Conservation of Linear Momentum - Ninth Grade Science
Among the grand conservation principles that structure modern physical science, the Law of Conservation of Linear Momentum occupies a preeminent position alongside the conservation of energy and electric charge. While individual forces can fluctuate wildly during complex collisions, explosions, or orbital maneuvers, the total momentum of an isolated system remains strictly constant. This law allows physicists to analyze chaotic, high-speed events—such as colliding automobiles, subatomic particle disintegrations, or ballistic impacts—without needing to track every intricate microsecond of the collision forces. In this chapter, you will derive the law of conservation of momentum directly from Newton's laws, classify elastic and inelastic collisions, and master quantitative collision calculations.
The Principle of Conservation of Momentum
The Law of Conservation of Linear Momentum states that if no net external unbalanced force acts upon a system of interacting bodies, the total linear momentum of the system remains constant over time.
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| Law of Conservation of Linear Momentum |
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| In an isolated system (net external force = 0), the total vector |
| momentum before an interaction equals the total vector momentum after |
| the interaction: |
| |
| Total Initial Momentum = Total Final Momentum |
| Sigma p_initial = Sigma p_final |
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What Defines an "Isolated System"?
An isolated system is a collection of interacting physical bodies that experiences zero net external force from outside the system boundary ($\Sigma F_{ ext{ext}} = 0$). Internal forces—the forces that the interacting bodies exert upon one another during collisions or explosions—can be arbitrarily large, but because they occur in equal and opposite Newton's third law pairs, their vector sum within the system is identically zero. Only an external force, such as friction or gravity from an outside body, can alter the total momentum of the system.
Derivation of the Conservation Law from Newton's Laws
The conservation of momentum is not an independent postulate; it is a direct mathematical consequence of Newton's second and third laws of motion.
Two Colliding Spheres (Body A and Body B):
Before Collision: [ m_A, u_A ] ------> [ m_B, u_B ] ------>
During Collision: [ m_A ] <-- F_AB -- F_BA --> [ m_B ]
After Collision: [ m_A, v_A ] ------> [ m_B, v_B ] ------>
Consider two spherical bodies, body A of mass $m_A$ and body B of mass $m_B$, moving along a straight line with initial velocities $u_A$ and $u_B$. Assume $u_A > u_B$ so that body A collides with body B. The collision lasts for a time duration $\Delta t$.
Step-by-Step Derivation
During the collision, body A exerts an action force $F_{BA}$ on body B. According to Newton's Second Law, this force equals the rate of change of momentum of body B:
F_{BA} = m_B \left( \frac{v_B - u_B}{\Delta t} \right)
Simultaneously, by Newton's Third Law, body B exerts an equal and opposite reaction force $F_{AB}$ on body A:
F_{AB} = m_A \left( \frac{v_A - u_A}{\Delta t} \right)
Because action and reaction forces are equal in magnitude and opposite in direction:
F_{AB} = -F_{BA}
Substitute the momentum rates into this equality:
m_A \left( \frac{v_A - u_A}{\Delta t} \right) = - m_B \left( \frac{v_B - u_B}{\Delta t} \right)
Because the collision duration $\Delta t$ is identical for both bodies, multiply both sides by $\Delta t$:
m_A (v_A - u_A) = - m_B (v_B - u_B)
Expand both sides:
m_A v_A - m_A u_A = - m_B v_B + m_B u_B
Rearrange the terms by collecting initial momentum values on the left and final momentum values on the right:
m_A u_A + m_B u_B = m_A v_A + m_B v_B
This mathematical identity demonstrates that the total initial momentum of the system before the collision is identical to the total final momentum after the collision.
Types of Collisions: Elastic versus Inelastic
Although linear momentum is conserved in all isolated collisions, mechanical kinetic energy behaves differently depending on the nature of the materials.
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| Elastic Collisions | Inelastic Collisions |
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| Linear momentum is conserved | Linear momentum is conserved |
| Total kinetic energy is conserved | Total kinetic energy is NOT |
| No permanent deformation or heat | conserved (converted to heat/sound)|
| Example: Subatomic particles, | Example: Car crashes, clay balls, |
| billiard balls (nearly) | ballistic pendulum |
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Perfectly Inelastic Collisions
A collision is called perfectly inelastic when the colliding bodies stick together upon impact and move forward as a single combined mass with a common final velocity ($v_f$). For a perfectly inelastic collision, the conservation equation simplifies to:
m_A u_A + m_B u_B = (m_A + m_B) v_f \implies v_f = \frac{m_A u_A + m_B u_B}{m_A + m_B}
In an inelastic collision, the "missing" kinetic energy is not destroyed; it transforms into thermal energy, acoustic sound waves, and internal work that causes permanent mechanical deformation.
