Mechanics Tutorial 11 - Momentum and Impulse



Momentum Basics


Impulse and Newton's Second Law

Kinetic Energy and Momentum


Momentum Basics

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.  Momentum is not a thing that we can see, but it does explain many things that go on in physics.


  Momentum (kg m/s) = mass (kg) x velocity (m/s)


 p = mv


Units are kilogram metres per second (kg m/s) or Newton seconds (N s).  We can show that the units are the same.


1 N = 1 kg m s-2 (Newton II)


1 kg m s-1 = 1 kg m s-2 × s = 1 N s


Question 1

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?



In the example above, we only looked at the value of the momentum.  This is why we used the word speed.   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.



Question 2

Show that the change in momentum is +2 mv.


Question 3

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?




The change in momentum is called the impulse and is given the physics code Dp.  We can define Newton’s Second Law in terms of change of momentum:


Force = mass × acceleration


Force = mass × change in velocity

                   time interval


Change in velocity = velocity at end - velocity at start


Change in momentum = mass × change in velocity




Force (N) = change in momentum (Ns)

                       time interval (s)     




   Þ Impulse (Ns) = Force (N) x time interval (s)


In code:


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. 


The graph below shows the effect of a force that is not constant:

We can work out the impulse to a first approximation by working out the areas of the rectangles and triangles and adding them together.  In calculus notation, we can write:


If the force is varying to a known function, we can work out the area under the graph using integration.


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:


F = p/t = Dmv/Dt = mv/t


The term v/t = change in velocity ÷ time interval = acceleration

So Force = mass × acceleration which is Newton II.  Remember that wherever there is a resultant force, there is always acceleration.  A change in momentum results in acceleration, which is caused by a force.   We saw another version of this in the previous tutorial.


Question 4

 Explain how the formula F = Dp/Dt is consistent with Newton II.



Impulse is the physics phenomenon that explains how a ball behaves when kicked or hit with a bat.  It also has important implications in road safety.  When a car is involved in a collision, we want the impulse to occur over a longer time interval to reduce the forces involved.


Question 5

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.



Kinetic Energy and Momentum (Edexcel Syllabus)

We know that for a mass m kg travelling at a velocity v m s-1, the momentum p kg m s-1 is given by:


p = mv


We also know that for kinetic energy:

We transpose the momentum equation for v:


We can now substitute this into the kinetic energy equation:


Tidying this up gives us:



Question 6

Use the change in momentum of the driver in Question 5, and his mass to calculate the work done on him.