You are here: Home Types of collisions

# Types of collisions

Two types of collisions are of interest:

• elastic collisions

• inelastic collisions

In both types of collision, total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions.

## Elastic collisions

Definition 1: Elastic Collisions

An elastic collision is a collision where total momentum and total kinetic energy are both conserved.

This means that in an elastic collision the total momentum and the total kinetic energy before the collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy.

### Before the collision

Figure 1 shows two balls rolling toward each other, about to collide:

Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. Ball 1 has a momentum which we call $pi1$ and ball 2 has a momentum which we call $pi2$, it means the total momentum before the collision is:

$pi=pi1+pi2$(1)

We calculate the total kinetic energy of the system in the same way. Ball 1 has a kinetic energy which we call $KEi1$ and the ball 2 has a kinetic energy which we call $KEi2$, it means that the total kinetic energy before the collision is:

$KEi=KEi1+KEi2$(2)

### After the collision

Figure 2 shows two balls after they have collided:

After the balls collide and bounce off each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. Ball 1 now has a momentum which we call $pf1$ and ball 2 now has a momentum which we call $pf2$, it means the total momentum after the collision is

$pf=pf1+pf2$(3)

Ball 1 now has a kinetic energy which we call $KEf1$ and ball 2 now has a kinetic energy which we call $KEf2$, it means that the total kinetic energy after the collision is:

$KEf=KEf1+KEf2$(4)

Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision. Therefore:

$InitialFinalpi=pfpi1+pi2=pf1+pf2andKEi=KEfKEi1+KEi2=KEf1+KEf2$(5)

#### Example 1: An elastic collision

##### Question

Consider a collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 m·s−1. The mass of each ball is 0,3 kg. After the balls collide elastically, ball 2 comes to an immediate stop and ball 1 moves off. What is the final velocity of ball 1?

###### Determine what is given and what is needed
• Mass of ball 1, .

• Mass of ball 2, .

• Initial velocity of ball 1, .

• Initial velocity of ball 2, .

• Final velocity of ball 2, .

• The collision is elastic.

All quantities are in SI units. We need to find the final velocity of ball 1, $vf1$. Since the collision is elastic, we know that

• momentum is conserved, $m1vi1+m2vi2=m1vf1+m2vf2$

• energy is conserved, $12(m1vi12+m2vi22)=12(m1vf12+m2vf22)$

###### Choose a frame of reference

Choose to the right as positive.

###### Solve the problem

Momentum is conserved. Therefore:

(6)

The final velocity of ball 1 is 2 m·s−1 to the right.

#### Example 2: Another elastic collision

##### Question

Consider 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. Edward rolls marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 m·s−1 in the positive x-direction. After they collide elastically, both marbles are moving. What is the final velocity of each marble?

###### Decide how to approach the problem

We are given:

• mass of marble 1,

• mass of marble 2,

• initial velocity of marble 1,

• initial velocity of marble 2,

• the collision is elastic

The masses need to be converted to SI units.

(7)

We are required to determine the final velocities:

• final velocity of marble 1, $vf1$

• final velocity of marble 2, $vf2$

Since the collision is elastic, we know that

• momentum is conserved, $pi=pf$.

• energy is conserved, $KEi=KEf$.

We have two equations and two unknowns ($v1$, $v2$) so it is a simple case of solving a set of simultaneous equations.

###### Choose a frame of reference

Choose to the right as positive.

###### Solve the problem

Momentum is conserved. Therefore:

$pi=pfpi1+pi2=pf1+pf2m1vi1+m2vi2=m1vf1+m2vf2(0,05)(0)+(0,1)(3)=(0,05)vf1+(0,1)vf20,3=0,05vf1+0,1vf2$(8)

Energy is also conserved. Therefore:

$KEi=KEfKEi1+KEi2=KEf1+KEf212m1vi12+12m2vi22=12m1vf12+12m2vf22(12)(0,05)(0)2+(12)(0,1)(3)2=12(0,05)(vf1)2+(12)(0,1)(vf2)20,45=0,025vf12+0,05vf22$(9)

Substitute Equation 8 into Equation 9 and solve for $vf2$.

