Conservation of momentum

AS · 12 min

When objects collide, push apart or explode, the forces between them are complicated and brief, but one quantity is perfectly predictable: the total momentum. This note covers the principle of conservation of momentum, elastic and inelastic collisions in one and two dimensions, and the special property of elastic collisions (relative speed of approach equals relative speed of separation). Collision calculations appear in almost every Paper 2, often combined with kinetic energy.

The principle

Definition

The principle of conservation of momentum: the total momentum of a system remains constant provided no resultant external force acts on the system.

A system is the collection of objects you choose to consider: for a collision, both colliding objects. Internal forces (the forces the objects exert on each other) cannot change the total momentum; only an external force can.

Why it works. When A and B collide, Newton's third law says the force of A on B is equal and opposite to the force of B on A at every instant, and the two forces act for the same time. So the change in momentum of B ( $F\Delta t$ ) is equal and opposite to the change in momentum of A. Whatever momentum A loses, B gains: the total is unchanged.

Key result

For two objects colliding in one dimension:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

where $u$ are the velocities before and $v$ the velocities after. Velocities carry signs: choose a positive direction and give velocities in the opposite direction a negative value.

Elastic and inelastic collisions

Momentum is always conserved in a collision (with no external force). Kinetic energy may or may not be.

Definition

In an elastic collision, the total kinetic energy is conserved.

In an inelastic collision, the total kinetic energy is not conserved: some is transferred to other forms (thermal energy, sound, energy of deformation). Total energy is still conserved.

A collision in which the objects stick together is inelastic, and loses the most kinetic energy consistent with conserving momentum. Collisions between gas molecules, and between subatomic particles that do not react, are elastic; collisions between everyday objects are always at least slightly inelastic.

Key result

For an elastic collision:

total kinetic energy before $=$ total kinetic energy after, and
the relative speed of approach equals the relative speed of separation:

u_1 - u_2 = v_2 - v_1

The second statement is much easier to use than the kinetic energy equation, because it is linear. Combined with conservation of momentum it gives two simultaneous linear equations.

Method

Solving a collision problem

Sketch before and after, with arrows and masses. Choose a positive direction.
Write conservation of momentum, with signs.
If the collision is elastic, add the relative-speed equation (or kinetic energy). If the objects stick, they share one final velocity.
Solve. A negative answer means the object moves in the negative direction.
If asked whether a collision is elastic, calculate the total kinetic energy before and after and compare.

Trolleys sticking together

A trolley of mass $2.0\ \text{kg}$ moving at $3.0\ \text{m s}^{-1}$ collides with a stationary trolley of mass $1.0\ \text{kg}$ . They stick together. Find (a) their common velocity and (b) the kinetic energy lost.

Solution

(a) Momentum before $=$ momentum after:

2.0 \times 3.0 + 1.0 \times 0 = (2.0 + 1.0)v \quad\Rightarrow\quad v = \frac{6.0}{3.0} = 2.0\ \text{m s}^{-1}

in the original direction.

(b) $E_K$ before $= \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$ . $E_K$ after $= \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}$ .

Loss $= 3.0\ \text{J}$ , transferred mainly to thermal energy and sound. The collision is inelastic.

Head-on collision

A $3.0\ \text{kg}$ ball moving to the right at $4.0\ \text{m s}^{-1}$ collides head-on with a $2.0\ \text{kg}$ ball moving to the left at $5.0\ \text{m s}^{-1}$ . They stick together. Find their velocity after the collision.

Solution

Right positive. Before: $3.0(4.0) + 2.0(-5.0) = 12 - 10 = 2.0\ \text{N s}$ .

2.0 = 5.0v \quad\Rightarrow\quad v = 0.40\ \text{m s}^{-1} \text{ to the right}

An elastic collision

A ball A of mass $0.50\ \text{kg}$ moving at $4.0\ \text{m s}^{-1}$ makes a head-on elastic collision with a stationary ball B of mass $0.30\ \text{kg}$ . Find the velocities of both balls after the collision.

Solution

Initial direction of A positive.

Momentum: $0.50(4.0) + 0 = 0.50v_A + 0.30v_B$ , so $2.0 = 0.50v_A + 0.30v_B$ .

Elastic, so relative speed of approach $=$ relative speed of separation: $4.0 - 0 = v_B - v_A$ , so $v_B = v_A + 4.0$ .

Substitute:

2.0 = 0.50v_A + 0.30(v_A + 4.0) = 0.80v_A + 1.2 \quad\Rightarrow\quad v_A = 1.0\ \text{m s}^{-1}

v_B = 5.0\ \text{m s}^{-1}

Both move in the original direction. Check kinetic energy: before $\tfrac{1}{2}(0.50)(16) = 4.0\ \text{J}$ ; after $\tfrac{1}{2}(0.50)(1.0) + \tfrac{1}{2}(0.30)(25) = 0.25 + 3.75 = 4.0\ \text{J}$ . Conserved, as it must be.

