impulse momentum theorem explained
It is hard to follow, moment by moment, what happens in the brief instant when two bodies collide, because the force shoots up and vanishes almost at once. So instead of the force itself, we look at what builds up while it acts: the impulse. The impulse equals exactly the change in the body's momentum (mv), so even without knowing how the force varied, the change in momentum tells you the impulse. If no force enters from outside, the total momentum of several bodies stays the same before and after they collide. If energy is force accumulated over distance, momentum is force accumulated over time.
Impulse is force multiplied by the time it acts, J = F·Δt. The graph is an F-t graph, with time on the horizontal axis and force on the vertical; the orange horizontal line is a constant force and the gold rectangle beneath it is the impulse. Increase Δt with the time slider and the rectangle widens to the right while the impulse in the readout grows in proportion to time. The same force pushing longer delivers more impulse. Even when the force varies with time, the impulse is always the area under the graph, the integral of force with respect to time.
Where does the impulse go? Give an object an impulse J, the gold segment, and its momentum p = mv grows by exactly that amount on top of the blue initial part. Written as an equation this is J = Δp, the impulse-momentum theorem. Increase J with the slider and the gold segment of the momentum bar grows; since the mass does not change, the blue velocity arrow lengthens too. Just as the work-energy theorem gave the work from the kinetic energy at the start and end, this theorem gives the impulse from the momentum at the start and end. It does not matter how wildly the force varies during the impact.
Here a spring squeezed between two carts pushes them apart. The spring's force is an internal force that the two carts exchange with each other, not a force from outside, so the total momentum of a system that started at rest stays 0 throughout. The momentum going left and the momentum going right must therefore be equal in size (m₁v₁ = m₂v₂), and the heavier cart backs away more slowly. Raise the mass m₁ of the left cart with the slider and its blue arrow v₁ shortens while the orange arrow v₂ of the right cart stays the same. It is the same reason a cannonball flies off fast while the heavy cannon recoils slowly.
The same impulse delivered over a short time makes a large force, and spread over a long time makes a small one (F = JΔt). The gold rectangle in the figure is drawn so that its area, the impulse, never changes: shorten the contact time with the slider and it grows taller as it narrows; lengthen it and it flattens. The rectangle's height is the force. Airbags, soft mats, and bending your knees when you jump down all lengthen the contact time so that the same impulse arrives as a smaller force. Conversely, a hammer driving a nail makes the contact time very short to get a large force.
Choose the type of collision with the buttons; the upper row is before the collision and the lower row after. Before, the blue cart runs to the right into an orange cart of equal mass at rest. Choose stick and the two carts become a single gold block moving on at half the speed; choose bounce and the blue cart stops while the orange cart runs off at the original speed. Either way, the numbers below show the total momentum unchanged from before the collision. What differs is the kinetic energy: when the carts stick, half of it disappears into heat, sound, and deformation (inelastic), and only in a perfectly bouncing collision does it all remain (perfectly elastic). Momentum is always conserved while kinetic energy depends on the case; that is the starting point for solving collisions.