Conservation of momentum

Forces and energy can solve most problems, but collisions are hard to handle with forces: the forces are huge, change rapidly and last only milliseconds. Momentum sidesteps all of that. If the system is closed, the total momentum before equals the total momentum after, whatever happened in between.

Momentum

The momentum of an object is its mass times its velocity:

$$\vec p = m\vec v$$

Momentum is a vector, in the direction of the velocity, with units kg m s⁻¹ (equivalent to N s). A 1500 kg car at 20 m s⁻¹ has $3.0 \times 10^4$ kg m s⁻¹; a 0.16 kg cricket ball at 40 m s⁻¹ has 6.4 kg m s⁻¹. Because it is a vector, direction matters: a ball moving left has negative momentum if right is positive.

Newton actually stated his second law in terms of momentum. The net force equals the rate of change of momentum:

$$\vec F_{net} = \frac{\Delta \vec p}{\Delta t}$$

For constant mass this is just $\vec F_{net} = m\dfrac{\Delta\vec v}{\Delta t} = m\vec a$.

Why momentum is conserved

Consider two objects A and B colliding. During the collision, A pushes B and B pushes A. By the third law these forces are equal and opposite at every instant, and they act for exactly the same time. So

$$\Delta\vec p_B = \vec F_{AB}\,\Delta t = -\vec F_{BA}\,\Delta t = -\Delta\vec p_A$$

Whatever momentum B gains, A loses. The total doesn't change. This holds for any closed system (also called isolated): one where no net external force acts.

$$\Sigma m\vec v_{before} = \Sigma m\vec v_{after}$$

In practice, systems are "closed enough" when external forces like friction are small compared with the collision forces, or when the collision is so brief that external forces have no time to change the momentum much.

Collisions and explosions in one dimension

Worked example: coupling trolleys

A 2.0 kg trolley moving at 3.0 m s⁻¹ runs into a stationary 1.0 kg trolley, and they couple together. Find their common velocity.

$\Sigma p = 2.0 \times 3.0 + 1.0 \times 0 = 6.0$ kg m s⁻¹.

before

$\Sigma p = (2.0 + 1.0)v = 3.0v$.

after

$3.0v = 6.0$, so $v = 2.0$ m s⁻¹ in the original direction.

conserve

Worked example: a head-on collision

A 1200 kg car travelling east at 15 m s⁻¹ collides head-on with a 1800 kg van travelling west at 10 m s⁻¹. They lock together. Find the velocity of the wreckage just after the collision.

$\Sigma p = 1200(15) + 1800(-10) = 18\,000 - 18\,000 = 0$.

east is positive

$(3000)v = 0$, so $v = 0$: the wreckage stops dead on impact.

after

Total momentum can be zero even when both objects are moving. All the kinetic energy (about $2.25 \times 10^5$ J) goes into deforming the vehicles.

lesson

In an explosion or recoil, a system at rest pushes itself apart. The total momentum is zero before, so it is zero after: the pieces move off with equal and opposite momenta.

Worked example: astronaut and wrench

An 85 kg astronaut (including suit) at rest relative to the station throws a 1.2 kg wrench away from the station at 8.0 m s⁻¹. How fast does the astronaut move towards the station?

Total momentum = 0.

before

$1.2(8.0) + 85v = 0$, so $v = -\dfrac{9.6}{85} = -0.11$ m s⁻¹.

after

0.11 m s⁻¹ towards the station: slow, but enough to cover 20 m in about three minutes.

interpret

Collisions in two dimensions

Momentum is a vector, so it is conserved separately in each direction. Resolve every momentum into $x$ and $y$ components and write two conservation equations:

$$\Sigma p_x \text{ before} = \Sigma p_x \text{ after} \qquad \Sigma p_y \text{ before} = \Sigma p_y \text{ after}$$

Alternatively, draw the momentum vectors after the collision tip to tail: they must add up to the total momentum before.

Worked example: a glancing collision

A 0.17 kg billiard ball moving at 2.0 m s⁻¹ strikes an identical stationary ball off-centre. Afterwards the first ball moves at 1.2 m s⁻¹ at 53° to its original direction. Find the velocity of the second ball.

$0.17(2.0) = 0.17(1.2\cos53°) + 0.17v_x$, so $v_x = 2.0 - 0.722 = 1.28$ m s⁻¹.

x (original direction)

$0 = 0.17(1.2\sin53°) + 0.17v_y$, so $v_y = -0.958$ m s⁻¹ (the other side of the original line).

y

$v = \sqrt{1.28^2 + 0.958^2} = 1.6$ m s⁻¹, at $\tan^{-1}(0.958/1.28) = 37°$ on the other side of the original direction.

combine

53° + 37° = 90°. For equal masses in an elastic glancing collision, the two balls always separate at right angles.

notice

Investigating collisions

InteractiveTwo-dimensional collisions on an air table

Try this

• With equal 1.0 kg pucks, elastic, aim offset 0.5, press Launch A. A leaves at 1.00 m/s and B at 1.73 m/s, exactly 90.0° apart. Momentum stays 2.00 kg m/s and KE stays 2.00 J.
• Set the offset to 0: head-on, A stops dead and B leaves at 2.00 m/s.
• Choose Stick together: they move off at 1.00 m/s and keep only 50% of the kinetic energy, but all of the momentum.
• Make A 2.0 kg and B 1.0 kg, head-on and elastic: A keeps going at 0.67 m/s and B leaves at 2.67 m/s.

Test yourself

1. A 0.050 kg ball moving at 4.0 m s⁻¹ east hits a wall and rebounds at 3.0 m s⁻¹ west. Its change in momentum is

   0.050 kg m s⁻¹ east0.35 kg m s⁻¹ west0.050 kg m s⁻¹ west0.35 kg m s⁻¹ east

2. A 60 kg skater at rest pushes off a 40 kg skater, who moves away at 1.5 m s⁻¹. The 60 kg skater moves at

   1.5 m s⁻¹2.3 m s⁻¹0.60 m s⁻¹1.0 m s⁻¹

3. Momentum is conserved in a collision because

   the forces the objects exert on each other are equal and opposite and act for the same timekinetic energy is conservedthe objects have equal massesfriction is always zero

4. A 3.0 kg object moving east at 4.0 m s⁻¹ collides with a 2.0 kg object moving north at 3.0 m s⁻¹. They stick together. The size of their momentum afterwards is

   18 kg m s⁻¹6.0 kg m s⁻¹13 kg m s⁻¹3.6 kg m s⁻¹

Try again

Label the momentum vectors for the glancing collision.

CheckStart again

Order the steps to solve a two-dimensional collision.

Draw before and after diagrams and choose x and y axes Resolve each known momentum into x and y components Write conservation of momentum for the x direction Write conservation of momentum for the y direction Solve for the unknown components and combine them with Pythagoras

CheckShuffle

Momentum is the product of mass and , and it is a quantity. In a system the total momentum stays constant. This follows from Newton's law. In two dimensions, momentum is conserved separately in each .

CheckStart again

6 marks

A 1500 kg car travelling north at 12 m s⁻¹ collides at an intersection with a 2500 kg ute travelling east at 9.0 m s⁻¹. The vehicles lock together. (a) Calculate the velocity of the wreckage immediately after the collision. (4 marks) (b) Explain why momentum can be treated as conserved even though friction from the road acts on the vehicles. (2 marks)

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