rd
3
Newton’s
Law
and Momentum
Specification Points 2.23 – 2.25
EdExcel 9-1 Physics Paper 1
2.23 Recall and apply Newton’s third law both to equilibrium situations
and to collision interactions and relate it to the conservation of
momentum in collisions
2.24 Define momentum, recall and use the equation: momentum
(kilogram metre per second, kg m/s) = mass (kilogram, kg) × velocity
(metre per second, m/s)
p = m× v
2.25 Describe examples of momentum in collisions
Newton’s Third Law
“For every action, there is an equal and opposite reaction.”
When we walk…
• We push against the floor with our feet.
• The floor pushes back on our feet.
When we drive…
• The tyres push against the road.
• The road pushes back against the tyres.
When we swim…
• We push the water backwards.
• The water pushes us forwards.
Interaction pairs
Forces are an interaction between objects. For every force, there are two objects involved:
• one object is causing the force
• the other object is experiencing the force.
We say that one object is exerting a force, and the force is acting on the other object.
But remember, for every force, there is really a pair of forces!
weight
weight
upthrust
normal
weight
weight
weight
normal
Force diagrams
thrust
friction
Interaction pairs
Newton’s third law states that:
For every action, there is an equal and opposite reaction.
This means that each force in an interaction pair
must be:
•
•
•
equal in size
opposite in direction
the same type of force
For each force pair, the object causing the force
and the object experiencing the force are swapped.
Object exerting
(causing the force)
Object force is acting on
(experiencing the force)
Normal force
The table
The book
Interaction
force
The book
The table
Object exerting
(causing the force)
Object force is acting on
(experiencing the force)
Tension
The tree
The rope
Interaction
force
The rope
The tree
Remember each force in an interaction pair must be:
•
•
•
equal in size
opposite in direction
the same type of force.
What is momentum?
Momentum is the tendency of an object to keep
moving in the same direction.
In other words, it is a measure of the difficulty of changing the direction
of an object.
What type of objects do you think would
be difficult to change direction?
What properties would they have?
Momentum is usually
represented by the
letter “p” because the
Latin word for
momentum is “petere”.
Which has more momentum?
Tip: think about which is
easiest to stop?
A – a charging rhino
The rhino has
about 2000 times
more momentum
than the bullet!
B – speeding bullet
Calculating momentum
Momentum = mass x velocity
What are the units of mass?
kg
What are the units of velocity?
m/s
Therefore, can you work out the units of momentum?
kg x m/s
but in maths, we don’t write multiplication signs between letters:
kgm/s
This rhino weighs 2000kg
and is travelling at 14m/s.
What is its momentum?
Momentum = mass x velocity
Momentum = 2000 x 14
Momentum = 28000 kgm/s
This bullet weighs 30g and
is travelling at 450m/s.
What is its momentum?
Which is correct, and why?
Momentum = mass x velocity
Momentum = mass x velocity
Momentum = 30 x 450
Momentum = 0.03 x 450
Momentum = 13,500 kgm/s
Momentum = 13.5 kgm/s
This bullet weighs 30g and
is travelling at 450m/s.
What is its momentum?
Which is correct, and why?
Momentum = mass x velocity
Momentum = mass x velocity
Momentum = 30 x 450
Momentum = 0.03 x 450
Momentum = 13,500 kgm/s
Momentum = 13.5 kgm/s
Vectors and scalars
Do you think momentum is a vector or a scalar? Why?
Momentum = mass x velocity
Momentum is a vector.
A 2kg object travelling at 5m/s upwards has the opposite momentum
of A 2kg object travelling at 5m/s downwards.
Calculate the momentum of:
1) A 730kg Smart car moving north at 20m/s
2) A 2500kg monster truck moving at a velocity of 8m/s south
Calculate the velocity of:
3) A 60kg student with a momentum of 3600kg m/s
4) A 1.5kg bag of sugar falling from a building with a momentum
of 150kg m/s
Calculate the momentum of:
1)
14600kgm/s north
2) 20000kgm/s south
Calculate the velocity of:
3)
60m/s
4) 10m/s down
momentum= mass x velocity
velocity= momentum / mass
mass= momentum / velocity
Conservation of Momentum
In any collision or explosion momentum is conserved (provided that there
are no external forces have an effect). Example question:
Two cars are racing around the M25. Car A collides with the back of car B
and the cars stick together. What speed do they move at after the
collision?
