A 0.4 kg basketball bounces off the ground.
The ball’s speed the moment it hits the
ground is 20 ms-1. The ball’s speed the
moment it leaves the ground is 20 ms-1.
a) Calculate the ball’s initial momentum
b) Calculate the ball’s final momentum
c) Calculate the change in momentum,
Δp = pf – pi (*remember - vector subtraction)
Note:
When you are
subtracting a vector,
SWAP the vector then
ADD it.
Momentum
Momentum is the amount of ‘oomph’
a moving object has.
“The more ‘oomph’ the object has, the
harder it is to stop”
p = mv
(vector quantity)
p = momentum (measured in ________)
m = mass (in kg)
v = velocity (in ms-1)
Example ONE
A cricket ball of mass 500 g is bowled
straight into the batter’s face at a speed of
35 ms-1. The batter blocks the ball with his
bat, then the ball flies directly back
towards the bowler’s face at 25 ms-1.
a) Calculate the ball’s initial momentum
b) Calculate the ball’s final momentum
c) Calculate the ball’s change in
momentum
Example TWO
A 120 kg man accidently falls off the roof of
a building. The man’s velocity, the moment
before he hits the ground, is 40 ms-1. He
comes to a complete stop after he has hit
the ground.
Calculate his change in momentum.
Impulse = “change in momentum”
Δp = pf – pi
= mvf – mvi
= m (vf – vi)
= m Δv
= m a Δt
= F Δt
F = ma
Δv
a=
Δt
Impulse = “change in momentum”
Δp = F Δt
“Change in momentum
is a result of force”
F = force
Δt = the duration of the force applied
Example THREE
A driver tries to slow down a runaway car by
pushing against its motion. The car’s mass is
2100 kg and its initial velocity is 3.0 ms-1.
Will the driver manage to stop the car if:
• the resultant force on the car is 600 N
(against the car’s motion) and,
• the force is applied for 9.0 seconds?
Example FOUR
A car travelling at 20 ms-1 crashes into a wall.
The mass of the car is 1500 kg. It takes 0.4
seconds, between the moment the car
comes into contact with the wall, and the
moment the car comes to a complete stop.
a) Calculate the change in momentum
b) Calculate the force exerted on the car
c) Calculate the force if; the wall was very
soft, and it took 1.6 seconds to stop,
instead of 0.4 seconds.
The battle between “F” and “t”
Δp = F Δt
You can achieve the exact same change
in momentum by:
• Applying a great force over a
short period of time or;
• Applying a small force over a
long period of time
Question
How do air bags work?
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With or without the airbag, the person would
experience the same change in momentum
However, with the presence of the airbag, the person
will stop over a longer period of time, because of the
cushioning. Since Δp = F Δt, a longer stopping time
means a smaller force is applied on the person, to
give the same change in momentum.
Without the airbag, the person will stop over a very
short period of time (almost instantaneously). Since
Δp = F Δt, short stopping time means a much greater
force is applied, to give the same change in
momentum.
A greater force means the person is more likely to
suffer greater injuries.
Example FIVE – “Total momentum”
Ball A has a mass of 400 g and is travelling
at 4 ms-1. Ball B has a mass of 250 g and is
travelling at 8 ms-1.
a) Calculate the momentum of ball A
b) Calculate the momentum of ball B
c) Calculate the total momentum
(*remember – vector addition)
Example FIVE – continued
A few moments later, ball A and ball B collide
then each ball moves in the opposite direction
afterwards. Ball A moves at 4.5 ms-1 and ball B
moves at 5.6 ms-1.
a)
b)
Calculate the total momentum
Which is greater – the total momentum
before the collision OR the total
momentum after the collision?
Conservation of Momentum
Total momentum before a collision or
explosion equals the total momentum
after the collision or explosion.
However, this is true only when no
resultant external force is present,
such as gravity or friction.
m1u1 + m2u2 = m1v1 + m2v2
u = initial velocity, v = final velocity
Collisions – elastic vs. inelastic
One may ask – “Couldn’t we just use the
conservation of energy principle, by
calculating the total kinetic energy?”
The answer is NO – because often kinetic
energy is lost due to friction, meaning that
the “total kinetic energy” is not conserved.
However, there are collisions where “total
kinetic energy” IS conserved. These
collisions are called “elastic collisions”.
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Activity 10A (green book, pg. 123)
Questions 1, 2, 3 and 6 only
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Homework Booklet
Worksheet TEN & ELEVEN
•
NCEA 2010 up to page 7
By Tuesday
th
26
June