Momentum and rocket propulsion

Two carts collide and stick together. The impact force changes greatly within a fraction of a second. Can we find their common velocity without reconstructing every instant of the collision? Comparing before and after will help.

An empty and a loaded cart moving at the same speed are not equally hard to stop. We need both velocity and mass: momentum is p⃗ = mv⃗, measured in kg·m/s. The momentum vector points along velocity; along an axis, pₓ = mvₓ. Doubling mass or velocity doubles momentum, and rest gives zero momentum. Opposite velocities give opposite momenta: adding only their magnitudes would lose an essential part of the description.

Pushing the same cart with the same force for longer changes its motion more. For constant mass, Newton's second law can be written J⃗ = F⃗Δt = Δp⃗: the net force's impulse equals the body's momentum change. If force varies, use its average F⃗ over the interval. N·s and kg·m/s are equivalent units. Doubling the duration at the same average force doubles the change; zero average net force gives zero net momentum change over the interval.

Catch a ball first with nearly stationary hands, then while moving your hands along with it. If the ball's initial and final velocities are the same in both cases, Δp⃗ is the same. The second stop takes longer, so the average force Fср = Δp/Δt is smaller. Doubling the stopping time halves it. Soft surfaces and airbags work this way: they spread the momentum change over more time rather than eliminate it.

Draw a boundary around both colliding carts. Their forces on each other are now internal to the chosen system. By Newton's third law they are equal and opposite at every instant, and act for the same time. The momentum gained by one cart therefore equals that lost by the other. These changes cancel in the total, although each cart's velocity may change greatly.

Can total momentum still change? Yes, if the system receives an external impulse during the chosen interval. When net external impulse is zero or negligible, p⃗₁ + p⃗₂ = p⃗′₁ + p⃗′₂. A prime marks the state after interaction. A brief impact sometimes lets us neglect external friction even when it exists. Conservation applies to a chosen system and interval, not automatically to every collision.

Send two identical carts towards each other at equal speeds. Their total momentum is zero even though both move: positive and negative components cancel. Choose an axis and assign signs before substituting numbers. The conserved quantity is the sum of pₓ, not a sum of positive mv values. A system at rest after impact can still conserve momentum, but its energy transformations need separate consideration.

After sticking, the carts share velocity u, so m₁v₁ₓ + m₂v₂ₓ = (m₁ + m₂)uₓ. Their combined mass carries the original total momentum. Making the initially stationary second cart heavier reduces the common speed. Kinetic energy usually decreases in such an inelastic collision: some goes into deformation and internal energy. We will study energy conservation next; conserving momentum does not mean conserving each form of energy.

Imagine a rocket far from Earth with no surrounding air. It ejects gas backwards, giving it momentum; in their interaction, the gas gives the rocket an opposite momentum change. The rocket plus expelled gas obeys the same total-momentum law. There is no need to push against air. Excluding the gas and then demanding constant momentum for the accelerating rocket alone would choose the wrong system boundary.

A 3 kg cart moves right at 2 m/s and sticks to a stationary 1 kg cart. Neglect external horizontal impulse.

Choose both carts as the system and right as positive. External horizontal impulse is negligible, so the total momentum component is conserved.

Before impact, pₓ = 3 · 2 + 1 · 0 = 6 кг·м/с. The stationary cart has mass but zero initial momentum.

After sticking, the moving pair has mass 3 + 1 = 4 кг. Its common velocity component is uₓ = 6/4 = 1,5 м/с.

Check: 4 · 1,5 = 6 кг·м/с. Speed is below the original 2 m/s because the same momentum is now carried by a larger moving mass; the sign remains positive.

The carts move together to the right at 1.5 m/s. Their total momentum remains 6 kg·m/s.

Momentum includes mass, velocity and direction; internal forces redistribute it between bodies.

Total momentum is conserved when the chosen system receives no significant external impulse.

Taking longer to stop with the same momentum change requires a smaller average force.

For motion through different heights and for braking, we will compare work and energy.