M1 – MOMENTUM AND IMPULSE
If you apply a force to an object, the effect it has depends on the mass of the object and for how long you exert the force.
You apply a constant force, F Newtons for a time t seconds in the direction of motion to a particle of mass m kg.
The particle accelerates at a ms -2 .
The velocity of the particle increases from u ms -1 to v ms -1 .
You have v = u + at
 a = v−u t
From Newton’s second law you know that F = ma.
 F = m(v−u) t or Ft = mv – mu
This relationship defines two quantities.
On the left hand side, Ft defines the impulse of the force.
Because force is a vector quantity, impulse is also a vector quantity.
The IMPULSE of a force F applied for a time t is defined as Ft.
Impulse = force x time = Ft
The SI unit of impulse is the newton second (Ns)
The right hand side, mv – mu, shows the change in value of the quantity “mass x velocity”.
This quantity is the momentum of the particle. Its units are the same as those of impulse.
For a particle of mass m moving with velocity v,
Momentum = mv
The SI unit of momentum is the newton second (Ns)
In general
Impulse = change in momentum
Ft = mv – mu

M1 – MOMENTUM AND IMPULSE
Example 1
A spacecraft of mass 120kg is travelling in a straight line at a speed of 4ms -1 . Its rocket is fired for
5 seconds, exerting a force of 150N. Find the new velocity of the spacecraft if the force was directed:
(a) in the direction of motion
(b) in the direction opposite to motion
The concept of impulse is most useful when the period of time for which the force acts is very small. For example when you hit a cricket ball with a bat, you cannot easily measure either the force (which is probably not constant anyway) or the time for which it acts, but you can still calculate the change in momentum and hence the impulse.
If you can measure both the impulse and the time, you can find an average force, F, using
Impulse = Ft.
Example 2
A batsman hits a ball of mass 0.15kg at a speed of 40ms -1 . The ball moves in a straight line with constant speed until it strikes a net at right angles and is brought to rest in 0.5 seconds. Find the magnitude of the impulse and the average force exerted by the netting on the ball.

M1 – MOMENTUM AND IMPULSE
Example 3
A particle of mass 2kg, travelling with a velocity of (3i + 5j)ms -1 , is given an impulse of (2i – 4j)Ns.
Find its new velocity.
Example 4
Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms -1 towards Monica. She kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms -1 and with an elevation of 40° to the horizontal.
Find the magnitude and direction of the impulse of Monica’s kick.

EXERCISE 1
M1 – MOMENTUM AND IMPULSE

M1 – MOMENTUM AND IMPULSE

M1 – MOMENTUM AND IMPULSE
CONSERVATION OF LINEAR MOMENTUM
We will now study the motion of systems comprising of two or more bodies, which are free to move separately.
A system consists of two bodies, A and B, which have momentum M
A
and M
B
respectively.
A
M
A
B
M
B
The total momentum of the system = (M
A
+ M
B
) Ns
Body A now exerts a force, F N on body B for a time t seconds.
Body B receives an impulse J = Ft Ns
By Newton’s third law, B exerts a force, -F N on A, also for time t seconds.
Body A receives an impulse –J Ns.
-J
We know that
Impulse = change in momentum
A B
J
The new momentum of A = (M
A
– J) Ns
The new momentum of B = (M
B
+ J) Ns
The total momentum of the system = (M
A
– J) + (M
B
+ J)
= M
A
+ M
B m
A
A v
A m
B
B v
B i.e. the total momentum has not been changed.
The forces described here are internal forces to the system. The total momentum of a system can only be changed by the application of an external force.
This is the principle of conservation of linear momentum.
The principle of conservation of linear momentum
The total momentum of a system in a particular direction remains constant unless an external force is applied in that direction
In the system below:
Before collision m
A
A u
A
After collision m
B u
B m
A v
A m
B
B A B
Total momentum before collision = Total momentum after collision v
B m
A u
A
+ m
B u
B
= m
A v
A
+ m
B v
B

M1 – MOMENTUM AND IMPULSE
Example 5
A particle of mass 4kg, travelling at a speed of 6ms -1 , collides with a second particle, of mass 3kg and travelling in the opposite direction at a speed of 2ms -1 . After the collision, the first particle continues in the same direction but with its speed reduced to 1ms -1 .
Find the velocity of the second particle after the collision.
Example 6
A body of mass 5kg, travelling with a speed of 6ms -1 , collides with a body B, of mass 3kg, travelling in the same direction with a speed of 4ms -1 . On colliding, the two bodies coalesce.
Find the velocity of the combined body after the collision.
Example 7
A railway truck of mass 4 tonnes, travelling along a straight horizontal trail at a speed of 4ms -1 , meets another truck, of mass 2 tonnes, travelling in the opposite direction at a speed of 5ms -1 .
They collide and become coupled together.
Find their velocity after the collision.

M1 – MOMENTUM AND IMPULSE
Example 8
A body of mass 4kg travelling with a velocity of (3i + 2j)ms -1 collides and coalesces with a second body of mass 3kg travelling with velocity (i – 3j)ms -1 .
Find their common velocity after impact.
Example 9
Two particles, A and B, of mass 3kg and 2kg respectively, lie at rest on a smooth horizontal table.
They are connected by a light inextensible string which is initially slack. B starts to move at a speed of 8ms -1 in the direction AB. Find the common velocity of the particles immediately after the string becomes taut, and the impulse received by each of the particles.
Example 10
A bullet of mass 50 grams is fired horizontally from a gun of mass 1kg, which is free to move. The bullet is fired with a velocity of 250ms -1 . Find the speed with which the gun recoils.

M1 – MOMENTUM AND IMPULSE
Example 11
A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms -1 . The gun rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary before the shot is fired. Find the distance the gun will move as a result of firing the shell.

EXERCISE 2
M1 – MOMENTUM AND IMPULSE

M1 – MOMENTUM AND IMPULSE

M1 – MOMENTUM AND IMPULSE