$m2vi22=m1vf12+m2vf22=m1m2m1(vi2-vf2)2+m2vf22=m1m22m12vi2-vf22+m2vf22=m22m1vi2-vf22+m2vf22vi22=m2m1vi2-vf22+vf22=m2m1vi22-2·vi2·vf2+vf22+vf220=m2m1-1vi22-2m2m1vi2·vf2+m2m1+1vf22=0.10.05-1(3)2-20.10.05(3)·vf2+0.10.05+1vf22=(2-1)(3)2-2·2(3)·vf2+(2+1)vf22=9-12vf2+3vf22=3-4vf2+vf22=(vf2-3)(vf2-1)$(10)

Therefore $vf2=1$ or $vf2=3$

Substituting back into Equation 8, we get:

(11)

But according to the question, marble 1 is moving after the collision, therefore marble 1 moves to the right at 4 m·s−1.

Therefore marble 2 moves to the right with a velocity of 1 m·s−1.

#### Example 3: Colliding billiard balls

##### Question

Two billiard balls each with a mass of 150 g collide head-on in an elastic collision. Ball 1 was travelling at a speed of 2 m·s−1 and ball 2 at a speed of 1,5 m·s−1. After the collision, ball 1 travels away from ball 2 at a velocity of 1,5 m·s−1.

1. Calculate the velocity of ball 2 after the collision.

2. Prove that the collision was elastic. Show calculations.

###### Decide how to approach the problem

Since momentum is conserved in all kinds of collisions, we can use conservation of momentum to solve for the velocity of ball 2 after the collision.

###### Solve problem
(12)

So after the collision, ball 2 moves with a velocity of 3 m·s−1.

###### Elastic collisions

The fact that characterises an elastic collision is that the total kinetic energy of the particles before the collision is the same as the total kinetic energy of the particles after the collision. This means that if we can show that the initial kinetic energy is equal to the final kinetic energy, we have shown that the collision is elastic.

###### Calculating the initial total kinetic energy
$EKbefore=12m1vi12+12m2vi22=12(0,15)(2)2+12(0,15)(-1,5)2=0.469....J$(13)
###### Calculating the final total kinetic energy
$EKafter=12m1vf12+12m2vf22=12(0,15)(-1,5)2+12(0,15)(2)2=0.469....J$(14)

So $EKbefore=EKafter$ and hence the collision is elastic.

## Inelastic collisions

Definition 2: Inelastic Collisions

An inelastic collision is a collision in which total momentum is conserved but total kinetic energy is not conserved. The kinetic energy is transformed into other kinds of energy.

So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is different. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy.

As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. The modification of the shape of an object requires energy and this is where the “missing” kinetic energy goes. A classic example of an inelastic collision is a motor car accident. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in Figure 3.

An asteroid is moving through space towards the Moon. Before the asteroid crashes into the Moon, the total momentum of the system is:

$pi=pim+pia$(15)

The total kinetic energy of the system is:

$KEi=KEim+KEia$(16)

When the asteroid collides inelastically with the Moon, its kinetic energy is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the Moon's surface that it hits, to melt into liquid rock! From the force of impact of the asteroid, the molten rock flows outwards to form a crater on the Moon.

After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), total kinetic energy is not the same as before the collision.

Momentum is conserved:

$pi=pf$(17)

But the total kinetic energy of the system is not conserved:

$KEi≠KEf$(18)

### Example 4: An inelastic collision

#### Question

Consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2 m·s−1 to the left. Both cars each have a mass of 500 kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal?

##### Determine how to approach the problem

We are given:

• mass of car 1,

• mass of car 2,

• initial velocity of car 1,

• initial velocity of car 2, to the left

• the collision is inelastic

All quantities are in SI units. We are required to determine the final velocity of the resulting mass, $vf$.

Since the collision is inelastic, we know that

• momentum is conserved, $m1vi1+m2vi2=m1vf1+m2vf2$

• kinetic energy is not conserved

##### Choose a frame of reference

Choose to the left as positive.

##### Solve problem

So we must use conservation of momentum to solve this problem.

(19)

Therefore, the final velocity of the resulting mass of cars is 1 m·s−1 to the left.

### Example 5: Colliding balls of clay

#### Question

Two balls of clay, 200 g each, are thrown towards each other according to the following diagram. When they collide, they stick together and move off together. All motion is taking place in the horizontal plane. Determine the velocity of the clay after the collision.

##### Analyse the problem

This is an inelastic collision where momentum is conserved.

The momentum before = the momentum after.

The momentum after can be calculated by drawing a vector diagram.

