Mistake Master
Mistake Master
Push a cart hard for a moment, or push it gently for a while. Both can deliver the same total kick. That total is the impulse, $\vec J = \vec F_{\text{avg}}\,\Delta t$, and it equals the change in the cart's momentum, $\Delta \vec p$. Impulse depends on how long the force acts, not just how strong it is. On a force-versus-time graph, it shows up as area, not slope or peak height.
When a net force $\vec F$ acts on a system for a time $\Delta t$, the system's momentum changes. The change equals the average net force times the duration. Because force has direction, so does this product. It is called the impulse, written $\vec J$:
$$\vec J = \vec F_{\text{avg}}\,\Delta t = \Delta \vec p.$$This isn't a new law. It's Newton's second law rewritten over an interval. Starting from $\vec F_{\text{net}} = \Delta \vec p / \Delta t$ and multiplying both sides by $\Delta t$ gives $\vec F_{\text{net}}\,\Delta t = \Delta \vec p$. The left-hand side is the impulse the net force delivered; the right-hand side is the resulting change in momentum.
The point of the rewrite is bookkeeping. When a force varies during a collision or a kick, tracking it instant by instant is hard. Tracking the total kick over the whole interval is easier, and that total is exactly what changes the momentum.
The change in a system's momentum vector is
$$\Delta \vec p = \vec p_f - \vec p_i = m\,\vec v_f - m\,\vec v_i.$$And the impulse delivered by the net force equals that change:
$$\vec J = \vec F_{\text{avg}}\,\Delta t = \Delta \vec p.$$Two units, one quantity. Momentum is in kilogram-meters per second ($\text{kg} \cdot \text{m/s}$). Impulse is in newton-seconds ($\text{N} \cdot \text{s}$). These are the same unit: $1\,\text{N} \cdot \text{s} = 1\,(\text{kg} \cdot \text{m/s}^2) \cdot \text{s} = 1\,\text{kg} \cdot \text{m/s}$. Either label is correct; the choice is cosmetic.
$\Delta \vec p$ is a vector, not a magnitude. In 1D, that means it carries a sign tied to the axis you picked. In 2D, it has components that subtract component-by-component: $\Delta p_x = m(v_{fx} - v_{ix})$ and the same for $y$.
A 2 kg cart starts at rest. A constant net force of $+6$ N acts on it for 0.5 s. Find $\vec J$, $\Delta \vec p$, and the cart's final velocity.
$\vec J = (+6)(0.5) = +3\,\text{N} \cdot \text{s}$. By the impulse-momentum relation, $\Delta \vec p = +3\,\text{kg} \cdot \text{m/s}$. Since $\Delta v = \Delta p / m$, the cart picks up $+3/2 = +1.5$ m/s, so $v_f = 0 + 1.5 = +1.5$ m/s.
Same number on both sides of $\vec J = \Delta \vec p$, same sign, same units once you remember $\text{N} \cdot \text{s}$ and $\text{kg} \cdot \text{m/s}$ are interchangeable. The positive sign means the cart ended up moving in the same direction as the force, which is what you'd expect.
Surprise one: impulse is the area under F versus t. When the force isn't constant, you can't just multiply $F \times \Delta t$. But the rectangle picture still works as a recipe: any varying force can be sliced into thin time slices, each one a narrow rectangle whose area is its own little impulse. Add up all those areas and you get the total impulse for the whole interval. That is the area under the force-versus-time curve between the start and end of the push.
A constant 4 N for 2 s delivers a rectangle of area $J = (4)(2) = 8\,\text{N} \cdot \text{s}$. A triangular pulse with the same peak height (4 N) and same width (2 s) delivers a triangle of area $J = \tfrac{1}{2}(4)(2) = 4\,\text{N} \cdot \text{s}$, exactly half. Same peak. Half the area. Half the impulse. Half the change in momentum.
Surprise two: $\Delta \vec p$ is vector subtraction, not magnitude subtraction. A ball bounces straight back off a wall at the same speed. Computing $|p_f| - |p_i|$ gives zero, and you'd conclude the wall delivered no impulse. It did. The momentum vector flipped direction. Picking $+x$ as the incoming direction,
$$\Delta p = p_f - p_i = (-m v) - (+m v) = -2 m v,$$a number twice the size of either initial or final momentum. The magnitudes alone hide that.
All three impulse traps below chase the same shortcut — reading the change in momentum off something quicker than force acting over time, whether that is a large force, a graph's slope, or a bare change in speed.
Impulse is force times time, not force alone. A 100 N force acting for 0.05 s delivers $J = 5\,\text{N} \cdot \text{s}$. A 10 N force acting for 1 s delivers $J = 10\,\text{N} \cdot \text{s}$, twice as much. The small steady push won because it acted twenty times longer than the big spike.
Fix. Before deciding which force delivered more impulse, ask the second question: for how long? The same impulse can come from a big force over a short time or a small force over a long time. Catching an egg in a cupped hand or on a concrete floor changes the same momentum; the cupped hand spreads the force over more time, so the force itself drops.
Two graphs, two operations, and they don't cross over. On a force-versus-time graph, the area between the curve and the t-axis is the impulse. On a momentum-versus-time graph, the slope is the net force. Reading the slope of an F-vs-t plot gives a "force change rate," which is rarely what's asked. Reading the area under a p-vs-t plot gives no standard quantity at all.
Fix. Match the operation to the graph. F-vs-t plot, want impulse? Find the area, splitting into rectangles and triangles. p-vs-t plot, want force? Find the slope, $F = \Delta p / \Delta t$. p-vs-t plot, want $\Delta \vec p$ between two moments? It's just the height difference between the two points, no slope and no area required.
This is the vector-subtraction trap. The ball's speed didn't change, so $|p_f| - |p_i| = 0$, so you might conclude no impulse was needed. The momentum vector reversed, though. Take "to the right" as positive and call the ball's incoming speed $v$. Then $p_i = +mv$ and $p_f = -mv$, so $\Delta p = -mv - (+mv) = -2mv$, not zero.
Fix. Always pick an axis, attach a sign to each velocity, then subtract. The magnitudes-only path looks shorter, but it throws away the direction flip that makes the bounce a bounce. $|\Delta \vec p|$ on a perfect bounce is twice the size of either initial or final momentum, never zero.