Transcript Chapter 3:Force, Momentum and Impulse

PHYSICS
CHAPTER 3
CHAPTER 3:
Momentum
and Impulse
(2 Hours)
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PHYSICS
CHAPTER 3
Learning Outcome:
3.1 Momentum and Impulse (1 hour)
At the end of this chapter, students should
be able to:
 Define momentum.


Define impulse J = Ft and use F-t graph
to determine impulse.
Use J   p
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CHAPTER 3
3.1 Momentum and Impulse
Momentum
 is defined as the product between mass and
velocity.
 is a vector quantity.


 Equation :
p  mv


The S.I. unit of linear momentum is kg m s-1.
The direction of the momentum is the same as
the direction of the velocity.
Momentum can be
resolve into
vertical (y)
component &
horizontal (x)
component.

p
py

p x  p cos θ  mv cos θ
p y  p sin θ  mv sin θ
px
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
Impulse, J

CHAPTER 3
Let a single constant force, F acts on an object in a short time
interval (collision), thus the Newton’s 2nd law can be written as


 dp
 F  F  dt  constant
 
 

J  Fdt  dp  p2  p1

where p : final momentum
2
p1 : initial momentum



F : impulsive force
is defined as the product of a force, F and the time, t
OR the change of momentum.
is a vector quantity whose direction is the same as the
constant force on the object.
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

CHAPTER 3
The S.I. unit of impulse is N s or kg m s1.
If the force acts on the object is not constant then


t2 
J   Fdt  Fav dt
t1

where Fav : average impulsive force

Since impulse and momentum are both vector quantities, then
it is often easiest to use them in component form :
consider 2-D
collision only
J x  Fav x dt  p2 x  p1x  mvx  u x 

J y  Fav y dt  p2 y  p1 y  m v y  u y

J z  Fav z dt  p2 z  p1z  mvz  u z 
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
CHAPTER 3
When two objects in collision, the impulsive force, F against
time, t graph is given by the Figure 3.1.
F
t
0 t1
t2
Figure 3.1
Shaded area under the Ft graph = impulse
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Example 3.1
CHAPTER 3
A car of mass 800 kg is travelling at 25
m/s. Find the constant force needed to
stop it in 4 seconds.
Solution
PHYSICS
CHAPTER 3
Example 3.2 :
A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100
ms1 and it bounces off with a speed of 70 m s1 in the opposite
direction.
a. Calculate the magnitude of impulse delivered to the ball by the wall,
b. If the ball is in contact with the wall for 10 ms, determine the
magnitude of average force exerted by the wall on the ball.
Solution :
m1  0.20 kg
u1  100 m s
1
1
Wall (2)
v1  70 m s
1
1
v2  u 2  0
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Solution :
a. From the equation of impulse that the force is constant,
J  dp  p2  p1
Therefore the magnitude of the impulse is 34 N s.
b. Given the contact time,
Fav  3400 N
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Exercise 3.1 :
F kN 
18
0 0.2
1.0
1.8
t ms 
Figure 3.2
An estimated force-time curve for a tennis ball of mass 60.0 g
struck by a racket is shown in Figure 3.2. Determine
a. the impulse delivered to the ball,
b. the speed of the ball after being struck, assuming the ball is
being served so it is nearly at rest initially.
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CHAPTER 3
Learning Outcome:
3.2 Conservation of linear momentum and impulse
(1 hour)
At the end of this chapter, students should be able
to:

State the principle of conservation of linear
momentum.

State the conditions for elastic and inelastic
collisions.

Apply the principle of conservation of
momentum in elastic and inelastic collisions.
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PHYSICS
3.2


CHAPTER 3
Principle of conservation of linear momentum
states “In an isolated (closed) system, the total
momentum of that system is constant.”
OR
“When the net external force on a system is zero,
the total momentum of that system is constant.”

