Study Material

Conservation of linear momentum: principle, examples, exercises.

The conservation of the linear momentum of a body establishes that the product of its mass and its velocity vector is a constant quantity, when the body is free of interaction with other bodies and with the velocity measured with respect to a fixed or non-accelerated reference frame.

When you have several bodies that interact only with each other, but not with the external environment, then theÂ linear momentumÂ of the set also remains constant through time.

MomentumÂ ,Â linear momentum or simplyÂ momentumÂ , is denoted by the letterÂ pÂ and is a vector quantity:

Momentum is not the same as speed, although the relationship is obvious: for example, a truck going 20 km / h has more linear momentum than a bicycle moving at the same speed.

For the linear momentum of a body to change, a net external force must be acting on it, otherwise it remains constant.Â Furthermore, the momentumÂ PÂ of a system formed byÂ n-bodiesÂ is the vector sum of the individual moments:

Principle of conservation of linear momentum

In a force-free body (or one in which all the forces on it cancel out) it happens that the linear moment remains constant.

The same happens in a system formed by several bodies that only interact with each other, but not with the external environment: the total linear momentum of the system remains fixed during the evolution of the movement of the whole.

This conservation principle is stated as follows:

The total linear momentum of a set of n-bodies that only interact with each other, but not with the external environment, is an invariable quantity in time.

And mathematically it is expressed as follows:

The previous equalities are fulfilled, if and only if theÂ nÂ -bodies interact with each other, butÂ notÂ with the external environment.Â Furthermore, individual moments must always be measured relative to an inertial reference frame.

Examples

Example 1

Two astronauts in space are holding hands and are held in a fixed position relative to the spacecraft.Â But if they push each other, they start to separate in opposite directions, when seen from the ship.

In this case, as the interaction between the astronauts is only between them through the contact force of their hands, the total momentum after pushing is still the initial value with respect to the spacecraft.Â That is, total momentum 0.

However, the momentum of each astronaut did change.Â Initially, each one had linear momentum 0 with respect to the ship, but after being pushed, one exits in one direction and the other in the opposite direction, with non-zero linear moments of equal magnitude and opposite directions.

Thus, when the individual moments are added vectorially, the initial total momentum is obtained, which is zero.

On the other hand, the conservation of the momentum quantity indicates that the astronaut with the lowest mass is the one that moves faster with respect to the spacecraft.Â But the result of multiplying its mass by its speed is equal to the product obtained by multiplying the mass of the other by the speed of the other.

Example 2

A puppy is on a floating platform in a calm lake and his owner watches him from a dock.Â At the beginning, both the platform and the puppy are at rest, but when the puppy wants to get closer to the owner, the platform moves away from the dock.

The explanation for this observation is precisely in the principle of conservation of the quantity of linear momentum.Â The system is made up of the puppy and the platform.

The puppy can walk on the platform thanks to the friction force between its legs and the surface, in this case the friction force is an internal force of interaction between it and the platform.

The whole is an isolated system, since the platform can move horizontally over the lake, free of any resistance to movement.Â On the other hand, in the vertical direction all the forces are balanced and compensated, and the assembly has no movement in that direction.

Therefore, in this situation all the hypotheses are fulfilled so that the principle of conservation of linear momentum applies.

Example 3

An Eskimo is trapped in the center of a frozen lake, the ice is so smooth that no matter how hard he tries, the Eskimo slips and always stays in the same place.

The only possible way for the Eskimo to get out of the lake is for him to throw in the opposite direction to which he wants to move some heavy object that he carries in his backpack (assuming he is carrying one).

Applications

Ship in space

Conservation of linear momentum is applied to propel a rocket in outer space where there are no external forces.Â In this case, the impulse of the ship is achieved by expelling gases at high speed, so that the rocket can move in the opposite direction to which they were ejected.

If originally the ship is at rest, when it burns and expels fuel, the force of the expulsion occurs against the ship itself.Â It is an internal force between the gases and the ship.Â There are no external forces and therefore the conservation of linear momentum applies.