Speed = 50m/s
Mass = 1000kg
Speed = 20m/s
Mass = 800kg
Mass = 1800kg
Speed = ??m/s
Momentum before = momentum after…
…so 1000 x 50 + 800 x 20 = 1800 x V…
…V = 36.7m/s
Momentum in different directions
What happens if the bodies are moving in opposite directions?
Speed = 50m/s
Mass = 1000kg
Speed = 20m/s
Mass = 800kg
Momentum is a VECTOR quantity, so the momentum of the
second car is negative…
Total momentum = 1000 x 50 – 800 x 20 = 34000 kgm/s
Speed after collision = 34000 kgm/s / 1800 = 18.9m/s
Another example
Consider the nuclear decay of Americium-241:
237
93
Np
241
95
Am
If the new neptunium atom moves away at
a speed of 5x105 m/s what was the speed
of the alpha particle?
2.96x107 m/s
4
2
α
More questions…
1.
A car of mass 1000kg heading up the M1 at 50m/s collides with a
stationary truck of mass 8000kg and sticks to it. What velocity
does the wreckage move forward at?
2. A defender running away from a goalkeeper at 5m/s is hit in the
back of his head by the goal kick. The ball stops dead and the
player’s speed increases to 5.5m/s. If the ball had a mass of 500g
and the player had a mass of 70kg how fast was the ball moving?
3. A white snooker ball moving at 5m/s strikes a red ball and pots it.
Both balls have a mass of 1kg. If the white ball continued in the
same direction at 2m/s what was the velocity of the red ball?
4. A gun has a recoil speed of 2m/s when firing. If the gun has a mass
of 2kg and the bullet has a mass of 10g what speed does the bullet
come out at?
5.6m/s
70m/s
3m/s
400m/s
Recap question on momentum
1. Bradley and Jack are racing against each other over 400m at Sports
Day. Brad is running at 8m/s and catches up with Jack who is running
at 6m/s. After the collision Brad stops and Jack moves slightly
faster. If Brad’s mass is 60kg and Jack’s is 70kg calculate how fast
Jack moves after the collision.
12.9m/s
2. Coryn is driving her 5kg toy car around. It is travelling at 10m/s
when it hits the back of Shannon’s (stationary) leg and sticks to it.
Assuming Shannon’s leg can move freely and has a mass of 10kg
calculate how fast it will move after the collision.
3.3m/s
Change in Momentum and Force
Instead of F=ma Newton actually said that the force acting on an
object is that object’s rate of change of momentum. In other words…
Force = Change in momentum (in kgm/s)
(in N)
Time (in s)
You DON’T need to learn this
equation!!
mv
F
T
For example, Rob Stocker scores from a free kick by kicking a stationary football with a
force of 40N. If the ball has a mass of 0.5kg and his foot is in contact with the ball for
0.1s calculate:
1) The change in momentum of the ball (its impulse),
2) The speed the ball moves away with
Example questions
1) Jack likes playing golf. He strikes a golf ball with a force of
80N. If the ball has a mass of 200g and the club is in contact
with it for 0.2s calculate a) the change in momentum of the golf
ball, b) its speed.
2) Chad thinks it’s funny to hit tennis balls at Illy. He strikes a
serve with a force of 30N. If the ball has a mass of 250g and
the racket is in contact with it for 0.15s calculate the ball’s
change in momentum and its speed.
3) Oli takes a dropkick by kicking a 0.4kg rugby ball away at 10m/s.
If his foot was in contact with the ball for 0.1 seconds calculate
the force he applied to the ball.
4) Paddy strikes a 200g golf ball away at 50m/s. If he applied a
force of 50N calculate how long his club was in contact with the
ball for.
160kgm/s,
80m/s
4.5kgm/s,
18m/s
40N
0.2s
2.23 Recall and apply Newton’s third law both to equilibrium situations
and to collision interactions and relate it to the conservation of
momentum in collisions
2.24 Define momentum, recall and use the equation: momentum
(kilogram metre per second, kg m/s) = mass (kilogram, kg) × velocity
(metre per second, m/s)
p = m× v
2.25 Describe examples of momentum in collisions