1. Unit 2
Momentum and Impulse
An Introduction to Momentum

2. When an object is in motion it is said to have momentum.
Momentum can be thought of as the quality of an object that makes it hard to stop. The higher the momentum the harder it is to stop. A stationary particle has no momentum since velocity is zero.
Momentum (p) = mass × velocity
Momentum is a vector quantity since it depends upon the velocity of the particle. We can therefore write the above formula as
p = mv
NB: mass must be in kg and velocity in m/s.
This gives the units of momentum as kgms-1 . Sometimes you will see unit of momentum given as Newton seconds (Ns). So clearly Ns = kgms-1.
Note that p and v are vectors.
Why? Newtons, the basic units of force, are given by N = kg ms–2 (F = ma), so
Newton seconds = N × s
= kg ms–2 × s
= kg ms–1

3. A force is something that has an effect on the object. It may:
If an object changes it’s momentum is because it has been acted upon by a Force.
To a physicist a force is recognised by the effect or effects that it produces
A force is something that has an effect on the object. It may:
Deform (i.e. change its shape)
Speed up/slow down
Slow Down
Mathematical proof:
=𝑚 𝑑𝒗 𝑑𝑡
= 𝑑 (𝑚𝒗) 𝑑𝑡
= 𝑑𝒑 𝑑𝑡
p = mv
By N II we have
F = ma
𝐅 = 𝑑𝒑 𝑑𝑡

4. Calculating momentum 1
A truck of mass 5 kg is travelling at 10 ms–1. Calculate the momentum of the truck and the direction in which it is acting.
10 ms–1
5 kg
Momentum = mass × velocity
= 5 × 10 Ns
= 50 Ns
The momentum of the truck is acting in the direction of motion.

5. Calculating momentum 2
Calculate the momentum of a bullet of mass 4 g travelling at 750 ms–1.
Mass of the bullet = 4 g = kg
Momentum = mass × velocity
= × 750 Ns
= 3 Ns
Again, the momentum is acting in the direction of the motion.

6. Deriving the Impulse Momentum Equation I = Ft = mv – mu
(This formula is for a constant Force.)
F = ma
Take N II
Multiply both sides by t
Ft = mat
Using v = u + at  a = 𝑣 −𝑢 𝑡
Ft = mt 𝑣 −𝑢 𝑡
Ft = m(v – u)
Ft = mv – mu
The value Ft is defined as the Impulse.
Impulse is : the change in the momentum of a body as a result of a force acting upon  it for a short period of time. Impulse is a vector with units of Ns.

7. Impulse The impulse on a body is defined as its change in momentum.
Impulse = change in momentum
= mv – mu
where u is the initial velocity and v is the final velocity.
Impulse is denoted by the vector I.
Note that the impulse I that a body A exerts on a body B is equal to the magnitude of the impulse that B exerts on A but in the opposite direction.
Since impulse is change in momentum it is also measured in Newton seconds.

8. Calculating impulse 1
A ball of mass 0.5 kg hits the floor with a speed of 8 ms–1. It rebounds with a speed of 6ms–1. Find the impulse exerted by the floor on the ball.
Taking the upwards direction to be positive:
Impulse (in Newton seconds)
= (0.5 × 6) – (0.5 × –8)
= 3 + 4
= 7

9. Calculating impulse 2
A ball of mass 0.2 kg falls into a lake. Its velocity before impact with the water was 15 ms–1 and after impact with the water 10 ms–1. Calculate the impulse exerted on the ball by the water.
Taking the downwards direction to be positive:
Impulse = 0.2 × 10 – 0.2 × 15
= 2 – 3
= –1 Ns
Therefore the impulse exerted on the ball by the water is 1 Ns in an upwards direction.

