Topic 4: Linear momentum

Linear momentum captures how hard a moving object is to stop, combining both its mass and its velocity. Forces acting over time change momentum, and when a system is isolated its total momentum stays fixed no matter how violently its parts interact. This unit ties Newton's laws to collisions and explosions and to the motion of a system's center of mass.

What momentum is

Linear momentum p is the product of an object's mass and its velocity: p = m*v. Because velocity is a vector, momentum is also a vector pointing in the same direction as the motion, with units of kg m/s. A heavy truck rolling slowly and a light bullet flying fast can carry the same momentum. Momentum is always measured relative to a chosen reference frame, so the same object can have different momentum values for different observers. For a system of several objects, the total momentum is the vector sum of the individual momenta; you add the x-components separately from the y-components, never just the speeds. Keeping signs consistent (for example, right positive, left negative in one dimension) is essential because momentum can cancel.

Impulse and the impulse-momentum theorem

Impulse J is the effect of a force applied over a time interval: J = F*t for a constant force, with units of N s. The impulse-momentum theorem states that the net impulse on an object equals its change in momentum: J = delta p = m*v_final - m*v_initial. Note that N s and kg m/s are the same unit, which is no accident since impulse is just accumulated change in momentum. This explains why air bags, crumple zones, and bending your knees on landing all help: by stretching the collision over a longer time, they lower the average force needed to produce the same momentum change. To stop a given object you must deliver a fixed impulse; you only get to choose whether it arrives as a large force over a short time or a small force over a long time.

Reading force-time graphs

When a force is not constant, the impulse equals the area under the force-versus-time graph. A spike of force that lasts a few milliseconds during a collision still delivers a measurable impulse equal to that shaded area. For an irregular curve, estimate the area by counting grid squares or by approximating the shape with triangles and rectangles. Once you have the area (the impulse in N s), apply delta p = J to find the change in velocity, dividing by the mass. The average force over the interval is the impulse divided by the total time, which is the height of a rectangle having the same area as the real curve. This graphical view is a favorite exam target because it links calculus-style reasoning to a simple geometric area.

Conservation of momentum

If the net external force on a system is zero, the system's total momentum cannot change. This follows from Newton's third law: during any interaction the two objects push on each other with equal and opposite forces for the same amount of time, so they receive equal and opposite impulses that cancel in the total. Internal forces, however strong, simply transfer momentum between members of the system without altering the sum. The key skill is drawing the system boundary correctly: choose a system over which outside forces (friction, gravity along the motion, applied pushes) are zero or negligible during the brief interaction. Then p_total before equals p_total after, written component by component. Conservation of momentum holds in every type of collision, even ones where kinetic energy is lost.

Elastic collisions

A collision is elastic when total kinetic energy is conserved in addition to total momentum. Perfectly elastic collisions are an idealization, closely approached by hard objects like billiard balls or gas molecules. Because two quantities are conserved, you get two equations to solve for two unknown final velocities. A useful shortcut for one-dimensional elastic collisions is that the relative velocity of approach equals the relative velocity of separation: v1_initial - v2_initial = -(v1_final - v2_final). Special cases worth memorizing: equal masses simply exchange velocities, and a very light object bouncing off a very heavy stationary one reverses its velocity while the heavy object barely moves. Always check your answer by confirming both momentum and kinetic energy balance.

Inelastic and perfectly inelastic collisions

In an inelastic collision momentum is still conserved but kinetic energy is not, because some energy converts to heat, sound, and permanent deformation. The extreme case is a perfectly inelastic collision, where the objects stick together and move with one common final velocity. Worked example: a 2.0 kg cart moving right at 3.0 m/s strikes and locks onto a 1.0 kg cart at rest. Momentum before = (2.0)(3.0) + (1.0)(0) = 6.0 kg m/s. After collision the combined 3.0 kg mass moves at v, so 3.0*v = 6.0 kg m/s, giving v = 2.0 m/s to the right. Checking energy: KE before = 0.5*2.0*3.0^2 = 9.0 J, KE after = 0.5*3.0*2.0^2 = 6.0 J, so 3.0 J was lost to the collision. Momentum balanced perfectly while kinetic energy did not, which is the signature of an inelastic event.

Two-dimensional collisions

Momentum is a vector, so in two dimensions it is conserved independently along each axis. Resolve every object's momentum into x- and y-components, then write one conservation equation for x and a separate one for y. Total p_x before equals total p_x after, and the same holds for p_y, but you must never combine them into a single scalar sum. After solving the two component equations for the unknown final components, reconstruct each object's speed with the Pythagorean theorem and its direction with the inverse tangent. A classic setup is a moving puck striking a stationary one and the two flying off at angles; if the collision is also elastic, add the kinetic-energy equation. Sketching the before and after vectors keeps the signs and angles straight.

Center of mass and its motion

The center of mass is the mass-weighted average position of a system, the single point that moves as if all the mass and all the external force were concentrated there. Its location is x_cm = (m1*x1 + m2*x2 + ...) / (m1 + m2 + ...), with a matching formula for y. The velocity of the center of mass is the total momentum divided by the total mass: v_cm = p_total / M_total. A powerful consequence is that internal forces, including collisions and explosions, never change the motion of the center of mass; only external forces do. So when a firework bursts in midair, the fragments scatter but their center of mass keeps following the same parabolic path the unexploded shell would have. Because total momentum is conserved in an isolated system, the center of mass moves at constant velocity throughout any internal interaction.