Transcript Document

Impulse
Notes p.29
During any collision some momentum is
transferred from one object to another.
So… momentum of one object decreases by Dp
and momentum of the other object increases by
Dp.
Dp = change in momentum of one object
Dp = the IMPULSE of the collision
Note: Impulse is just “change in momentum” so has momentum units!
Now if the velocity of one object, before the
collision is “u” and the velocity of the same object,
after the collision, is “v”
then impulse = Dp = m
v-mu
Dp = m (v – u)
Then consider that during the collision the object
experiences an average force, F, over a contact
time, t, causing an acceleration, a.
So Dp = m(v-u)
= m(at)
IMPULSE = Dp = F t
or
m(v – u) = F t
Note: Impulse units can therefore also be “Ns”.
Impulse Example 1
A 0.8 kg trolley travelling at 0.7 ms-1 hits a wall.
a) The trolley rebounds at 0.3 ms-1. Calculate the
impulse of the force on it.
Dp = m (v – u)
m = 0.8kg
u = 0.7ms-1
v=
-0.3ms-1
Dp = ?
= 0.8 (-0.3 – 0.7)
= - 0.8 kg ms-1
Impulse = 0.8 kgms-1
from wall onto trolley
Impulse Example 1 (cont.)
b) The trolley is in contact with the wall for 50ms.
Calculate the average force the wall exerts on
the trolley during this time.
Dp = - 0.8 kgm/s
t = 50 x 10-3 s
F =
?
Dp
F
= Ft
= Dp / t
= - 0.8 / 50 x 10-3
= - 16 N
Average Force is 16N
from wall onto trolley
Problems 89 – 91.
Notes p.31
IMPULSE from a Force-Time Graph
A force-time graph can be plotted showing how
the force on an object changes during the contact
time of a collision.
The equation, m(v-u) = Ft, refers to average
force! In reality, the force on an object would
vary like this:
Force (N)
time (s)
Force (N)
F
time (s)
IMPULSE = F t
= AREA UNDER F-t GRAPH
Impulse Example 2
During a collision the force experienced by a
700kg stationary object was recorded as follows:
800
Force
(N)
0
2
time (s)
Determine the speed of the object after the
collision.
800
Force
(N)
0
2
time (s)
1st IMPULSE = AREA = ½ x 2 x 800 = 800 kgms-1
2nd m = 700 kg
Dp
= mv (as u = 0ms-1)
u = 0 ms-1
v= ?
Dp = 800kgms-1
= m (v – u)
v
= Dp / m
= 800 / 700
= 1.14 ms-1
Key Questions on Impulse
A car travelling at 20 ms-1 crashes into a wall
and is brought to rest by the wall.
An identical car crashes into the same wall at
the same speed, but it has a cushioned bumper
so it takes longer to come to rest.
1. In which collision is the impulse greater?
Explain your answer.
Answer
The impulse is the same in each collision!
As the object masses and initial and final
velocities are unchanged, then the change in
momentum (Dp = m(v – u)) is unchanged.
Key Questions on Impulse
2. Explain why the second car experiences
less damage.
Answer
The second car takes longer to come to rest.
The IMPULSE = F t, and the impulse is
unchanged.
So, as t has increased, then F must decrease
to keep the impulse of the collision the same.
This reduced force means less damage to the car!
3. On the one set of axes, sketch graphs of
force against contact time for each
collision.
Force (N)
First car – big force,
short time.
0
Second car – longer time
so smaller force
BOTH SAME
AREA !!
time (s)
area under the graph is equal to the
impulse.
The
N.B.
If the impulse stays constant but contact time
increases, then the maximum force has to
decrease to allow the area under the graph to
remain the same. (Think … air bags!)
One practical use of impulse is to find the force
exerted on one object by another. For example, you
could find the force exerted by a club on a golf ball.
Newton’s Third Law
From National 5 you might remember Newton’s
Third Law …
“Every force has an equal and opposite reaction
force.”
This law is the basis of Impulse theory.
Think outside the box ???!!!
Try to make the connection
between Newton’s Third Law and
Impulse theory.
In a collision between 2 objects, each object
experiences an equal but opposite force from the
other object.
As Impulse = av Force
x
time
then, during a collision, each object must
experience equal but opposite impulses.
This means that as one object experiences a
POSITIVE change in momentum, the other will
experience the same size but NEGATIVE change
in momentum.
So momentum is “transferred” .
Problems to do
Q. 76- 88
76a) 20 kgms-1, right b) 500kgms-1, down c) 9kgms-1, left
77 0.75 ms-1,
Ek before = 2.25J, Ek after = 1.125J so INELASTIC
78 2.4 ms-1
79 3 kg
81
82a) 23ms-1
8.6 ms-1
80 a) 2.7ms-1 b) 0.2J
82 b) INELASTIC … Ek before = 730000J,
Ek after = 702000J .
83
8.67 ms-1
84 0.6ms-1 in initial direction of 2nd vehicle
85
16.7 ms-1
86 4kg
88
1.3 ms-1
87
0.8 ms-1 in opp direction
Problems 89 – 99.
89. 1.58 ms-1
90. 100N
92. 2.67 ms-1
91. 0.03s
93. a) -0.39 kgms-1
b) 15.6N
94. a) 3.96 ms-1, 2.97 ms-1 b) 0.1s
95.a)
AB = const acceleration towards wall, BC = in contact with wall slowing
down towards wall, CD = in contact with wall speeding up away from wall, DE
slowing down to rest moving away from wall.
b) 0.2s
96. 1250N
c) –20N
d) 4.0 J
97. 9x104 N
98. See notes
99.a) (i) 4 ms-1 (ii) 4 kgms-1 (iii) 4 kgms-1
b) 8N
EXTRA PRACTICE!
Revision questions for Higher Physics, page 23,
Q. 1 - 16