# Collisions and Impulse

## Collisions and Momentum

### In Physics, a collision occurs when two (or more) objects come into contact with each other.

### In order to discuss collisions, first we must define the Physics concept of Momentum.

### In everyday language, Momentum means "how hard it is to change the Speed of an object". A more Massive object or faster travelling object has more Momentum.

### In Physics, Momentum is defined mathematically as :-

### p = m x v

### Where :-

### m = Mass of the Object (kg)

### v = Velocity of Object (ms^{-1})

### p = Momentum (kgms^{-1})

### As the vector quantity Velocity is used to calculate Momentum, Momentum is also a vector and as such, direction does matter in these calculations. Therefore if an object is :-

### Moving Right - Positive Velocity

### Moving Left - Negative Velocity

### Similar to the conservation of Energy, the total Momentum before a collision is equal to the total Momentum after a collision if no external Forces act. This is referred to as 'The conservation of Momentum'.

## Momentum Calculations

### When any two objects collide, Momentum can be transferred between the objects, causing them to change Speed or direction.

### In order to understand how the two objects behave due to the collision, the Momentum of each before the collision and after the collision must be found.

## Example 1 -

### In Curling, heavy stones are slid across a surface of ice. In one game, a stone with a mass of 18 kg travelling at 3 ms^{-1} collides with a stationary stone of mass 20 kg.

### After the collision, the 18 kg stone continues in the same direction at 1 ms^{-1}. What is the velocity of the 20 kg stone after the collision ?

### Step 1 - Before the collision diagram

### Total Momentum before = m_{1}v_{1} + m_{2}v_{2}

### Total Momentum before = ( 18 x 3 ) + ( 20 x 0 )

### Total Momentum before = 54 + 0

### Total Momentum before = 54 kgms^{-1}

### Step 2 - After the collision diagram

### Total Momentum after = m_{1}v_{1} + m_{2}v_{2}

### Total Momentum after = (18 x 1 + (20 x v_{2})

### Total Momentum after = 18 + 20v_{2}

### Step 3 - Conservation of Momentum

### Total Momentum Before = Total Momentum after

### 54 = 18 + 20 v_{2}

### v_{2} = ^{36 }/ _{20 }

### v_{2} = 1.8 ms^{-1 }(to the right)

## Types of Collision

### In a collision, Momentum is always conserved. The same does not apply to Energy, however, as Energy can be 'lost' by the system through sound, heat, light etc. This gives rise to the definition of the three types of collision :-

### 1. Kinetic Energy conserved - Elastic Collision

### 2. Kinetic Energy lost - Inelastic Collision

### 3. Kinetic Energy gained - Inelastic Explosion

### In order to identify which type of collision occurs, the Kinetic Energy of each object is found both before and after the collision and compared to each other.

### Note - Energy is a scalar quantity, therefore the direction of each object does not matter. All energy values will be positive.

## Example 2 -

### What type of collision is the Curling stone question above?

### Step 1 - Before the collision diagram

### Total Kinetic Energy before = ^{1}/_{2}m_{1}v_{1}^{2} + ^{1}/_{2}m_{2}v_{2}^{2}

### Total Kinetic Energy before = (0.5x18x9) + (0.5x20x0)

### Total Kinetic Energy before = 81 J

### Step 2 - After the collision diagram

### Total Energy after =

### = (0.5 x 18 x 1) + ( .5 x 20 x 3.24)

### = 9 + 32.4

### = 41.4 J

### Total Energy Before > Total Energy After

### Therefore this is an Inelastic Collision

### Note - In real life situations, an elastic collision is very rare. Most collisions will cause energy to be 'lost' from the system in some way, heat, light, sound etc. Therefore most collisions in real life are inelastic. The main example of a real life elastic collision is collisions between molecules within a gas, as they don't physically collide, their electrostatic charges repel each other.

## Change in Momentum and Impulse

### In the above section, it was shown that the total Momentum was conserved in a collision. This does not mean however that the Momentum of each was unchanged. In the above example, this is obvious, as the 20 kg stone gained Momentum in the collision.

### This change in Momentum is caused by Momentum being transferred from one object to another. In order to do this, a Force must be applied by one object onto another. The longer the Force is applied, the more Momentum is transferred.

### By combining these two variables together, the change in Momentum can be found :-

### F t = mv - mu

### Where :-

### F = Force applied to object (N)

### t = time Force acts (s)

### m = Mass of object (kg)

### v = Final Speed of the object (ms^{-1})

### u = Initial Speed of the object (ms^{-1})

### The Force multiplied by the time taken (F x t) is referred to as the impulse applied to the object.

## Example 3 -

### A golf ball of Mass 0.045 kg is struck by a golf club head with a Mass of 0.5 kg. If the club applies a Force of 2500 N over a period of 0.8 ms, at what Velocity does the ball leave the club ?

### F t = mv - mu

### 2500 x 0.8x10^{-3} = (0.045 x v) - (0.045 x 0)

### v = 44 ms^{-1}

## Force-Time Graphs

### If a graph of Force with respect to Time is generated for a collision the following shape would generally be found :-

### The graph above shows the Force-Time graph for a squash ball being hit by a squash racquet.

### As was shown by Velocity-Time graphs in National 5 Physics, the area under the graph can have meaning. In this case, the area under the graph is equal to F x t, the Impulse.

### For a collision with a fixed change of Momentum, If the time of contact is increased, then the maximum Force will be reduced, in order to conserve the area under the graph.

### The following graph shows the effect of hitting a softer squash ball with a squash racquet :-

### As can be seen, the softer ball remains in contact with the racquet for longer, reducing the maximum Force required, by conserving area under the graph.

### This increasing of contact time to reduce overall Force is the Physics principle behind the function of car airbags.

### Note - In the Higher Physics course, all Force-Time graphs will be made up of geometric shapes to allow the calculation of the area under them. In order to find the area under the graphs above, Integration would be required, which is beyond the scope of this course.

## Example 4 -

### A car is travelling at 13 ms^{-1} when it collides with a tree. Due to a fault , the driver's airbag deploys but the passenger airbag fails. The driver collides with the airbag and is brought to rest in 120 ms, whereas the passenger collides with the dashboard and is brought to rest in 20 ms. If both the driver and passenger have masses of 75 kg, calculate the Force acting on :-

### 1. The Driver

### 2. The Passenger

### The Driver -

### Ft = mv - mu

### F = (mv - mu) / t

### F = (75( 0 - 13)) / 120x10^{-3}

### F = - 8125 N

### F = 8.1 kN (in the opposite direction to motion)

### The Passenger -

### Ft = mv - mu

### F = (mv - mu) / t

### F = (75( 0 - 13)) / 20x10-3

### F = - 48750 N

### F = 48.8 kN (in the opposite direction to motion)

### Note - As can be seen in the above example, even though both occupants experience the same Impulse, the Force experienced is much greater without the airbag, meaning that the passenger would be at a much greater risk of injury.

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