# Inelastic shocks: in one dimension and examples

The **inelastic collisions** or inelastic collisions are a short and intense interaction between two objects, in which the amount of movement is maintained, but the kinetic energy, which is transformed percentage some other form of energy.

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

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. Objects now have kinetic energy associated with the motion they have. What can happen to this energy when it hits?

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

More possible uses for kinetic energy: frictional heat and, of course, the inevitable deformation of objects when colliding, like the car bodies in the figure above.

**Examples of Inelastic Collisions**

– Two plasticine masses that collide and stay together, moving like one piece after the accident.

– A rubber ball that hits 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 hold a certain amount of this energy. Later we will see how to calculate the percentage.

When colliding pieces are joined 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*, moving towards each other with velocities

_{2}*v*and

_{i1}*v*respectively. Everything happens horizontally, that is, it is a collision in one dimension, the easiest to study.

_{i2}Objects collide and then stick together by moving to the right. It’s a perfectly inelastic collision, so keep the momentum:

*P ** _{o}* =

*P*

_{f}Momentum is a vector whose units in SI are Ns. In the situation described, vector notation can be dispensed with in the case of one-dimensional collisions:

*mv _{o} = mv _{f}*

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

*m _{1} v _{i1} + m _{2} v _{i2} = (m _{1}* +

*m*

_{2}) v_{f}The final speed is given by:

*v _{f} = (m _{1} v _{i1} + m _{2} v _{i2} ) / (m _{1}* +

*m*

_{2})**Return Coefficient**

There is a quantity that can indicate the elasticity of a collision. This is the *restitution coefficient* , which is defined as the negative ratio 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 particle velocities initially. And let v _{1} and v _{2 be} the respective final speeds. Mathematically, the refund coefficient can be expressed as follows:

– If ε = 0 is equivalent to saying that v _{2} = v _{1} . This means that the final velocities are the same and the collision is inelastic as described in the previous section.

– When ε = 1 it means that the relative speeds before and after the crash do not change, in this case the crash is elastic.

– And if 0 <ε <1 part of the kinetic energy of the collision is transformed into another of the energies mentioned above.

__How to determine the refund coefficient?__

__How to determine the refund coefficient?__

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

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

In 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 objects that came with movement collide and join together to form a single object that suddenly rests? This is not so easy to imagine.

However, let’s imagine that it happens backwards, like in a movie seen backwards. So the object was initially at rest and then exploded, fragmenting into various parts. This situation is entirely possible: it’s an explosion.

Therefore, an explosion can be seen as a perfectly inelastic collision seen in time. The amount of movement is also preserved and it can be said that:

*P _{o} = P _{f}*

__Solved Examples__

__Solved Examples__

**-Exercise 1**

It is known from measurements that the return coefficient of steel is 0.90. A 7 m high steel ball is thrown onto a fixed plate. Calculate:

a) How high will he jump.

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

**Solution**

a) The equation deduced earlier in the section on determining the return coefficient is used:

The height *h _{2}* is clean :

0.90 ^{2} . 7 m = 5.67 m

b) To climb 5.67 meters, a speed given by:

t _{max} = V _{or} / g = (10.54 / 9.8) s = 1.08 s.

The time taken 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 wooden block of mass M hanging the remaining length of string in pendulum mode. This is called a ballistic pendulum and serves to measure the velocity v of the entry of a bullet of mass m. The faster the bullet hits the block, the greater the oh will increase.

The bullet in the image is embedded in the block, so it’s a totally inelastic shock.

Suppose a 9.72 g bullet hits the 4.60 kg mass block and then the mount rises to 16.8 cm from the equilibrium position. What is the speed *v* of the bullet?

**Solution**

During the collision, the amount of movement is conserved and *u _{f} *is the velocity of the set, since the bullet has been incorporated into the block:

*P _{o} = P _{f}*

The block is initially at rest, while the marker hits the target with velocity *v* :

*mv + M.0 = (m + M) u _{f}*

*u _{f} *is not known yet, but after the collision mechanical energy is conserved, being the sum of gravitational potential energy and kinetic energy UK:

Initial mechanical energy = final mechanical energy

*E _{mo} = E _{mf}*

*U _{o} + K _{o} = U _{f} + K _{f}*

The gravitational potential energy depends on the height at which the set reaches. For the equilibrium position, the initial height is that considered as the reference level, therefore:

*U _{o} = 0*

Thanks to the bullet, the set has kinetic energy *K _{o}* , which becomes gravitational potential energy when the set reaches its maximum height

*h*. Kinetic energy is given by:

*K = ½ mv ^{2}*

Initially the kinetic energy is:

*K _{o} = (1/2) (M + m) u _{f }^{2}*

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

*L _{f} = (M + m) gh*

Therefore:

*K _{o} = U _{f}*

*(1/2) (M + m) u _{f }^{2} = (m + M) gh*

**-Exercise 3**

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

**Solution**

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

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

*M = m + m + 2m = 4m*

The momentum is preserved on the x-axis and the y-axis, and is declared separately:

*4m u*_{x}= mv_{3}*4m u*_{e}= m. 2v_{1}– 2m. v_{1}

Note that the large fragment moves down with velocity v1 to indicate this fact that a minus sign has been placed.

From the second equation it immediately follows that *u _{y} = 0* and from the first equation ux immediately: