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4.3Conservation of momentum

Consider a system with momentum p→.

The net force on the system is F→net =Δ p→Δ t.

If F→net =0, it follows that Δ p→=0.

KEY POINT

This means that when the net force on a system is zero the momentum does not change (it stays the same); that is, it is conserved.

This is the law of conservation of momentum when the net force on a system is zero, the total momentum of the system remains constant.

Notice that ‘system’ may refer to a single body or a collection of many different bodies. Look at Figure 4.10. Consider the blue block of mass 4.0 kg moving at speed 6.0 m s–1 to the right. It will collide with the red block of mass 8.0 kg that is initially at rest. Suppose that, after the collision, the bodies stick together and move as one.

Figure 4.10: In a collision (or explosion) with no external forces acting the total momentum of the system stays the same.

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As the blocks collide, each exerts a force on the other. The magnitude of the force on each block is the same by Newton’s third law. There are no forces that come from outside the system (no external forces). You might say that the weights of the blocks are forces that come from the outside. That is correct, but the weights are cancelled by the normal forces from the table. So, the net external force on the system is zero. Hence we expect that the total momentum will stay the same.

The total momentum before the collision is

4.0×6.0+8.0×0=24 N s

The total momentum after the collision is (4.0 + 8.0) × v where v is the common speed of the two blocks.

So, 12v = 24 ⇒ v = 2.0 m s–1.

The kinetic energy before the collision is 1 2×4.0×6.0 2=72 J.

The kinetic energy after the collision is 1 2×12×2.0 2=24 J.

It appears that 48 J has been ‘lost’ (as other forms of energy, such as thermal energy in the blocks themselves and the surrounding air, or energy to deform the bodies during the collision and some to sound generated in the collision).

Now consider the outcome of the collision of these two blocks in which the blue block rebounds with speed 2.0 m s–1, as in Figure 4.11.

Figure 4.11: An outcome of the collision in which total kinetic energy stays the same.

What is the speed of the red block in Figure 4.11?

The total momentum before the collision is 24 N s.

The total momentum after the collision is –4.0 × 2.0 + 8.0 × v

(Note the minus sign.)

Equating the total momentum before and after the collision we find –8.0 + 8.0 × v = 24.

This gives

v=4.0 m s−1.

The total kinetic energy after the collision is

1 2×4.0×(−2.0)2+1 2×8.0×4.0 2=72 J

which is the same as the initial kinetic energy.

So, in a collision, the momentum is always conserved, but kinetic energy may or may not be conserved. There is more discussion of momentum and kinetic energy in Section 4.4.

It all depends on the system

Consider a ball that you drop from rest from a certain height. As the ball falls, its speed and momentum increase so the momentum of the ball does not stay the same, Figure 4.12.

Figure 4.12: As the ball falls, an external force acts on it (its weight), increasing its momentum.

This is to be expected—there is an external force on the ball, namely its weight. So the momentum of the system that consists of just the falling ball is not conserved. If we include the earth as part of the system, then there are no external forces anymore and the total momentum will be conserved. This means that the earth moves upward a bit as the ball falls!

THEORY OF KNOWLEDGE

Predicting outcomes

Physics is supposed to be able to predict outcomes. So, why is there more than one outcome in the collision of Figure 4.11?

Physics does predict what happens, but more information about the nature of the colliding bodies is needed. We need to know if they are soft or hard, deformable or not, sticky or breakable, and so on. If this information is given, physics will uniquely predict what will happen.

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EXTENSION

The rocket equation

One example of motion with varying mass is the rocket (Figure 4.13). The engines exert a force on the burnt gases pushing them out of the rocket and, by Newton’s third law, the gases exert an equal and opposite force on the engines (that is, the rocket).

Figure 4.13: Exhaust gases from the booster rockets propel this space shuttle during its launch.

This is quite a complex topic and is included here only as optional material. The rocket moves with speed v. All speeds are measured by an observer at rest on the ground. The engine is turned on, and gases leave the rocket with speed u relative to the rocket. The initial mass of the rocket including the fuel is M. After a short time δt the rocket has ejected fuel of mass δm. The mass of the rocket is, therefore, reduced to M – δm and its speed increased to v + δv (see Figure 4.14).

Figure 4.14: Deriving the rocket equation: The velocities are all relative to an observer ‘at rest on the ground’.

Applying the law of conservation of momentum gives:

M v=(M−δ m)(v+δ v)−δ m(u−v−δ v⏟speed relative to ground )

M v=M v+M δ v−v δ m−δ m δ v−u δ m+v δ m+δ m δ v

M δ v=u δ m

δ v=δ m M u

(Note how the terms that have been written in the same colour cancel each other out.)

This gives the change in speed of the rocket as a result of gases leaving with speed u relative to the rocket. At time t the mass of the rocket is M. Dividing by δt and taking the limit as δt goes to zero gives the rocket differential equation

M d v d t=μ u

where μ is the rate at which mass is being ejected.

TEST YOUR UNDERSTANDING

11A rocket in space where gravity is negligible has a mass (including fuel) of 5000 kg. It is desired to give the rocket an average acceleration of 15.0 m s−2 during the first second of firing the engine. The gases leave the rocket at a speed of 1500 m s−1 (relative to the rocket). Estimate how much fuel must be burned in that second.