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```					Momentum
Newton’s 2nd Law tells us that

F  ma

From here we get ...
F  ma
F  ma
v
F m
t
F  ma
v
F m
t
F t  mv
Imagine trying to throw a tennis ball and
a bowling ball at the same speed.

Since   F t  mv ,
the product F t would have to be
greater for the bowling ball to get the
same v .

(The bowling ball has the greater mass.)
To get the same speed we would have to
throw harder (more force) or longer
(more time). Either way, the product F t
is important. It is called the impulse of
the force.
Imagine this time that we give the tennis
ball and the bowling ball the same
impulse. Since the mass of the bowling
ball is greater, its velocity will be less.

So the product mv is quite different
from just v on its own. We call this
product momentum.
p  mv
Velocity tells us how fast and in which
direction an object moves, it tell us
nothing about the effort required to get it
moving or to get it to stop.

Momentum tells us the impulse required
to get it moving or to get it to stop, but
does not tell us the speed of the object.
Two people on ice skates are at rest.
Two people on ice skates are at rest.
The boy pushes the man.
The boy pushes the man.
They move off in opposite directions, but
their speeds are inversely proportional to
their masses. If the 50 kg boy pushes the
80 kg man so that he moves off at 0.25 m/s,
we find that the boy moves off at - 0.40 m/s.
The momentum of the man is
mv  80  0.25  20 kgm/s
The momentum of the man is
mv  80  0.25  20 kgm/s

The momentum of the boy is
mv  50  0.40  20 kgm/s
This shows that momentum is conserved.
This shows that momentum is conserved.

We can express this as
This shows that momentum is conserved.

We can express this as
m1v1  m2 v2  (m1  m2 )v3
This shows that momentum is conserved.

We can express this as
m1v1  m2 v2  (m1  m2 )v3
or
ptotal  p1  p2  p3  ....
This shows that momentum is conserved.

We can express this as
m1v1  m2 v2  (m1  m2 )v3
or
ptotal  p1  p2  p3  ....
or
p  0
In a 60km/hr zone, a volvo skids 14.3 m
before colliding with a parked BMW. The
cars become locked together and skid another
6.6 m before stopping. The coefficient of
friction between the road and the tyres is
0.75. The mass of the volvo (and driver) is
1470 kg and the mass of the BMW is 1260 kg.
In a 60km/hr zone, a volvo skids 14.3 m
before colliding with a parked BMW. The
cars become locked together and skid another
6.6 m before stopping. The coefficient of
friction between the road and the tyres is
0.75. The mass of the volvo (and driver) is
1470 kg and the mass of the BMW is 1260 kg.

Was the volvo speeding?
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
ma  mg
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
ma  mg
 is the coefficient
a  g              of friction. It is
negative because
friction is in the opposite
direction to the
movement of the car.
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
ma  mg
 is the coefficient
a  g               of friction. It is
negative because
a  0.75  9.8      friction is in the opposite
direction to the
 7.35 m/s   2
movement of the car.
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
ma  mg
 is the coefficient
a  g                of friction. It is
negative because
a  0.75  9.8       friction is in the opposite
direction to the
 7.35 m/s   2
movement of the car.

The deceleration of the two cars is -7.35 m/s2
When the two cars became locked together and
skidded for 6.6 m, the only horizontal force
on the cars was due to friction. Balancing
forces gives us
ma  mg
 is the coefficient
a  g                of friction. It is
negative because
a  0.75  9.8       friction is in the opposite
direction to the
 7.35 m/s   2
movement of the car.

The deceleration of the two cars is -7.35 m/s2
v  u  2as
2    2

u  v  2as
2      2

 0  2  7.35  6.6
 97.02
u  97.02
 9.85 m/s
v  u  2as
2    2

u  v  2as
2      2

 0  2  7.35  6.6
 97.02
u  97.02
 9.85 m/s

The initial velocity of the two cars is 9.85 m/s
The momentum of the two cars, after the
collision, is given by
pvolvo& BMW  mvolvo& BMW vvolvo& BMW
The momentum of the two cars, after the
collision, is given by
pvolvo& BMW  mvolvo& BMW vvolvo& BMW
pvolvo&BMW  2730  9.85
 26890.5 kgm/s
The momentum of the two cars, after the
collision, is given by
pvolvo& BMW  mvolvo& BMW vvolvo& BMW
pvolvo&BMW  2730  9.85
 26890.5 kgm/s
The momentum of the two cars, before the
collision, is given by
The momentum of the two cars, after the
collision, is given by
pvolvo& BMW  mvolvo& BMW vvolvo& BMW
pvolvo&BMW  2730  9.85
 26890.5 kgm/s
The momentum of the two cars, before the
collision, is given by
pvolvo  pBMW  mvolvo vvolvo  mBMW vBMW
pvolvo  pBMW  mvolvo vvolvo  mBMW vBMW
pvolvo  0  1470  vvolvo  1260  0
pvolvo  1470vvolvo
pvolvo  pBMW  mvolvo vvolvo  mBMW vBMW
pvolvo  0  1470  vvolvo  1260  0
pvolvo  1470vvolvo
Since momentum is conserved,
pvolvo  pBMW  mvolvo vvolvo  mBMW vBMW
pvolvo  0  1470  vvolvo  1260  0
pvolvo  1470vvolvo
Since momentum is conserved,
pvolvo (before)  pvolvo& BMW (after)
1470vvolvo  26890.5
26890.5
vvolvo   
1470
vvolvo    18.3 m/s
v  u  2as
2    2

u  v  2as
2      2

 18.3  2  7.35  14.3
2

 545.1
u  545.1
 23.4 m/s
 23.4  3.6 km/h
 84.0 km/h
Yes, the volvo was speeding.

```
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