Lesson 5 · 35 min
Conservation of Linear Momentum
When two bodies push on each other, each gets an impulse, and the two impulses are equal and opposite. Add the momenta of the bodies, and those impulses cancel. If nothing outside the bodies gives them a significant impulse, their total momentum does not change, however violent the push. That one fact settles explosions, recoils, couplings and, in the next lessons, collisions.
Learning objectives
- Apply the principle of impulse and momentum to a system of particles, and explain why internal forces drop out.
- Decide when, and in which direction, the momentum of a system is conserved.
- Tell impulsive forces from nonimpulsive ones during a short interaction.
- Apply conservation of momentum to bodies that push apart or join together, including speeds given relative to a moving body, and check what happens to the kinetic energy.
A system of particles
Write the principle of impulse and momentum for each particle of a system and add the equations. The forces on each particle are of two kinds: external forces, from outside the system, and internal forces, from the other particles. By Newton's third law, internal forces come in equal, opposite, collinear pairs, and they act for the same time, so their impulses cancel in pairs. Only the external forces remain:
Impulse and momentum for a system of particles
\[ \textstyle\sum \colL{m_i(\vvec_i)_1} + \sum\displaystyle\int_{t_1}^{t_2}\Fvec_i\,dt = \textstyle\sum \colL{m_i(\vvec_i)_2} \]The sum of the impulses includes the external forces only.
The total momentum is the total mass times the velocity of the mass center: \(\sum m_i\vvec_i = m\vvec_G\). So the external impulses change the motion of the mass center, and internal forces never can. A person standing on a cart cannot move the pair's mass center by pushing on the cart.
When momentum is conserved
If the sum of the external impulses on a system is zero, or small enough to neglect, in some direction, then the system's momentum in that direction does not change:
Conservation of linear momentum
\[ \textstyle\sum \colL{m_i(\vvec_i)_1} = \sum \colL{m_i(\vvec_i)_2} \]Use it one direction at a time: momentum can be conserved horizontally while it changes vertically.
Two situations make the external impulse negligible:
- No external force in that direction. On a smooth floor nothing pushes horizontally, so the horizontal momentum of carts, skaters or pucks is conserved for as long as you like.
- A very short interaction. Forces that stay finite, such as weights, springs and ordinary friction, are nonimpulsive: over a few milliseconds their impulse is tiny. Forces that are very large and very brief, such as those between colliding bodies, the push of a blast or the pull of a cord that suddenly goes taut, are impulsive. During a collision only the impulsive forces count.
| Usually nonimpulsive (neglect) | Impulsive (keep) |
|---|---|
| Weights | Contact forces between colliding bodies |
| Spring forces (a spring cannot change its force instantly) | The push of a rigid floor, wall or stop that is struck |
| Ordinary friction on a floor | The pull of an inextensible cord that jerks taut |
| Air drag | The gases of an explosion or a gun's charge |
Bodies that push apart
Explosions, recoils and people jumping off carts have the same structure: the bodies start together, internal forces push them apart, and the total momentum stays what it was.
Example 5.1 — Jumping off a cart
A \(60\ \text{kg}\) person stands on a \(40\ \text{kg}\) cart at rest on a smooth floor, then jumps off horizontally at \(3\ \text{m/s}\) relative to the ground. (a) How fast does the cart move? (b) What if the person jumps at \(3\ \text{m/s}\) relative to the cart?
Show solution
(a) No horizontal external force acts on the person and the cart together, so their horizontal momentum, zero to start with, stays zero. Positive in the direction of the jump:
\[ 0 = 60(3) + 40\,v_c \quad\Rightarrow\quad v_c = -4.5\ \text{m/s} \]The cart rolls the other way at \(4.5\ \text{m/s}\). Meanwhile the kinetic energy rose from \(0\) to \(\tfrac12(60)(3)^2 + \tfrac12(40)(4.5)^2 = 675\ \text{J}\), supplied by the person's muscles. Momentum is conserved and energy is not: the two principles are independent.
(b) Now the person's velocity is \(v_p = v_c + 3\). Convert to ground velocities before you add momenta:
\[ 0 = 60(v_c + 3) + 40\,v_c \quad\Rightarrow\quad v_c = -1.8\ \text{m/s},\qquad v_p = 1.2\ \text{m/s} \]The cart's recoil takes some of the "3 m/s", so the person leaves more slowly than in (a).
In Figure 5.1 the spring's force on \(A\) and its force on \(B\) are internal to the pair, equal and opposite at every instant, so the two impulses are too. Make \(B\) four times as heavy as \(A\): \(A\) leaves four times as fast, and with four times the kinetic energy, since \(T = (mv)^2/2m\) and the two momenta are equal in size.
