Download Safety Devices in Vehicles and Change in Momentum

Momentum and Impulse: Lesson 3
SAFETY DEVICES IN VEHICLES AND CHANGE IN
MOMENTUM
It`s not how fast you are going: it is how fast
you stop
 To reduce the impact force, increase the
stopping distance or increase the stopping
time

F (t )  mv
constan
t
Eg) The driver of a car has a mass of 100 kg. The car is
vf  0
travelling at 100 km/h.
Impact
The driver is not wearing a seat belt. It takes 0.10s for
the driver to impact onto a concrete wall and stop
Determine the impact force on the driver.
F (t )  mv
v  v f  vi  27.778m / s
F (0.10s)  100kg (27.8m / s)
F  27800 N  very l arg e force exerted onto body, result  death
What if the driver wears a seatbelt and the car has an
airbag? The airbag inflates in 3/100 s so the driver
comes to rest in 1.0 s. Determine the impact force now.
F (t )  mv
F (1.0s )  100kg (27.8m / s )
F  2780 N less impact force, more likely for driver to survive
STOPPING DISTANCE
ball
Very short
stopping distance
F(up)
ball
Concrete
floor
-High impact
force
Greater stopping distance, less impact force
foam
If you increase the stopping distance or increase the
stopping time, there is less impact force resulting in
less injury being sustained
LAW OF CONSERVATION OF MOMENTUM


Definition: in an
isolated system the total
momentum before a
collision or interaction is
equal to the total
momentum after the
collision
Total momentum
remains constant


Isolated system – is a
condition in which there
is no exchange of matter
of energy
Collision – interaction or
contact of two or more
objects during a short
period of time
Conservation of p
 momentum before  momentum after
p

p
 before  after
APPLICATIONS OF THE LAW OF CONSERVATION
OF MOMENTUM
Recoil Problems
Eg) A 500 kg cannon fires an 800 g shell at a velocity of
700 east. Determine the recoil velocity of the cannon.
1.
Before
After
Eg) A 500 kg cannon fires an 800 g shell at a
velocity of 700 east. Determine the recoil velocity
of the cannon.
p
before
  p after
cannon shell
cannon shell
m1v1  m2 v2  m v  m2 v2
' '
1 1
'
'
00
 500kg (v1' )  .800kg (700m / s)
0
 500kg (v )  560kgm / s
'
1
560kgm / s  500kg (v1' )
v1 '  1.12m / s
2ND METHOD
pwest  peast
cannon shell
m1v1  m2 v2
500kg (v1 )  0.800kg (700m / s)
500kg (v1 )  560kgm / s
v1  1.12m / s
 negative because it ' s west 
RECOIL PROBLEMS

Eg) A bomb bursts into two fragments, X and Y.
Fragment X has a mass of 0.50 kg and is directed
at 240 m/s left. Fragment Y has a mass 3.5 times
that of X and moves in the opposite direction.
Determine the velocity of Fragment Y.
pleft  pright
x
y
m1v1  m2 v2
0.80kg (240m / s )  3.5(0.80kg )(v2 )
v2  69m / s