In a Closed system,

F 0

From the Newton’s second law, thus

 dp
F
0
dt

dp  0
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
Therefore p  constant then
p
p

x
 constant
y
 constant
According to the principle of conservation of
linear momentum, we obtain
The total of initial momentum = the total of final momentum
OR


 pi   p f
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Elastic collision
 is defined as one in which the total kinetic energy
(as well as total momentum) of the system is the
same before and after the collision.
 Figure 3.4 shows the head-on collision of two billiard
balls.
m1u1 m2 u 2
Before collision
1
At collision
After collision
2
1
m1v1
2
1
2
Figure 3.3
m2 v2
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
CHAPTER 3
The properties of elastic collision are
a. The coefficient of restitution, e = 1
b. The total momentum is conserved.


 pi   p f
c. The total kinetic energy is conserved.
1
 Ki   K f OR 2
2
m1u1

1
2
2
m2 u2

1
2
2
m1v1

1
2
2
m2 v2
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Inelastic (non-elastic) collision
 is defined as one in which the total kinetic energy of the
system is not the same before and after the collision
(even though the total momentum of the system is
conserved).
 Figure 3.4 shows the model of a completely inelastic
collision of two billiard balls.
u2  0
m1u1
Before collision
1
2
m2
At collision
1
2
After collision
(stick together)
1
2
Figure 3.4
v
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

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Caution:
 Not all the inelastic collision is stick together.
 In fact, inelastic collisions include many situations in which
the bodies do not stick.
The properties of inelastic collision are
a. The coefficient of restitution, 0  e < 1


b. The total momentum is conserved.
 pi   p f
c. The total kinetic energy is not conserved because some
of the energy is converted to internal energy and some of it is
transferred away by means of sound or heat. But the total
energy is conserved.
E  E
i
f
OR
K
i
  K f  losses energy
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Elastic versus inelastic collision
Elastic collision
e=1
Inelastic collision
0  e<1
Coefficient of
resituition


 pi   p f
Momentum
 Ki   K f
Kinetic
energy


 pi   p f
K
i
  K f  losses energy
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Linear momentum in one dimension collision
Example 3.3 :
uB  3 m s
1
uA  6 m s
1
A
B
Figure 3.5
Figure 3.5 shows an object A of mass 200 g collides head-on with object B of
mass 100 g. After the collision, B moves at a speed of 2 m s-1 to the left.
Determine the velocity of A after Collision.
Solution :
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Linear momentum in two dimension collision
Example 3.4 :
m1

u1
m2
50

m1
Before collision
Figure 3.6

v1
After collision
A tennis ball of mass m1 moving with initial velocity u1
collides with a soccer ball of mass m2 initially at rest. After
the collision, the tennis ball is deflected 50 from its initial
direction with a velocity v1 as shown in figure 3.6.
Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s1 and v1
= 4 m s1. Calculate the magnitude and direction of soccer
ball after the collision.
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Solution : m1  0.250 kg; m2  0.900 kg; u1  20 m s
1
u2  0; v1  4 m s ; θ1  50
1

From the principle of conservation of linear momentum,


 pi   p f
The x-component of linear momentum,


pix 


p fx
m1u1 x  m2 u2 x  m1v1 x  m2 v2 x
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Solution :
The y-component of linear momentum,


piy 


p fy
Magnitude of the soccer ball,
v2 
v2 x 
2
Direction of the soccer ball, v
1
 
 v2 y
θ2  tan 
 v2 x
2y
2




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Exercise 3.2 :
1. An object P of mass 4 kg moving with a velocity 4
ms1 collides elastically with another object Q of mass
2 kg moving with a velocity 3 ms1 towards it.
a. Determine the total momentum before collision.
b. If P immediately stop after the collision, calculate
the final velocity of Q.
c. If the two objects stick together after the collision,
calculate the final velocity of both objects.
ANS. : 10 kg ms1; 5 ms1 to the right; 1.7 m s1 to the right
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Exercise 3.2 :
2. A ball moving with a speed of 17 m s1 strikes an identical ball
that is initially at rest. After the collision, the incoming ball has
been deviated by 45 from its original direction, and the struck
ball moves off at 30 from the original direction as shown in
Figure 3.17. Calculate the speed of each ball after the
collision.
ANS. : 8.80 m s 1; 12.4 m s1
Figure 3.7
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THE END…
Next Chapter…
CHAPTER 4 :
FORCE
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