10. Variable Force
There may be time when we can define a variable force F as a function of t, during the time of impact. Suppose such a force acts on a particle of mass m, then from Newton’s second law,
=𝑚 𝑑𝒗 𝑑𝑡
F = ma
𝑡1 𝑡2 𝐅 dt = t1 t2 m d𝐯 dt dt
Integrating this with respect to time we get
Where:
v1 is the velocity at time t1 and
v2 is the velocity at t2
𝑡1 𝑡2 𝐹 𝑑𝑡 = 𝑡1 𝑡2 𝑚 𝑑𝑣 𝑑𝑡 𝑑𝑡
I = 𝑡1 𝑡2 𝐹 𝑑𝑡 = 𝑡1 𝑡2 𝑚 𝑑𝑣 𝑑𝑡 𝑑𝑡
This is another form of the
impulse momentum equation

11. Example
A baseball of mass 45g is hit by a bat. Find the speed of the ball 0.005s after impact if the contact force is
𝐅=800 sin π t
and sketch a graph of the force F against time.
Using
I = 𝑡1 𝑡2 𝐹 𝑑𝑡
So I = 8 𝜋 and I=mv −mu F
= 𝑡1 𝑡2 800 sin π t dt 80
Since u = 0  I = mv
0.005
= 800 (− cos π t ) x 𝜋
 8 𝜋 =0.0045𝐯
0.005
0.005 t
= −4 𝜋 𝑐𝑜𝑠 π t
 8 𝜋 x =𝐯
The area under the (F,t) graph gives the magnitude of the impulse
= ( −4 𝜋 𝑐𝑜𝑠 π x ) −( −4 𝜋 𝑐𝑜𝑠 π x 0)
 𝐯=56.6 𝑚𝑠-1
= 4 𝜋 + 4 𝜋 = 8 𝜋

12. The principle of conservation of momentum
The principle of conservation of linear momentum states that if two objects moving in a straight line collide, the total momentum before the collision is equal to the total momentum after the collision.
We can prove this as follows:
Let the forces acting on two particles of masses m1 and m2 in an isolated system be F1(t) and F2(t) respectively, where the forces are variable and functions of time.
(Note that in an isolated system there are no external forces such as air resistance and friction)

13. By Newton’s Third Law:
F1 = –F2
 F1 + F2 = 0
From Newton’s second Law:
m1a1 + m2a2= 0
Integrating with respect to time we get:
m1v1 + m2v2= constant
Therefore, the total momentum of the system is conserved.
In general, for two particles of mass m1 and m2 we have,
m1u1 + m2u2 = m1v1 + m2v2

14. Modelling collisions

15. Conservation of momentum 1
Two smooth spheres A and B are projected towards each other with speeds of 5 ms–1 and 7 ms–1 respectively. Sphere A has mass 0.3 kg and sphere B has mass 0.4 kg. After they collide sphere A rebounds with a speed of 6 ms–1. Calculate the speed with which sphere B rebounds.
Taking the positive direction of motion to be from left to right:
(5 × 0.3) + (–7 × 0.4) = (–6 × 0.3) + (v × 0.4)
1.5 – 2.8 = – v
0.3 kg
Before impact
0.4 kg
5 ms–1
7 ms–1 A B
4v = 0.5
v = 0.125
Therefore, sphere B moves to the right with a speed of ms–1.
0.3 kg
After impact
0.4 kg
v ms–1
6 ms–1 A B

16. Conservation of momentum 2
A particle A of mass 0.3 kg collides directly with a particle B of mass 0.1 kg. Immediately before the collision A has a velocity of 5 ms–1 and B is at rest. After the collision the particles coalesce into a single particle C and it continues to move in the same direction as A. Find the velocity of C immediately after the collision.
Total momentum before collision = total momentum after
0.3 × 5 = 0.4 × v
0.4v = 1.5
 v = 3.75
Therefore C moves with velocity 3.75 ms–1 immediately after the collision.

17. Example – modelling a given situation
Two satellites of masses 1000Kg and 500Kg are moving with speeds of 10ms and 15m/s at 900 to each other. They collide and after impact move as one body.
Calculate:
The speed of the combined body
The angle through which the 1000Kg satellite is deflected.
Using conservation of momentum
Choose and create a set of appropriate system.
m1u1 + m2u2 = (m1 + m2)v
After impact – moving as one body
Before impact
1000(10i) + 500(15j) =1500v
Speed = 𝐯
speed=
m1 = 1000kg
10000i j =1500v
20 3 𝑖+5𝑗
v =
𝐯 = m/s
1500kg
m2= 500kg b) v
tan = 𝜃 5
First write the initial velocities in terms of i and j 𝜃
20 3
u1 = 10i
u2 = 15j
𝜃=36.9∘ North of East