Explosions and Recoil Mechanics
The conservation of momentum applies with equal power to explosions and firearm recoil, where a single resting system splits into multiple fragments due to internal chemical energy.
Recoil of a Rifle:
System Before Firing (At Rest): Total Momentum = 0
System After Firing:
Bullet: [ m_bullet, v_bullet ] ------------------------> Forward
Rifle: <------------------------ [ M_gun, V_recoil ] Backward
Total Momentum = M_gun * V_recoil + m_bullet * v_bullet = 0
Calculating Recoil Velocity
Before a firearm is discharged, both the rifle (mass $M$) and the bullet (mass $m$) are at rest, so the initial total momentum is zero. When the gunpowder detonates, internal chemical forces drive the bullet forward at high velocity $v$. To keep the total system momentum at zero, the rifle must recoil backward with velocity $V$:
0 = M V + m v \implies V = -\frac{m v}{M}
The negative sign indicates that the rifle moves in the opposite direction to the bullet. Because the mass of the rifle $M$ is vastly greater than the mass of the bullet $m$, the recoil velocity $V$ is much smaller than the bullet's velocity $v$, allowing a shooter to absorb the recoil safely.
Chapter Exercises
Exercise 1: A freight railcar of mass 12000 kilograms rolls along a straight frictionless track at 2.5 meters per second. It collides with and automatically couples to a stationary second railcar having a mass of 8000 kilograms. Calculate the common velocity of the coupled railcars after the collision.
Exercise 2: A hunting rifle with a mass of 4.5 kilograms fires a 0.015-kilogram lead bullet with a muzzle velocity of 600 meters per second. Assuming the rifle is free to move horizontally without external restraint, calculate the recoil velocity of the rifle, and explain why the shooter's shoulder experiences a force during the discharge.
Exercise 3: Two ice skaters stand facing each other on frictionless ice at complete rest. Skater 1 has a mass of 50 kilograms, and Skater 2 has a mass of 75 kilograms. Skater 1 pushes Skater 2 away, causing Skater 2 to glide backward with a velocity of 2.0 meters per second. Determine the resulting velocity and direction of motion of Skater 1.
Solutions and Explanations
Solution 1: This interaction is a perfectly inelastic collision because the two railcars latch together and move as a single combined mass. Identify the known parameters: $m_1 = 12000 ext{ kg}$, $u_1 = 2.5 ext{ m/s}$, $m_2 = 8000 ext{ kg}$, and $u_2 = 0 ext{ m/s}$. The total momentum before the collision is $m_1 u_1 + m_2 u_2 = (12000 imes 2.5) + (8000 imes 0) = 30000 + 0 = 30000 ext{ kg}\cdot ext{m/s}$. By the law of conservation of momentum, the total momentum after the collision must also be $30000 ext{ kg}\cdot ext{m/s}$. The combined mass after coupling is $12000 + 8000 = 20000 ext{ kg}$. The common final velocity $v_f$ is $30000 / 20000 = 1.5$ meters per second in the original direction of motion.
Solution 2: The rifle-bullet system is initially at rest, so initial total momentum is zero. Let the forward direction of the bullet be positive. Knowns: rifle mass $M = 4.5 ext{ kg}$, bullet mass $m = 0.015 ext{ kg}$, bullet velocity $v = +600 ext{ m/s}$. The momentum of the bullet after firing is $m imes v = 0.015 imes 600 = +9.0 ext{ kg}\cdot ext{m/s}$. By conservation of momentum, $M V + mv = 0$, so $4.5 V + 9.0 = 0$, giving $4.5 V = -9.0$, which simplifies to $V = -9.0 / 4.5 = -2.0$ meters per second. The negative sign indicates that the rifle recoils backward toward the shooter at 2.0 meters per second. The shooter's shoulder experiences a force because it must exert an external stopping force over a short time interval to decelerate the recoiling rifle back to rest.
Solution 3: The two skaters constitute an isolated system on frictionless ice with zero initial total momentum ($p_{ ext{initial}} = 0$). When Skater 1 pushes Skater 2, internal muscular forces act between them. Skater 2 (mass $m_2 = 75 ext{ kg}$) acquires a velocity of $v_2 = +2.0 ext{ m/s}$, gaining a momentum of $m_2 v_2 = 75 imes 2.0 = +150 ext{ kg}\cdot ext{m/s}$. By conservation of momentum, the total final momentum must equal zero: $m_1 v_1 + m_2 v_2 = 0$, which gives $50 v_1 + 150 = 0$. Solving for $v_1$ yields $50 v_1 = -150$, resulting in $v_1 = -3.0$ meters per second. Therefore, Skater 1 moves backward in the opposite direction at a speed of 3.0 meters per second. Skater 1 moves faster than Skater 2 because Skater 1 has less mass.