Elastic shocks: in one dimension, special cases, exercises

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

Collisions between billiard balls can be considered elastic. Source: PixabayIn this case, it is true that:

P _o = P _f

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

Now, the following must be considered: during a collision, objects experience a certain amount of deformation. When shock is elastic, objects quickly regain their original shape.

Kinetic energy conservation

Normally, during an accident, part of the objects’ energy is spent on heat, deformation, sound and sometimes even producing light. Therefore, the kinetic energy of the system after the collision is less than the original kinetic energy.

When the kinetic energy K then is conserved:

K _o = K _f

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

Perfectly elastic collisions are infrequent, although billiard balls are a pretty 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 pass without vector notation, as movement is performed along the x-axis, however, the (-) and (+) signs indicate the direction of movement. Left is negative and right is positive by convention.

-Formulas for elastic collisions

for the amount of movement

m ₁ u ₁ + m ₂ u ₂ = m ₁ v ₁ + m ₂ v ₂

for kinetic energy

½ m ₁ u ² ₁ + ½ m ₂ u ² ₂ = ½ m ₁ v ² ₁ + ½ m ₂ v ² ₂

Whenever the initial masses and velocities are known, it is possible to regroup the equations to find the final velocities.

The problem is that, in principle, it is necessary to do some very tedious algebra, because the equations for kinetic energy contain the squares of the velocities, which makes the calculation a little complicated. Ideally, find expressions that do not contain them.

The first thing is to do without the factor ½ and reorder the two equations so that a minus sign appears and the masses can be factored:

m ₁ u ₁ – m ₁ v ₁ = m ₂ v ₂ – m ₂ u ₂

m ₁ u ² ₁ – m ₁ v ² ₁ = + m ₂ v ² ₂ – m ₂ u ² ₂

Being expressed in this way:

m ₁ (u ₁ – v ₁ ) = m ₂ (v ₂ – u ₂ )

m ₁ (u ² ₁ – v ² ₁ ) = m ₂ (v ² ₂ – u ² ₂ )

Simplification to eliminate speed squares

Now, we must make use of the remarkable sum of the product of its difference in the second equation, which gives an expression that does not contain the squares, as originally intended:

m ₁ (u ₁ – v ₁ ) = m ₂ (v ₂ – u ₂ )

m ₁ (u ₁ – v ₁ ) (u ₁ + v ₁ ) = m ₂ (v ₂ – u ₂ ) (v ₂ + u ₂ )

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

m ₂ (v ₂ – u ₂ ) (u ₁ + v ₁ ) = m ₂ (v ₂ – u ₂ ) (v ₂ + u ₂ )

And when the term m ₂ (v ₂ – or ₂ ) is repeated on both sides of the equality, this term is canceled and it goes like this:

(u ₁ + v ₁ ) = (v ₂ + u ₂ )

Or better yet:

u ₁ – u ₂ = v ₂ – v ₁

Final velocities v ₁ and v ₂ of the particles

Now there are two linear equations that are easier to work with. Let’s put them one below the other:

m ₁ u ₁ + m ₂ u ₂ = m ₁ v ₁ + m ₂ v ₂

u ₁ – u ₂ = v ₂ – v ₁

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

m ₁ u ₁ + m ₂ u ₂ = m ₁ v ₁ + m ₂ v ₂

m ₁ u ₁ – m ₁ u ₂ = m ₁ v ₂ – m ₁ v ₁

———————————————

2 m ₁ u ₁ + (m ₂ – m ₁ ) u ₂ = (m ₂ + m ₁ ) v ₂

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

Special cases in elastic collisions

Now that the equations are available for the final velocities of both particles, it’s time to look at some special situations.

two identical masses

In this case, m ₁ = m ₂ = me :

v ₁ = u ₂

v ₂ = u ₁

Particles simply change their speed after the collision.

Two identical masses, one of which was initially at rest

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

v ₁ = u ₂

v ₂ = 0

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

Two different masses, one of them initially at rest

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

What happens if m ₁ is much larger than m ₂ ?

It turns out that m _1 is still at rest and m ₂ is returned as quickly as reached.

Huygens-Newton Refund Coefficient or Rule

Previously, the following relationship between velocities for two objects in elastic collision was deduced: 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:

u ₁ – u ₂ = – (v ₁ – v ₂ )

The concept of relative velocity is best appreciated if the reader imagines that he is in one of the particles and, from that position, observes how fast the other particle is moving. The previous equation is rewritten as follows:

solved exercises

Exercise -Fixed 1

A billiard ball moves to the left at 30 cm / s, colliding head-on with another identical ball that moves to the right at 20 cm / s. The two balls have the same mass and the shock is perfectly elastic. Find the speed 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, so the velocities are swapped.

v ₁ = +20 cm / s

v ₂ = -30 cm / s

– Exercise solved 2

The restitution coefficient of a ball that bounces on 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 hits a firm surface and loses height with each bounce. Source: own elaboration.

Solution

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

With that speed, it jumps:

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

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

This achieves a maximum height of:

 

successive jumps

Each time the ball bounces and goes up, the speed must be multiplied again by 0.82:

At this point, h ₃ is approximately 30% of h _o . How high would the sixth rebound be without having to make calculations as detailed as the previous ones?

It will be H ₆ = 0.82 ¹² h _or = 0.092h _or or only 9% of H _or .

– Exercise solved 3

A 300 g block moves north at 50 cm / s and hits a 200 g block that moves south at 100 cm / s. Suppose the shock is perfectly elastic. Find speeds after impact.

Data

m ₁ = 300 g; u ₁ = + 50 cm / s

m ₂ = 200 g; u ₂ = -100 cm / s

– Exercise solved 4

A mass of m ₁ = 4 kg is released from the point indicated on the track without friction, until it collides with m ₂ = 10 kg at rest. What is the height m ₁ after the collision?

Solution

Since there is no friction, the mechanical energy is conserved to find the speed u ₁ with which m ₁ affects m _2. Initially, the kinetic energy is 0, since m _{1 is} part of the remainder. When it moves on the horizontal surface, it has no height; therefore, the potential energy is 0.

mgh = ½ mu ₁ ²

 

u ₂ = 0

The minus sign means it was returned. With that speed, it ascends and the mechanical energy is conserved again to find h ‘ , the height at which it can ascend after the accident:

½ mv ₁ ² = mgh ‘

Note that it does not return to the starting point at 8 m high. It doesn’t have enough energy because part of its kinetic energy produced mass m _1.