Topic 4.3 Conservation of Linear Momentum: apply conservation of momentum to an isolated system, where the total momentum before equals the total momentum after an interaction.

What this topic is asking

The College Board (Topic 4.3) wants you to apply conservation of linear momentum: for an isolated system (no net external force), the total momentum is constant, so the total momentum before an interaction equals the total momentum after. You should be able to choose the system, identify internal versus external forces, and use the conservation law to find unknown velocities in recoil, explosions and collisions.

The conservation law

This is one of the most powerful principles in mechanics. It holds exactly when external forces are absent or negligible during the interaction, and it lets you relate the velocities before and after without knowing the complicated internal forces.

Why internal forces cannot change total momentum

This is the deep reason momentum is conserved. During a collision, object A pushes on B and B pushes back on A with an equal and opposite force for the same time, so the impulses cancel in the total. The internal forces can be enormous and complicated, but they redistribute momentum within the system without changing the sum. Choosing the system wisely, so that the large interaction forces are internal, is the key strategic move, exactly the systems thinking from Topic 2.1.

Choosing the system and external forces

The art of applying conservation of momentum is choosing the system so that the forces you do not want to track become internal. If you include both colliding carts in the system, the collision forces are internal and cancel; if you included only one cart, the other's force would be external and momentum would not be conserved for that single object. External forces such as gravity, friction or a normal force can still act, but over a brief collision their impulse is often negligible compared with the huge internal forces, so momentum is conserved to a good approximation during the impact. This is why momentum methods work for collisions even on a frictional floor: the collision lasts a tiny time, so friction's impulse during it is small. The general routine is: define the system, take a direction as positive, write the total momentum before (with signs), write the total momentum after, set them equal, and solve for the unknown. Because momentum is a vector, in two dimensions you conserve each component separately. This single principle handles recoil, explosions and every kind of collision, and it pairs with energy conservation (Topic 3.4) to fully analyze what happens when objects interact.

Try this

Q1. A $40$ kg child on frictionless ice, initially at rest, throws a $2.0$ kg ball at $6.0$ m/s. Calculate the child's recoil speed. [2 points]

• Cue. $0 = (2.0)(6.0) + (40)v$, so $v = -12/40 = -0.30$ m/s (opposite to the ball).

Q2. A $1.0$ kg trolley moving at $3.0$ m/s couples to a $2.0$ kg stationary trolley. Calculate their common speed afterward. [2 points]

• Cue. $p_i = (1.0)(3.0) = 3.0$; $p_f = (3.0)v$, so $v = 3.0/3.0 = 1.0$ m/s.