Tip

Two results worth recognising in Paper 1: in a head-on elastic collision between equal masses, with one at rest, the moving object stops and the other moves off with its velocity. If a light object hits a very heavy stationary one elastically, it bounces back with (almost) the same speed.

Explosions and recoil

When a system starts at rest and pushes itself apart (a gun firing, two trolleys springing apart, a nucleus emitting a particle), the total momentum is zero before, so it is zero after: the parts move off with equal and opposite momenta.

Recoil of a rifle

A rifle of mass $4.0\ \text{kg}$ fires a bullet of mass $10\ \text{g}$ at $400\ \text{m s}^{-1}$ . Find the recoil velocity of the rifle.

Solution

Total momentum before $= 0$ :

0 = 0.010 \times 400 + 4.0v \quad\Rightarrow\quad v = -\frac{4.0}{4.0} = -1.0\ \text{m s}^{-1}

The rifle recoils at $1.0\ \text{m s}^{-1}$ in the opposite direction to the bullet.

Kinetic energy is created here (from chemical energy): the bullet has $\tfrac{1}{2}(0.010)(400)^2 = 800\ \text{J}$ , the rifle only $2.0\ \text{J}$ . Equal momenta, very unequal kinetic energies: the lighter object gets almost all of it.

Since $E_K = \tfrac{1}{2}mv^2 = \dfrac{p^2}{2m}$ , two objects with equal-sized momenta share kinetic energy in the inverse ratio of their masses. This explains why an $\alpha$ -particle emitted from a heavy nucleus carries about $98\%$ of the energy released (see Radioactive decay).

Momentum and energy together

A classic question combines momentum (for the collision) with energy (before or after it). Treat each stage separately: momentum is conserved during the collision; mechanical energy is conserved before or after it if there is no friction. Kinetic energy is generally not conserved during the collision.

Bullet into a hanging block

A bullet of mass $20\ \text{g}$ is fired horizontally into a wooden block of mass $2.0\ \text{kg}$ hanging from long strings. The bullet stays in the block, which swings up through a vertical height of $0.15\ \text{m}$ . Calculate (a) the speed of the block just after impact, (b) the speed of the bullet and (c) the fraction of the bullet's kinetic energy that is lost in the impact.

Solution

(a) After the impact, kinetic energy becomes gravitational potential energy (no friction):

\tfrac{1}{2}Mv^2 = Mgh \quad\Rightarrow\quad v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.15} = 1.72\ \text{m s}^{-1}

(b) During the impact, momentum is conserved (the strings are vertical, so there is no horizontal external force):

0.020u = (2.0 + 0.020)(1.72) \quad\Rightarrow\quad u = \frac{2.02 \times 1.716}{0.020} = 173\ \text{m s}^{-1}

(c) $E_K$ of bullet $= \tfrac{1}{2}(0.020)(173.3)^2 = 300\ \text{J}$ . $E_K$ just after $= \tfrac{1}{2}(2.02)(1.716)^2 = 2.97\ \text{J}$ . Fraction lost $= (300 - 2.97)/300 = 0.99$ : about $99\%$ becomes thermal energy and deformation of the wood.

You cannot use energy conservation across the impact; that is why momentum is needed for part (b).

Collisions in two dimensions

Momentum is a vector, so in two dimensions it is conserved separately in each of two perpendicular directions. Resolve every momentum into components along $x$ and $y$ , and write a conservation equation for each.

Glancing collision of pucks

On a frictionless air table, puck A of mass $0.20\ \text{kg}$ moving at $5.0\ \text{m s}^{-1}$ along the $x$ -axis strikes a stationary puck B of mass $0.30\ \text{kg}$ . After the collision A moves at $3.0\ \text{m s}^{-1}$ at $60^\circ$ above the $x$ -axis. (a) Find the velocity of B. (b) Determine whether the collision is elastic.

Solution

(a) Before: $p_x = 0.20 \times 5.0 = 1.0\ \text{N s}$ , $p_y = 0$ .

A after: $p_x = 0.20 \times 3.0\cos 60^\circ = 0.30\ \text{N s}$ , $p_y = 0.20 \times 3.0\sin 60^\circ = 0.52\ \text{N s}$ .

Conservation in $x$ : $1.0 = 0.30 + 0.30v_{Bx}$ , so $v_{Bx} = 2.33\ \text{m s}^{-1}$ .

