INELASTIC CRASHES: IN ONE DIMENSION AND EXAMPLES - PHYSICAL - 2024

The inelastic collisions or inelastic collisions are a short and intense interaction between two objects in which the amount of movement is retained, but not the kinetic energy, which is transformed percentage some other form of energy.

Crashes or collisions are frequent in nature. Subatomic particles collide at extremely high speeds, while many sports and games consist of continuous collisions. Even galaxies are capable of colliding.

Figure 1. Test car collision. Source: Pixabay

In fact, momentum is conserved in any type of collision, as long as the colliding particles form an isolated system. So in this sense there is no problem. Now, objects have kinetic energy associated with the movement they have. What can happen to that energy when it hits?

The internal forces that take place during the collision between objects are intense. When it is stated that kinetic energy is not conserved, it means that it is transformed into other types of energy: for example, into sound energy (a spectacular collision has a distinctive sound).

More possibilities of use for kinetic energy: heat by friction, and of course the inevitable deformation that objects undergo when they collide, such as the bodies of the cars in the figure above.

Examples of inelastic collisions

- Two masses of plasticine that collide and remain together, moving as one piece after the collision.

- A rubber ball that bounces off a wall or floor. The ball deforms when it hits the surface.

Not all kinetic energy is transformed into other types of energy, with few exceptions. Objects can keep a certain amount of this energy. Later we will see how to calculate the percentage.

When the colliding pieces stick together, the collision is called perfectly inelastic, and the two often end up moving together.

Perfectly inelastic collisions in one dimension

The collision in the figure shows two objects of different masses m ₁ and m ₂, moving toward each other with velocities v _i1 and v _i2 respectively. Everything happens on the horizontal, that is, it is a collision in one dimension, the easiest to study.

Figure 2. Collision between two particles of different masses. Source: self made.

The objects collide and then stick together moving to the right. It is a perfectly inelastic collision, so we just have to keep the momentum:

The momentum is a vector whose SI units are Ns. In the situation described, the vector notation can be dispensed with when dealing with collisions in one dimension:

The momentum of the system is the vector sum of the momentum of each particle.

The final speed is given by:

Coefficient of restitution

There is a quantity that can indicate how elastic a collision is. It is the coefficient of restitution, which is defined as the negative quotient between the relative velocity of the particles after the collision and the relative velocity before the collision.

Let u _{1 and} u _{2 be} the respective velocities of the particles initially. And let v ₁ and v _{2 be} the respective final velocities. Mathematically the coefficient of restitution can be expressed as:

- If ε = 0 it is equivalent to affirming that v ₂ = v ₁. It means that the final speeds are the same and the collision is inelastic, like the one described in the previous section.

- When ε = 1 it means that the relative velocities both before and after the collision do not change, in this case the collision is elastic.

- And if 0 <ε <1 part of the kinetic energy of the collision is transformed into some other of the energies mentioned above.

How to determine the coefficient of restitution?

The coefficient of restitution depends on the class of materials involved in the collision. A very interesting test to determine how elastic a material is to make balls is to drop the ball on a fixed surface and measure the rebound height.

Figure 3. Method to determine the coefficient of restitution. Source: self made.

In this case, the fixed plate always has speed 0. If it is assigned index 1 and the ball index 2 is:

At the beginning it was suggested that all kinetic energy can be transformed into other types of energy. After all, energy is not destroyed. Is it possible that moving objects collide and join together to form a single object that suddenly comes to rest? This is not so easy to imagine.

However, let's imagine it happens the other way around, like in a movie seen in reverse. So the object was initially at rest and then explodes fragmenting into various parts. This situation is perfectly possible: it is an explosion.

So an explosion can be thought of as a perfectly inelastic collision viewed backwards in time. The momentum is also conserved, and it can be stated that:

Worked Examples

-Exercise 1

It is known from measurements that the coefficient of restitution of steel is 0.90. A steel ball is dropped from a height of 7 m onto a fixed plate. Calculate:

a) How high it will bounce.

b) How long it takes between the first contact with the surface and the second.

Solution

a) The equation that was deduced previously in the section on determining the coefficient of restitution is used:

The height h ₂ is cleared:

0.90 ². 7 m = 5.67 m

b) For it to rise 5.67 meters, a speed is required given by:

t _max = v _o / g = (10.54 / 9.8 s) = 1.08 s.

The time it takes to return is the same, therefore the total time to climb the 5.67 meters and return to the starting point is twice the maximum time:

t _flight = 2.15 s.

-Exercise 2

The figure shows a block of wood of mass M hanging at rest by strings of length in pendulum mode. This is called a ballistic pendulum and is used to measure the velocity v of entry into a bullet of mass m. The faster the bullet hits the block, the higher h it will rise.

The bullet in the image is embedded in the block, therefore it is a totally inelastic shock.

Figure 4. The ballistic pendulum.

Suppose a 9.72-g bullet hits the block of mass 4.60 kg, then the assembly rises 16.8 cm from equilibrium. What is the velocity v of the bullet?

Solution

During the collision, the momentum is conserved and u _f is the velocity of the whole, once the bullet has embedded itself in the block:

The block is initially at rest, while the bullet is aimed at the target with velocity v:

U _f is not yet known, but after the collision the mechanical energy is conserved, this being the sum of the gravitational potential energy U and the kinetic energy K:

Initial mechanical energy = Final mechanical energy

The gravitational potential energy depends on the height to which the set reaches. For the equilibrium position, the initial height is the one taken as the reference level, therefore:

Thanks to the bullet, the set has kinetic energy K _o, which is converted into gravitational potential energy when the set reaches its maximum height h. The kinetic energy is given by:

Initially the kinetic energy is:

Remember that the bullet and the block already form a single object of mass M + m. The gravitational potential energy when they have reached their maximum height is:

Thus:

-Exercise 3

The object in the figure explodes into three fragments: two of equal mass and a larger one of mass 2m. The figure shows the velocities of each fragment after the explosion. What was the initial velocity of the object?

Figure 5. The stone that explodes in 3 fragments. Source: self made.

Solution

This problem requires the use of two coordinates: x and y, because two of the fragments have vertical velocities, while the remainder has horizontal velocity.

The total mass of the object is the sum of the mass of all the fragments:

The momentum is conserved both in the x-axis and in the y-axis, it is stated separately:

1. 4m. u _x = mv ₃
2. 4m. u _y = m. 2v ₁ - 2m. v ₁

Note that the large fragment moves down with speed v1, to indicate this fact a negative sign has been placed on it.

From the second equation it follows immediately that u _y = 0, and from the first one we solve for ux immediately:

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