ELASTIC SHOCKS: IN ONE DIMENSION, SPECIAL CASES, EXERCISES - PHYSICAL - 2025

The elastic collisions or elastic collisions are brief but intense interactions between objects, in which both the momentum and kinetic energy are conserved. Crashes are very frequent events in nature: from subatomic particles to galaxies, to billiard balls and bumper cars at amusement parks, they are all objects capable of colliding.

During a collision or collision, the forces of interaction between objects are very strong, much more than those that can act externally. In this way it can be stated that during the collision, the particles form an isolated system.

Billiard ball collisions can be considered elastic. Source: Pixabay.

In this case it is true that:

The momentum P _o before the collision is the same as after the collision. This is true for any type of collision, both elastic and inelastic.

Now consider the following: during a collision, objects undergo a certain deformation. When the shock is elastic, objects quickly revert to their original shape.

Conservation of kinetic energy

Normally during a crash, part of the energy of objects is spent on heat, deformation, sound and sometimes even on producing light. So the kinetic energy of the system after the collision is less than the original kinetic energy.

When the kinetic energy K is conserved then:

Which means that the forces acting during the collision are conservative. During the collision, the kinetic energy is briefly transformed into potential energy and then back to kinetic energy. The respective kinetic energies vary, but the sum remains constant.

Perfectly elastic collisions are rare, although billiard balls are a fairly good approximation, as are collisions that occur between ideal gas molecules.

Elastic shocks in one dimension

Let's examine a collision of two particles of this in a single dimension; that is, the interacting particles move, say, along the x-axis. Suppose they have masses m ₁ and m ₂. The initial velocities of each are u _{1 and} u ₂ respectively. The final velocities are v ₁ and v ₂.

We can dispense with the vector notation, since the movement is carried out along the x axis, however, the signs (-) and (+) indicate the direction of the movement. On the left is negative and on the right positive, by convention.

-Formula for elastic collisions

For the amount of movement

For kinetic energy

As long as the masses and initial velocities are known, the equations can be regrouped to find the final velocities.

The problem is that in principle, it is necessary to carry out a bit of quite tedious algebra, since the equations for kinetic energy contain the squares of the speeds, which makes the calculation a bit cumbersome. The ideal would be to find expressions that do not contain them.

The first thing is to do without the factor ½ and reorder both equations in such a way that a negative sign appears and the masses can be factored:

Being expressed in this way:

Simplification to eliminate the squares of the velocities

Now we must make use of the notable product sum by its difference in the second equation, with which we obtain an expression that does not contain the squares, as originally wanted:

The next step is to substitute the first equation in the second:

And since the term m ₂ (v ₂ - u ₂) is repeated on both sides of the equality, said term is canceled and remains like this:

Or even better:

Final speeds v

Now you have two linear equations that are easier to work with. We will put them back one under the other:

Multiplying the second equation by m ₁ and adding term to term is:

And it is already possible to clear v ₂. For example:

Special cases in elastic collisions

Now that equations are available for the final velocities of both particles, it is time to analyze some special situations.

Two identical masses

In that case m ₁ = m ₂ = my:

The particles simply exchange their velocities after the collision.

Two identical masses, one of which was initially at rest

Again m ₁ = m ₂ = m and assuming u ₁ = 0:

After the collision, the particle that was at rest acquires the same speed as the particle that was moving, and this in turn stops.

Two different masses, one of them initially at rest

In this case suppose that u ₁ = 0, but the masses are different:

What if m ₁ is much larger than m ₂ ?

It happens that m _{1 is} still at rest and m ₂ is returned with the same speed with which it impacted.

Coefficient of restitution or Huygens-Newton rule

Previously the following relationship between the velocities for two objects in elastic collision was derived: u ₁ - u ₂ = v ₂ - v ₁. These differences are the relative speeds before and after the collision. In general, for a collision it is true that:

The concept of relative velocity is best appreciated if the reader imagines that he is on one of the particles and from this position he observes the speed with which the other particle is moving. The above equation is rewritten like this:

Solved exercises

-Solved exercise 1

A billiard ball is moving to the left at 30 cm / s, colliding head-on with another identical ball that is moving to the right at 20 cm / s. The two balls have the same mass and the collision is perfectly elastic. Find the velocity of each ball after impact.

Solution

u ₁ = -30 cm / s

u ₂ = +20 cm / s

This is the special case where two identical masses collide in one dimension elastically, therefore the speeds are exchanged.

v ₁ = +20 cm / s

v ₂ = -30 cm / s

-Solved exercise 2

The coefficient of restitution of a ball that bounces off the ground is equal to 0.82. If it falls from rest, what fraction of its original height will the ball reach after bouncing once? And after 3 rebounds?

A ball bounces off a firm surface and loses height with each bounce. Source: self made.

Solution

The soil can be object 1 in the equation for the coefficient of restitution. And it always remains at rest, so that:

With this speed it bounces:

The + sign indicates that it is an ascending speed. And according to it, the ball reaches a maximum height of:

Now it returns to the ground again with a speed of equal magnitude, but opposite sign:

This achieves a maximum height of:

Get back to the ground with:

Successive bounces

Every time the ball bounces and rises, multiply the speed again by 0.82:

At this point h ₃ is about 30% of h _o. What would be the height to the 6th bounce without the need to make such detailed calculations as the previous ones?

It would be h ₆ = 0.82 ¹² h _o = 0.092h _o o just 9% of h _o.

-Solved exercise 3

A 300-g block is moving north at 50 cm / s and collides with a 200-g block heading south at 100 cm / s. Assume that the shock is perfectly elastic. Find the velocities after impact.

Data

m ₁ = 300 g; u ₁ = + 50 cm / s

m ₂ = 200 g; u ₂ = -100 cm / s

-Solved exercise 4

A mass of m ₁ = 4 kg is released from the indicated point on the frictionless track until it collides with m ₂ = 10 kg at rest. How high does m ₁ rise after the collision?

Solution

Since there is no friction, the mechanical energy is conserved to find the velocity u ₁ with which m ₁ hits m _2. Initially the kinetic energy is 0, since m ₁ starts from rest. When it moves on the horizontal surface it has no height, so the potential energy is 0.

Now the velocity of m ₁ after the collision is calculated:

The negative sign means that it has been returned. With this speed it ascends and the mechanical energy is conserved again to find h ', the height to which it manages to ascend after the collision:

Note that it does not return to the starting point at 8 m height. It does not have enough energy because the mass m ₁ gave up part of its kinetic energy .

References

1. Giancoli, D. 2006. Physics: Principles with Applications. 6 ^th. Ed Prentice Hall. 175-181
2. Rex, A. 2011. Fundamentals of Physics. Pearson. 135-155.
3. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9 ^na Cengage Learning. 172-182
4. Tipler, P. (2006) Physics for Science and Technology. 5th Ed. Volume 1. Editorial Reverté. 217-238
5. Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. MacGraw Hill. 185-195