Conservation in $y$ : $0 = 0.52 + 0.30v_{By}$ , so $v_{By} = -1.73\ \text{m s}^{-1}$ .

v_B = \sqrt{2.33^2 + 1.73^2} = 2.91\ \text{m s}^{-1}, \qquad \tan\theta = \frac{1.73}{2.33} \Rightarrow \theta = 37^\circ \text{ below the } x\text{-axis}

(b) $E_K$ before $= \tfrac{1}{2}(0.20)(5.0)^2 = 2.50\ \text{J}$ . After $= \tfrac{1}{2}(0.20)(3.0)^2 + \tfrac{1}{2}(0.30)(2.91)^2 = 0.90 + 1.27 = 2.17\ \text{J}$ . Kinetic energy is lost, so the collision is inelastic.

The momentum vectors form a closed triangle: the initial momentum of A equals the vector sum of the two final momenta.

(0, 0) -> (1, 0) (0, 0) -> (0.3, 0.52) (0.3, 0.52) -> (1, 0)

The horizontal arrow is A's initial momentum ( $1.0\ \text{N s}$ ). Adding A's final momentum (up to the right) and B's final momentum (down to the right) head to tail returns to the same point.

Watch out

Do not add speeds or momenta in two dimensions as if they were in a line. "Total momentum after $= 0.20 \times 3.0 + 0.30v_B$ " is wrong unless both move along the same line. Resolve first.

Momentum is conserved; kinetic energy may not be

The syllabus stresses this: in every interaction between objects (with no external force), momentum is conserved, while some change in kinetic energy may take place. Kinetic energy can decrease (inelastic collision, energy to heat and sound) or increase (explosion, energy from chemical or nuclear stores). Total energy is always conserved.

Exam tip

State the principle with its condition: "...provided no resultant external force acts". Omitting the condition loses the mark.
"Show that the collision is inelastic" needs numerical kinetic energies before and after, compared. "Because the objects stick together" alone is not enough unless the question allows it.
Set out momentum equations with a clear sign convention and units: "taking right as positive, $3.0 \times 4.0 + 2.0 \times (-5.0) = 5.0v$ ".
For "explain why momentum is conserved", use Newton's third law: equal and opposite forces, for equal times, give equal and opposite changes of momentum.
Two-dimensional problems: resolve in two perpendicular directions and write two equations. A labelled vector diagram earns credit.

Summary

Total momentum of a system is constant provided no resultant external force acts.
It follows from Newton's third law: equal and opposite forces for equal times.
Elastic collision: total kinetic energy conserved, and relative speed of approach $=$ relative speed of separation.
Inelastic collision: kinetic energy not conserved (total energy still is); sticking together is the extreme case.
Explosions from rest: equal and opposite momenta; the lighter part gets more kinetic energy ( $E_K = p^2/2m$ ).
In two dimensions, conserve momentum separately in two perpendicular directions.
Use momentum across a collision; use energy before or after it.

Practice

Question

State the principle of conservation of momentum.
A $1200\ \text{kg}$ car moving at $15\ \text{m s}^{-1}$ hits a stationary $800\ \text{kg}$ car and they move off together. Find their common velocity and the kinetic energy lost.
A $60\ \text{kg}$ skater at rest pushes a $40\ \text{kg}$ skater, who moves off at $1.5\ \text{m s}^{-1}$ . Find the velocity of the first skater, and the total kinetic energy produced.
A $0.10\ \text{kg}$ ball moving at $6.0\ \text{m s}^{-1}$ collides head-on with a $0.20\ \text{kg}$ ball moving at $3.0\ \text{m s}^{-1}$ in the opposite direction. After the collision the $0.10\ \text{kg}$ ball moves back at $4.0\ \text{m s}^{-1}$ . Find the velocity of the $0.20\ \text{kg}$ ball and determine whether the collision is elastic.
Two identical gliders on an air track approach each other at $2.0\ \text{m s}^{-1}$ and $1.0\ \text{m s}^{-1}$ and collide elastically. Find their velocities after the collision.
Explain, using Newton's laws, why momentum is conserved when two objects collide.
A stationary nucleus of mass $3.90 \times 10^{-25}\ \text{kg}$ emits an $\alpha$ -particle of mass $6.64 \times 10^{-27}\ \text{kg}$ at $1.50 \times 10^{7}\ \text{m s}^{-1}$ . Calculate the recoil speed of the remaining nucleus and the ratio of the kinetic energy of the $\alpha$ -particle to that of the nucleus.
A ball of mass $0.40\ \text{kg}$ moving at $5.0\ \text{m s}^{-1}$ due east collides with a ball of mass $0.60\ \text{kg}$ moving at $2.0\ \text{m s}^{-1}$ due north. They stick together. Find the magnitude and direction of their common velocity.
A $2.0\ \text{kg}$ trolley A moving at $3.0\ \text{m s}^{-1}$ collides elastically and head-on with a stationary trolley B of mass $m$ . After the collision A moves at $1.0\ \text{m s}^{-1}$ in its original direction. Find $m$ and the velocity of B.
A firework of mass $0.60\ \text{kg}$ moving vertically upwards at $12\ \text{m s}^{-1}$ explodes into two pieces. A $0.20\ \text{kg}$ piece moves horizontally at $30\ \text{m s}^{-1}$ immediately after the explosion. Find the velocity (magnitude and direction) of the $0.40\ \text{kg}$ piece.

