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8/6/2019 TF015 Ch 3 Momentum and Impulse
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3.1 Conservation of Linear Momentum and Impulse
a) Define momentum.
b) State the principle of conservation of linear momentum.
c) State the conditions for elastic and inelastic collisions.
d) Apply the principle of conservation of momentum in
elastic and inelastic collisions.
e) Define and use the coefficient of restitution,
to determine the types of collisions.
f) Define impulse, and use F-tgraph to
determine impulse.
pJTT
(!
LEARNING OUTCOMES
!12
12
uu
vvek
CHAPTER 3
MOMENTUM & IMPULSE(2 Hours)
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2
Principle of Conservation of Linear Momentum
When the net external force acting on a system is
zero (isolated system), the total momentum of that
system is constant (or conserved).
constant00 system !!!7 pdt
dpFnet
T
Hence, for an isolated system:
! finalinitial ppTT
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3
We often use the principle of conservation of
momentum to solve problems where objectscollide or explode apart.
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Collision
an isolated event in which 2 or more bodies (thecolliding bodies) exert relatively strong forces on
each other for a relatively short time.
In a collision, the change in the momentum of the
system is zero because no external forces are acting
on the system.
Momentum is conserved for any collision
!finalinitial
pp
TT
There are 2 types of collision :
(1) Elastic
(2) Inelastic
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In an inelastic collision,
The momentum of a system is conserved.
But its kinetic energy is not conserved.
( part of the initial kinetic energy would have been
transformed to other forms of energy like sound & heat. )
! finalinitial ppTT
finalintial KK 7{7
In an elastic collision,
Momentum and kinetic energy are conserved.
! finalinitial ppTT
finalintial KK 7!7
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For completely / perfectly inelastic collision,
objects stick together after impact. The
objects have the same final velocity.
vmmumum )( 212211 !Apply Principle Conservation of Momentum,
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Elastic Collision in one dimension motion
Consider 2 particles that undergo an elastic head-on
collision.
In this case, both momentum and kinetic energy are
conserved.According to the Principle of Conservation of
Momentum:
22112211 vmvmumum !
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Knowing that kinetic energy is conserved for the
system:
222222112211
2
1
2
1
2
1
2
1vmvmumum !
Elastic Collision in two dimension motion
- can be analyzed by using the fact that momentum is
a vector quantity.- consider an elastic collision between 2 objects in
which the collision is oblique (not head-on)
collision.
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before after
The initial velocity for object m1 is u1 along the x axis
and the object with mass m2 is initially at rest.
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After the collision, two objects move off in different
direction.
According to Principle of Conservation ofMomentum:
Momentum along x-axis:
afterxbeforex pp 7!722211111 coscos UU vmvmum !
Momentum along y-axis:
afterybeforey pp 7!7222111 sinsin0 UU vmvm !
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ExampleA 200 g tennis ball moving with a velocity of 15 m s1
along x axis collides with a stationary ball of 800 g asshown in figure below. After the collision, tennis ball is
scattered at an angle of 45 from its original direction with
velocity of 5 m s1. Find the final velocity (magnitude and
direction) of the struck ball.
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Solution
Conservation of Momentum along x-axis:
)1(8661.2cos
2929.2cos8.0
cos8.07071.03
cos)8.0(45cos5)2.0()15()2.0(
22
22
22
22
.!
!
!
!
U
U
U
U
v
v
v
v
afterxbeforex pp 7!7
22211111 coscos UU vmvmum !
Conservation of Momentum along y-axis:
afterybeforey pp 7!7
222111 sisi vmvm !
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)2(8839.0sin
7071.0sin8.0
sin)8.0(45sin5)2.0(0
22
22
22
.!
!
!
U
U
U
v
v
v
8661.2
8839.0
cos
sin:
)1(
)2(
22
22
!U
U
v
v
r!
!
14.17
3084.0tan
2
2
U
U
Substitute 2 into equation (2):
1sm32947.0
8839.0
8839.014.17sin
2
2
!!
!r
v
v
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Example ( PYQ Session 2004/05 )
Object A of mass 8 kg moving at 4 m s1 collides with
another object B of mass 6 kg moving at 5 m s1 in theopposite direction. After the collision, object A moves
opposite to its initial direction at 0.1 m s1.
(i) What is the velocity of B after the collision?
