In Physical Science you met momentum as mass times velocity. Physics 1 adds one important idea: momentum is a vector. It points the way the object moves, so on a line it carries a sign, and in a plane it has components, just like the vectors you add by components.
Key idea
Momentum is $\vec{p} = m\vec{v}$, measured in $\text{kg·m/s}$. On a line, pick a positive direction and give each momentum a sign. The momentum of a system is the vector sum of the momenta of its parts: $\vec{p}_{\text{tot}} = m_1\vec{v}_1 + m_2\vec{v}_2 + \dots$
Why it works
Why multiply mass by velocity? Because that product is what a force changes. Newton's second law says $F = ma = m\dfrac{\Delta v}{\Delta t}$, so $F\Delta t = \Delta(mv)$: a push lasting a while changes $mv$, whatever the mass. A loaded truck and a bicycle at the same speed are very different to stop, and $mv$ measures exactly that difference. Since velocity has a direction, so does $mv$, and two momenta in opposite directions cancel when you add them.
Worked examples
Worked example: One object, with a sign
A $1200\ \text{kg}$ car drives west at $15\ \text{m/s}$. Taking east as positive, find its momentum.
- West is the negative direction, so $v = -15\ \text{m/s}$.
- $$p = mv = (1200\ \text{kg})(-15\ \text{m/s}) = -18{,}000\ \text{kg·m/s}.$$
- The size is $18{,}000\ \text{kg·m/s}$; the minus sign says "west".
Answer: $p = -1.8 \times 10^{4}\ \text{kg·m/s}$
Worked example: Two objects on a line
Cart A ($2.0\ \text{kg}$) moves right at $3.0\ \text{m/s}$. Cart B ($4.0\ \text{kg}$) moves left at $1.0\ \text{m/s}$. Find the total momentum (right is positive).
- $$p_A = (2.0\ \text{kg})(+3.0\ \text{m/s}) = +6.0\ \text{kg·m/s}.$$
- $$p_B = (4.0\ \text{kg})(-1.0\ \text{m/s}) = -4.0\ \text{kg·m/s}.$$
- $p_{\text{tot}} = 6.0 + (-4.0) = +2.0\ \text{kg·m/s}$, pointing right.
Answer: $+2.0\ \text{kg·m/s}$
Worked example: Two objects at right angles
On ice, puck 1 ($0.20\ \text{kg}$) slides east at $6.0\ \text{m/s}$ and puck 2 ($0.20\ \text{kg}$) slides north at $8.0\ \text{m/s}$. How big is their total momentum?
- East component: $p_x = (0.20)(6.0) = 1.2\ \text{kg·m/s}$. North component: $p_y = (0.20)(8.0) = 1.6\ \text{kg·m/s}$.
- $|\vec{p}| = \sqrt{1.2^2 + 1.6^2} = 2.0\ \text{kg·m/s}$, pointing $53^\circ$ north of east.
- Adding the sizes ($2.8$) would be wrong because the momenta are not in the same direction.
Answer: $2.0\ \text{kg·m/s}$
Common mistake
Adding momenta as plain numbers. Two $3\ \text{kg·m/s}$ momenta can total $6$, $0$ or anything in between, depending on their directions. Before you add, write every velocity with its sign (or split it into components). Quick check: if two objects move toward each other, the total must be smaller than the bigger of the two momenta.
A small, fast object can carry as much momentum as a big, slow one: a $0.15\ \text{kg}$ baseball at $40\ \text{m/s}$ and a $6.0\ \text{kg}$ bowling ball at $1.0\ \text{m/s}$ both have $6.0\ \text{kg·m/s}$. Try collisions with momentum bars in the Momentum and Collisions lab of the Science Lab.