Answers

The total momentum of a system remains constant provided no resultant external force acts on it.
$1200 \times 15 = 2000v$ , so $v = 9.0\ \text{m s}^{-1}$ . $E_K$ before $= \tfrac{1}{2}(1200)(225) = 135\,000\ \text{J}$ ; after $= \tfrac{1}{2}(2000)(81) = 81\,000\ \text{J}$ ; lost $= 5.4 \times 10^{4}\ \text{J}$ .
$0 = 40(1.5) + 60v$ , so $v = -1.0\ \text{m s}^{-1}$ : $1.0\ \text{m s}^{-1}$ in the opposite direction. $E_K = \tfrac{1}{2}(40)(2.25) + \tfrac{1}{2}(60)(1.0) = 45 + 30 = 75\ \text{J}$ .
Original direction of the $0.10\ \text{kg}$ ball positive. Before: $0.10(6.0) + 0.20(-3.0) = 0$ . After: $0.10(-4.0) + 0.20v = 0$ , so $v = 2.0\ \text{m s}^{-1}$ (in the positive direction). $E_K$ before $= 1.8 + 0.9 = 2.7\ \text{J}$ ; after $= 0.8 + 0.4 = 1.2\ \text{J}$ . Inelastic.
For equal masses colliding elastically head-on, the velocities are exchanged: the first moves back at $1.0\ \text{m s}^{-1}$ and the second moves back at $2.0\ \text{m s}^{-1}$ (each reverses with the other's speed). Check: momentum $m(2.0 - 1.0) = m(-1.0 + 2.0)$ ; relative speeds $3.0 = 3.0$ .
During the collision, A exerts a force on B and B exerts an equal and opposite force on A (Newton's third law), for the same time. Force $=$ rate of change of momentum (second law), so the changes in momentum of A and B are equal in size and opposite in direction. The total change is zero, so total momentum is unchanged.
Recoiling nucleus mass $= 3.90 \times 10^{-25} - 6.64 \times 10^{-27} = 3.834 \times 10^{-25}\ \text{kg}$ . $v = \dfrac{6.64 \times 10^{-27} \times 1.50 \times 10^{7}}{3.834 \times 10^{-25}} = 2.60 \times 10^{5}\ \text{m s}^{-1}$ . Equal momenta, so $E_{K\alpha}/E_{K\text{N}} = m_\text{N}/m_\alpha = 3.834 \times 10^{-25}/6.64 \times 10^{-27} = 57.7$ .
$p_x = 2.0\ \text{N s}$ , $p_y = 1.2\ \text{N s}$ . $p = 2.33\ \text{N s}$ ; $v = 2.33/1.00 = 2.3\ \text{m s}^{-1}$ at $\tan^{-1}(1.2/2.0) = 31^\circ$ north of east.
Relative speeds: $3.0 - 0 = v_B - 1.0$ , so $v_B = 4.0\ \text{m s}^{-1}$ . Momentum: $2.0(3.0) = 2.0(1.0) + 4.0m$ , so $m = 1.0\ \text{kg}$ .
Before: $p_y = 0.60 \times 12 = 7.2\ \text{N s}$ , $p_x = 0$ . Small piece: $p_x = 0.20 \times 30 = 6.0\ \text{N s}$ . Large piece: $p_x = -6.0\ \text{N s}$ , $p_y = 7.2\ \text{N s}$ , so $v_x = -15\ \text{m s}^{-1}$ , $v_y = 18\ \text{m s}^{-1}$ . Speed $= \sqrt{15^2 + 18^2} = 23.4\ \text{m s}^{-1}$ at $\tan^{-1}(18/15) = 50^\circ$ above the horizontal, on the opposite side to the small piece.

The principle

Elastic and inelastic collisions

Explosions and recoil

Momentum and energy together

Collisions in two dimensions

Momentum is conserved; kinetic energy may not be

Summary

Practice

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