake Master
Mistake Master
Momentum is mass times velocity: $\vec p = m\vec v$. Velocity has direction, so momentum does too. The new term is linear momentum, and it's a vector that points the same way as $\vec v$. The trap: dropping the direction and reporting $\vec p$ as a single number, or reading a minus sign on $p$ as "smaller" or "stopped" instead of direction along your chosen axis.
Momentum tells you how hard something is to stop. A heavy truck creeping forward and a light bullet flying fast both take real effort to bring to rest, and momentum is the one number that captures that effort for either.
The formula is $\vec p = m\vec v$. Since $m$ is a positive scalar, $\vec p$ always points the same way as $\vec v$. That's why momentum is a vector: it inherits direction from velocity. Drop the direction and you've thrown away half the information.
The job through this lesson is keeping that direction visible: hold the sign on $p$, and when more than one object is in play, add momenta as vectors.
For mass $m$ moving with velocity $\vec v$:
$$\vec p = m\vec v.$$Units: $\text{kg} \cdot \text{m/s}$. No shortcut name; you carry the compound. No $\tfrac{1}{2}$, no square: $\vec p$ is exactly mass times velocity. (The $\tfrac{1}{2}$ belongs to kinetic energy.)
In 1D you'll mostly write the signed component $p$ and let the sign carry the direction. Pick a positive axis, write velocities with signs, multiply by mass: that's $p$ for each object. Once a sign convention is chosen, every momentum follows it through the rest of the problem.
A $3.0$ kg cart moves west at $4.0$ m/s. Take east as positive. Find $p$.
West is negative, so $v = -4.0$ m/s. Then $p = mv = (3.0)(-4.0) = -12$ kg$\cdot$m/s.
The minus sign means west, not "less momentum." Pick west as positive instead and the same motion gives $p = +12$ kg$\cdot$m/s. Sign tracks convention; magnitude tracks the cart.
The sign on $p$ is direction along your chosen axis. Three reads are wrong:
For two objects on the same axis, the system momentum is the vector sum:
$$\vec p_{\text{tot}} = \vec p_1 + \vec p_2 = m_1 \vec v_1 + m_2 \vec v_2.$$Add the signed momenta, not the speeds. Two carts headed at each other pick up opposite signs once an axis is chosen, so their momenta partially or fully cancel. Equal masses and equal speeds, opposite directions: cancels to zero. Otherwise the net points the way of the bigger $|mv|$.
Three traps catch most students on Topic 4.1. Each comes from forgetting a piece of what makes $\vec p$ a vector.
This treats $\vec p$ as a scalar, like mass or temperature. The magnitude is fine; what's missing is the direction. Any problem that mixes objects moving different ways (head-on, system momentum, before-and-after a collision) needs the direction to come out right.
Fix. $\vec p = m\vec v$, and $\vec v$ has direction, so $\vec p$ does too. In 1D, write $p$ as a signed number with the axis up front: "$p = +20$ kg$\cdot$m/s, east as positive." The sign carries direction compactly; drop it and you've lost half the information.
This confuses the sign of $p$ with the cart's state of motion. Negative numbers on a number line can feel like "less than nothing," but on a signed-axis convention a minus is just the opposite direction. A cart with $p = -8$ kg$\cdot$m/s is moving just as fast as one with $p = +8$; the only difference is which way.
Fix. The sign on $p$ is direction, period. To check whether a cart is stopped, look at the magnitude: $|\vec p| = 0$ means at rest, $|\vec p| > 0$ means moving, whatever sign $p$ carries.
This ranks momenta by one factor and ignores the other. The same trap runs in reverse ("the bullet has more, since it's faster"). $\vec p = m\vec v$ uses both; a small change in either can flip the ranking.
Fix. Compute $|\vec p| = m|\vec v|$ for each object and compare the products, not the factors. A heavy slow object and a light fast object can carry the same momentum, and often do.