(ii) Show that the collision is inelastic.
Solution
Given: Before collision :
mA= 8 kg , uA = 4 m s
1
mB = 6 kg , uB = 5 m s1 ( opposite direction
with object A )
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after collision :
vA = 0.1 m s1 ( opposite to its initial direction ) vB = ?
(i) Using Principle of Conservation of Momentum:
1sm467.0
8.26
6)1.0(8)5(6)4(8
!
!
!
!
7!7
B
B
B
BBAABBAA
afterbefore
v
v
v
vmvmumum
pp
(ii) For an inelastic collision, kinetic energy of the
system is conserved
22
2
1
2
1BBAAbefore uuKE !
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J139
7564
)5)(6(2
1)4)(8(
2
1 22
!
!
!
22
21
21 BBAAafter vvKE !
J6943.0
6543.004.0
)467.0)(6(2
1)1.0)(8(
2
1 22
!
!
!
KE before KE after inelastic collision
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Example
ub
A stationary 2.50 kg block of wood suspended by a wire
of negligible mass. A 0.01 kg bullet is fired into the
block. The block (with the bullet in it) swings to a
maximum height of 0.65 m above the initial position.
Find the initial speed the bullet is fired.
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Solution
Given: m1 = 0.01 kg, m2 = 2.50 kg
hf= 0.65 m
before collision, block is stationary u2 = 0
find u1 = ?
Completely inelastic collision
fvmmumum )( 212211 !
fvu )50.201.0(0)(01.0 1 !)1(2511 .fvu !
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Apply Conservation of Mechanical energy
ff ghmmvmm )()(2
121
2
21 !
)65.0)(81.9(51.2)51.2(2
1 2
!fv-1sm57.3!fv
Substitute into (1) :
)57.3(2511 !u-1sm07.8961 !u
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Coefficient of RestitutionIs the ratio of the difference in velocities before
and after the collision.
!
12
12
uu
vvek
where
v1 & v2 velocity of objects after collision
u1 & u2 velocity of objects before collision
The value ofek is always between 0 and 1.
A perfectly elastic collision has ek= 1
A perfectly inelastic collision has ek= 0
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Example
A ball with the mass of 1 kg moves to the right at 4 m s1
and collides with a stationary ball with the mass of 2 kg.
Calculate the final velocity of each ball if the collision is
elastic.
Solution
A B
uA= 4 m s1
before after
A B
vA= ? vB = ?
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Conservation of momentum
fi pp 7!7
BBAABBAA vmvmumum !
BA vv 2)0(2)4(1 !
)1(24 .BA vv !
The coefficient of restitution
! 12
12
uu
vv
ek
! 40AB vv
)2(4 .! BA vv
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Equating (1) & (2) :
424 ! BB vv-1sm67.2!Bv
( Ball B moves to the right )
4! BA vv
467.2 !Av
-1
sm33.1!Av
From:
( Ball A moves to the left )
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Impulse
When 2 objects such as baseball and a bat, a hammer
and a nail or even two cars collide, they can exert
large forces on one another in a short period of time,t.
Forces of this type, which exist only over a very shorttime during the impact are often called impulsive force
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Impulse of the force equals to the change inthe
momentum of the particles.
JT
FT
pJTT
(!
initialfinal ppTT
!before during after
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From Newtons 2nd law of motion
dtdp!
dtdp !
dtdp
t
t
p
p
f
i
!2
1
Integrating :
where pi momentum of body at time ti
pf momentum of body at time tf
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Impulse isequaltotheareaunderthe Ft graph.
Iftheforce is constant (does not vary withtime)
ortaking theaverageforcefortime interval t(
tJ (!T
dtFpp
t
t
if
!
2
1
fi ttJ tofromgrapht-Funderthearea!T
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Solution
Taking motion to the right as +
(a) Impulse = area under the graph F-t
(b)
20003.02
1vv!
sN3!JT
pJTT
(!
mumv !
))22((06.03 ! v-1
sm2
8!v( to the right )
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Follow Up exercise
A karate student tries to break a board.
How can the abrupt stop of the hand generate so much force
to break the board?
Assume that the hand has a mass of 0.35 kg and that thespeeds of the hand just before and just after hitting the
board are 10 m s1 and 0 m s1 respectively. What is the
average force exerted by the fist on the board if the contact
time is only 0.30 ms?