Example 5.2 — A cannon's recoil
A \(1500\ \text{kg}\) cannon, free to roll on smooth level rails, fires a \(10\ \text{kg}\) shell at \(600\ \text{m/s}\) relative to the ground, at \(30^\circ\) above the horizontal. Find the cannon's recoil velocity just after firing.
Show solution
Vertically, the rails push up on the cannon with a large, brief force while the shell is fired: an impulsive external force. So vertical momentum is not conserved. Horizontally, the rails are smooth, and the only horizontal forces are the internal ones between shell and cannon. Positive in the direction the shell travels:
\[ 0 = 10(600\cos 30^\circ) + 1500\,v_c \quad\Rightarrow\quad v_c = -3.46\ \text{m/s} \]The vertical momentum the shell carries away, \(10(600\sin 30^\circ) = 3000\ \text{kg·m/s}\) upward, is supplied by the rails' impulse.
Bodies that join
When bodies stick together, the final state has one unknown velocity, and conservation of momentum gives it directly. Kinetic energy is always lost: it goes into deforming the couplers, and into sound and heat.
Example 5.3 — Railcars coupling
A \(20\ \text{Mg}\) (\(20\,000\ \text{kg}\)) railcar \(A\) rolling at \(1.5\ \text{m/s}\) runs into a \(15\ \text{Mg}\) railcar \(B\) at rest, and the two couple together. (a) What is their speed after coupling? (b) How much kinetic energy is lost? (c) If the coupling takes \(0.5\ \text{s}\), what is the average force between the cars?
Show solution
(a) The rolling resistance is nonimpulsive and the track is level, so the momentum of the two cars is conserved through the coupling:
\[ 20\,000(1.5) + 0 = 35\,000\,v \quad\Rightarrow\quad v = 0.857\ \text{m/s} \](b) \(T_1 = \tfrac12(20\,000)(1.5)^2 = 22.5\ \text{kJ}\) and \(T_2 = \tfrac12(35\,000)(0.857)^2 = 12.9\ \text{kJ}\), so \(9.6\ \text{kJ}\) is lost: \(43\%\) of the original. With \(B\) at rest, the fraction lost is \(m_B/(m_A + m_B) = 15/35\).
(c) The coupling force is internal to the pair but external to car \(B\). Impulse and momentum for \(B\) alone:
\[ 0 + F_{\text{avg}}(0.5) = 15\,000(0.857) \quad\Rightarrow\quad F_{\text{avg}} = 25.7\ \text{kN} \]Example 5.4 — A block sliding onto a cart
A \(2\ \text{kg}\) block slides at \(5\ \text{m/s}\) onto an \(8\ \text{kg}\) cart at rest on a smooth floor. The coefficient of kinetic friction between the block and the cart is \(0.3\). (a) What is their common velocity once the block stops sliding on the cart? (b) How long does the block slide on the cart? (c) How much energy does friction dissipate?
Show solution
(a) Friction between block and cart is internal to the pair, and the floor is smooth. Momentum is conserved horizontally, however long the sliding takes:
\[ 2(5) + 0 = (2 + 8)\,v \quad\Rightarrow\quad v = 1\ \text{m/s} \](b) For the block alone, friction is external: \(\colF{F_f = 0.3(2)(9.81) = 5.886\ \text{N}}\), backward.
\[ 2(5) \colF{{} - 5.886\,t} = 2(1) \quad\Rightarrow\quad t = 1.36\ \text{s} \](c) \(T_1 = \tfrac12(2)(5)^2 = 25\ \text{J}\) and \(T_2 = \tfrac12(10)(1)^2 = 5\ \text{J}\): friction turns \(20\ \text{J}\) into heat. That equals \(F_f\) times the distance the block slides relative to the cart, \(3.40\ \text{m}\).
Check your understanding
Key takeaways
- For a system, \(\sum m_i(\vvec_i)_1 + \sum\int\Fvec\,dt = \sum m_i(\vvec_i)_2\) with external forces only: internal impulses cancel in pairs.
- Momentum is conserved in any direction in which the external impulse is zero or negligible. Check each direction separately.
- During a short interaction, neglect nonimpulsive forces (weights, springs, ordinary friction); keep impulsive ones (contact, rigid supports, taut cords).
- Convert relative velocities to ground velocities before adding momenta.
- Conservation of momentum says nothing about energy: bodies that push apart gain kinetic energy, bodies that join lose it.
- Next, Lesson 6 analyzes collisions with the coefficient of restitution.