(20)
##### Calculate the momentum after the collision.

Here we need to draw a diagram:

$p1+2(after)=(0,8)2+(0,6)2=1$(21)
##### Calculate the final velocity

First we have to find the direction of the final momentum:

$tanθ=0,80,6θ=53,13°$(22)

Now we have to find the magnitude of the final velocity:

(23)

### Exercise 1: Collisions

A truck of mass 4500 kg travelling at 20 m·s−1 hits a car from behind. The car (mass 1000 kg) was travelling at 15 m·s−1. The two vehicles, now connected carry on moving in the same direction.

1. Calculate the final velocity of the truck-car combination after the collision.

2. Determine the kinetic energy of the system before and after the collision.

1. ${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=\left({m}_{1}+{m}_{2}\right){v}_{}$

$v=\frac{\left(4500\right)\left(20\right)+\left(1000\right)\left(15\right)}{4500+1000}$

$v=19,09\text{m}\cdot {\text{s}}^{-1}$

2. $KE=\frac{1}{2}m{v}^{2}$

$K{E}_{i1}=\frac{1}{2}\left(4500\right)\left({20}^{2}\right)=900000\text{J}$

$K{E}_{i2}=\frac{1}{2}\left(1000\right)\left({15}^{2}\right)=112500\text{J}$

$K{E}_{f}=\frac{1}{2}\left({m}_{1}+{m}_{2}\right)\left({v}^{2}\right)=\frac{1}{2}\left(1000+4500\right)\left(19,{09}^{2}\right)=819963.225\text{J}$

$K{E}_{T}=1012500\text{J}$

3. The energy is used up during the collision.

4. Inelastic. The two bodies move in contact with each other.

Two cars of mass 900 kg each collide and stick together at an angle of 90°. Determine the final velocity of the cars if

car 1 was travelling at 15 m·s−1 and

car 2 was travelling at 20 m·s−1.

${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=\left({m}_{1}+{m}_{2}\right)v$

$\left(900\right)\left(15\right)+\left(900\right)\left(20\right)=\left(900+900\right)v$

$v=17,5\text{m}\cdot {\text{s}}^{-1}$

### Extension — Tiny, violent collisions:

Author: Thomas D. Gutierrez

Tom Gutierrez received his Bachelor of Science and Master degrees in Physics from San Jose State University in his home town of San Jose, California. As a Master's student he helped work on a laser spectrometer at NASA Ames Research Centre. The instrument measured the ratio of different isotopes of carbon in $CO2$ gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he received his PhD in physics from the University of California, Davis where he performed calculations for various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, New York.

High Energy Collisions

Take an orange and expand it to the size of the earth. The atoms of the earth-sized orange would themselves be about the size of regular oranges and would fill the entire “earth-orange”. Now, take an atom and expand it to the size of a football field. The nucleus of that atom would be about the size of a tiny seed in the middle of the field. From this analogy, you can see that atomic nuclei are very small objects by human standards. They are roughly 10−15 meters in diameter – one-hundred thousand times smaller than a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or microscopes. So how do scientists study the structure of very small objects like atomic nuclei?

The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other protons or electrons) at very high energies, one hopes to learn about what they are made of and how they work. The American physicist Richard Feynman once compared this process to slamming delicate watches together and figuring out how they work by only examining the broken debris. While this analogy may seem pessimistic, with sufficient mathematical models and experimental precision, considerable information can be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature of the forces at work and also about the sub-structure of such systems.

The experiments are in the category of “high energy physics” (also known as “subatomic” physics). The primary tool of scientific exploration in these experiments is an extremely violent collision between two very, very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the more detail of the original system you are able to resolve. These experiments are operated at laboratories such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the collisions are roughly the size of towns. For example, the RHIC collider at BNL is a ring about 1 km in diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a ring 9 km in diameter!

#### Case study 1: Atoms and its Constituents

Questions

1. What are isotopes? (2)

2. What are atoms made up of? (3)

3. Why do you think protons are used in the experiments and not atoms like carbon? (2)

4. Why do you think it is necessary to find out what atoms are made up of and how they behave during collisions? (2)

5. Two protons (mass 1,67 × 10−27 kg) collide and somehow stick together after the collision. If each proton travelled with an initial velocity of 5 × 107 m·s−1 and they collided at an angle of 90°, what is the velocity of the combination after the collision. (9)