1. Momentum and Impulse
Elliott

2. Momentum
Momentum is the product between mass and velocity.  Being a vector quantity, it has a direction, and the direction is very important when doing momentum calculations.
Units are kilogram metres per second (kg m/s) or Newton seconds (N s). 

3. Check Your Progress
What is the value of the momentum of a 10 kg ball running down a bowling alley at a speed of 5 m s-1?

4. Answer
What is the value of the momentum of a 10 kg ball running down a bowling alley at a speed of 5 m/s?
Formula first: p = mv
p = 10 kg x 5 m s-1 = 50 kg m s-1

5. Signs
It is very important to make sure that you pay attention to the signs when doing momentum calculations. Think of a ball bouncing off a wall. It leaves the wall at the same speed as before. Let’s call going from right to left negative, and going from left to right positive.

6. Check Your Progress
Show that the change in momentum is +2 mv. The ball has a mass of 200 g, and the value of its velocity throughout remains 6 m s-1. What is the change in momentum?

7. Answer Change in momentum = momentum after - momentum before.
Change in momentum = + mv - - mv = + 2mv
Change in momentum = momentum after - momentum before. Change in momentum = + mv - - mv = + 2mv Change in momentum = kg x 6 m s-1  kg x 6 m s-1 = +2.4 kg m s-1

8. Impulse
The change in momentum is called the impulse and is given the physics code Δp.
If we plot a force time graph, we can see that impulse is the area under the graph.
In this graph, both impulses are the same.  The forces and time intervals are different.  In these cases, the force is constant.

9. Impulse and Newton’s Second Law
Consider an object of mass m which is subject to a force F for a time period of t seconds.  We can say that there is an impulse on the object:
change in velocity ÷ time interval = acceleration 
Force = mass × acceleration

10. Check Your Progress
A car is involved in a collision in which it is brought to a standstill from a speed of 24 m s-1.  The driver of mass 85 kg is brought to rest by his seat belt in a time of 400 ms.
a)    Calculate the average force exerted on the driver by his seat belt.                            
      
b)    Compare this force to his weight and hence work out the “g- force”                        
       
c)   Comment on the likelihood of serious injury.

11. Answer
a)      Calculate the average force exerted on the driver by his seat belt.  
First we need to work out the change in momentum (impulse):
Dp = mv - mu = kg x 24 m s- 1 = 2040 N s
Now we can work out the force:
F = Dp/Dt = 2040 N s ÷ s = 5100 N
b)      Compare this force to his weight and hence work out the “g-force”  
The weight of the driver = 85 kg x 9.81 N kg-1 =  834 N
The g-force = 5100 N ÷ 834 N =  6.1 g (i.e.  about 60 m s-2)
 c)   Comment on the likelihood of serious injury.
This will reduce the likelihood of serious injury as the body withstand accelerations of up to 8 g.  Aerobatic pilots regularly pull 5 to 6 g in their manoeuvres