As the linear momentum of the gases is the same and opposite to that of the ship, it manages to come out of rest, and by continuing to expel gases, it increases its amount of movement and therefore its speed.

Daily life

Another case of application of the conservation of linear momentum in everyday life is to drive a nail into the wood taking advantage of the amount of movement or linear momentum of the hammer.

It could be argued that in this case the principle does not apply, because there is an external force: the resistance that the wood offers to the nail.

However, at the moment of contact, the force that the hammer imposes on the nail is an internal force (between the system that is the nail and the hammer) much greater than the resistance that the wood opposes, and therefore the latter is negligible.

All the momentum of the hammer, which is quite large due to its large mass and speed, is transferred to the nail just after the collision.Â Note that all the moment is transferred, but not all the kinetic energy of the hammer, since part of this is transformed into thermal energy in the nail and in the hammer, which raise their temperature after the impact.

Training

Exercise 1

Astronauts Andrew and Berenice are outside the space station holding both hands and at rest relative to the station.Â They are propelled by pushing the hands of one against the other and they are released.Â If Andrew, 70 kg of mass, moves at 1 m / s with respect to the station, what is the speed of Berenice with 49 kg of mass?

Solution

In this case, the hypotheses of the conservation of linear momentum clearly apply, since there are no external forces in outer space.Â The force with which both astronauts push their hands is an internal force.

Suppose that Andrew’s mass is MÂ aÂ and Berenice’s is MÂ bÂ .Â Similarly, the velocities of both of after the impulse areÂ VÂ aÂ for Andrew andÂ VÂ bÂ Â for Berenice.Â Then the conservation of linear momentum applies like this:

MÂ aÂ âˆ™Â 0Â + MÂ bÂ âˆ™Â 0Â = MÂ aÂ âˆ™Â VÂ aÂ + MÂ bÂ âˆ™Â VÂ b

Solving for Berenice’s velocity we have:

VÂ bÂ = – (MÂ aÂ / MÂ bÂ ) âˆ™Â VÂ a

Placing the numerical values:

VÂ bÂ = – (Â 70/49Â ) âˆ™ (1m / s)Â Ã»Â = -1.43m / sÂ Ã»

In other words, Berenice is moving at a speed of 1.43 m / s in the opposite direction from Andrew.

Exercise 2

A 5 kg puppy is at rest on a 15 kg platform that floats, also at rest, on a still lake.Â If the puppy begins to walk on the platform at a rate of 0.5 m / s with respect to it.Â How fast will the puppy and the platform have with respect to an observer fixed to the ground?

Solution

The inertial reference system will be taken as the dock where the owner of the puppy is.Â Initially, both the puppy and the floating platform are at rest with respect to the dock.

When the dog owner decides toÂ walk into the rapidlyÂ v ‘Â with respect to the platform, then the platform moves away from the dock with speedÂ + VÂ .Â The speed of the puppy with respect to the spring is obtained by the vector sum of its speed with respect to the platform plus the speed of the platform and we denote it by:

vÂ = –Â vÂ ‘+Â V

As the resistance of the water to the movement of the platform is practically nil due to its low speed, then it can be stated that the system made up ofÂ the puppy + the platformÂ is an isolated system and the principle of conservation of linear momentum is applied:

0 = m âˆ™ v + M âˆ™ V

Remembering that v = v ‘+ V we have:

0 = -m âˆ™ v ‘+ m âˆ™ v + M âˆ™ V

That is: m âˆ™ v ‘= (m + M) âˆ™ V

Therefore V = [m / (m + M)] v ‘and v = – (M / m) V = – [M / (m + M)] v’

Substituting the numerical values â€‹â€‹we have:

V = [5 / (5 +15)] âˆ™ 0.5m / s = 0.125 m / s

This is how fast the deck is moving away from the dock.

V = – (15/20) âˆ™ 0.5m / s = -0.375 m / s

And this is the speed with which the puppy approaches the dock.

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