PHYSICS-I/Chapter 9
LINEAR MOMENTUM AND COLLISIONS
Lecturer: Dr. Emine YILDIRIM
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Vocabulary of Chapter 9
interact with=ile etkilesmek
impulse=impuls
collision=carpisma
collide=çarpismak
Elastic collision=elastik(esnek) çarpisma
Inelastic collision=elastik olmayan çarpisma
head-on collision=kafa kafaya çarpisma
bounce off=geri sekmek
rebound=geri semek
momentum=momentum
momenta=momentumlar
repel=itelemek
stick together=birbirine yapismak
explosion=patlama
hit=vurmak
crash=çarpmak
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Linear Momentum
●
The linear momentum of an object or a particle is defined as:
⃗p =m ⃗v
Momentum is a vector quantity
momentum
●
●
SI units of momentum=mass X velocity=kg. m/s=N.s
By using Newton’s second law, we can show the relationship between the
net force acting on a particle and its linear momentum.
This equation says that the time
rate of change of the linear
momentum of a particle is equal
to the net force acting on the
particle.
●
The momentum of any particle CANNOT change UNLESS a net
external force changes it.
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Analysis Model: Isolated System (Momentum)
Here, we will discuss the conservation of momentum of an isolated system.
●
Let’s consider the isolated system of two particles with own mass and velocity in
figure below.
●
According to Newton’s third law, if two particles
interact with each other, they exert a force on each
other equal in magnitude but opposite in direction.
●
F⃗12=− F⃗21
F⃗12 + F⃗21=0
●
Replace force with ma, and then the above equation becomes;
F21 =the force exerted on m1 by m2
F12 =the force exerted on m2 by m1
●
Rewrite the above equation as follows
Remember!!!
There are no external forces
acting on the isolated system
●
Using the definition of momentum
(9.1)
●
Total momentum of isolated system of many particles must be constant.
p⃗total =constant
●
Δ p⃗total =0 (9.2)
Eq. (9.2) can be rewritten as follows
Law of conservation of linear momentum
P1i and P2i =initial momentum vectors
P1f and P2f =final momentum vectors
Linear momentum is conserved separately along each axis.
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Problem 11: Two blocks of masses m and 3m are placed on a
frictionless, horizontal surface. A light spring is attached to the more
massive block, and the blocks are pushed together with the spring
between them . A cord initially holding the blocks together is burned;
after that happens, the block of mass 3m moves to the right with a
speed of 2 m/s. (a) What is the velocity of the block of mass m? b)
Find the system’s original elastic potential energy, taking m =0.35 kg.
Solution: NO EXTERNAL FORCE IN THE SYSTEM
a) The block’s initial speed=0
●
We apply law conservation of linear momentum.
vi=0
Thus ∆p=0, pi=pf
b) Elastic potential energy stored in the spring is converted to kinetic
energy in the blocks.
Ui=?
∆U+∆K=0
Uf – Ui=-(Kf-Ki)
Ui=Kf
Uf=0 ,
Ki =0
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Problem 11: Two blocks of masses m and 3m are placed on a
frictionless, horizontal surface. A light spring is attached to the more
massive block, and the blocks are pushed together with the spring
between them . A cord initially holding the blocks together is burned;
after that happens, the block of mass 3m moves to the right with a
speed of 2 m/s. (c) Is the original energy in the spring or in the cord?
(d) Explain your answer to part (c). (e) Is the momentum of the system
conserved in the bursting-apart process? Explain how that is possible
considering (f) there are large forces acting and (g) there is no motion
beforehand and plenty of motion afterward?
Solution: c) The original energy is in the spring.
d) Compressed spring does work within the system. The cord provide
force, but not displacement thus the cord does not do work. Thus,
only spring provide energy within the system. The original energy is
in the spring.
e) The forces acting on the blocks are internal forces thus the
system is isolated . Momentum is conserved due to isolated system.
f) The forces acting on the blocks are internal so the system is
isolated and the momentum conserved. No matter how large the forces
are.
g) Initial motion of the system is zero thus pi=0, and then system gains
lots of motion thus pf=3m(2) +m(-6)=0. As we see pi=pf momentum is
conserved.
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Problem: A 20 kg body is moving through space in the positive
direction of an x axis with a speed of 200 m/s when, due to an internal
explosion, it breaks into three parts. One part, with a mass of 10 kg,
moves away from the point of explosion with a speed of 100 m/s in the
positive y direction. A second part, with a mass of 4 kg, moves in the
negative x direction with a speed of 500 m/s. (a) In unit-vector notation,
what is the velocity of the third part? (b) How much energy is released
in the explosion? Ignore effects due to the gravitational force.
Solution: a) Linear momentum is conserved independently in x and y
axis.
p =p
xi
xf
M=20kg body breaks into 3 parts
M=m1+m2+m3 20=10+4+m3
m3=6 kg
p xi = p1 xf + p2 xf + p3 xf
M v xi=m1 v 1 xf +m2 v 2 xf +m3 v 3 xf
(20)200 ^i =0+(4)(−500 ^i )+(6) v 3 xf
Solving for v 3 xf v 3 xf =1000 m/ s
pxi=Total initial momentum in x-axis
Pxf= Total final momentum in x-axis
pyi= Total initial momentum in y-axis
p yi = p yf
p yi = p1 yf + p2 yf + p3 yf
M v yi=m1 v 1 yf +m2 v 2 yf +m3 v 3 yf
0=10(100 ^j)+0+(6) v 3 yf
Solving for v 3 yf v 3 yf =−167 m/ s
v⃗3 =1000 ^i−167 ^j
vi=200
v1=100
v2=500
V3=?
vxi=200î
v1xf=0
v2xf=-500î
v3xf=?
vyi=0
v1yf=100j
v2xf=0
V3yf=?
b) Fg is neglected so that ∆Ug=0. ∆K is transferred into energy released
in the explosion .
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Analysis Model: Nonisolated System (Momentum)
Here, we will discuss linear momentum of a nonisolated system.
●
Let’s assume a net force (internal+external forces) acting on a particle
may vary with time.
●
According to Newton’s second law net force ∑F is given by
●
⃗
d
p
∑ F⃗ = dt
⃗ dt
d ⃗p =∑ F
●
If both sides of above equation is integrated, it becomes
The left side of this equation gives us the change in momentum.
●
The change in momentum is called “impulse”
●
The impulse of net force over the time interval ∆t is given
mathematically by
●
Vector quantity
●
●
Thus, the change in the total momentum of the system is equal to the
total impulse on the system.
Impulse theorem can also be written in such component forms as
Δ p x =I x ,
●
Δ p y =I y ,
Δ p z =I z ,
If the average force acting on the particle and time interval ∆t is
known, the impulse can be defined as follows
⃗I =(∑ F
⃗) Δt
avg
Area of the rectangle=area under the curve =impulse
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●
●
●
●
How seat belts and airbags work together
during the collision?
Seat belts and airbags increase the
“collision time” (i.e., the time your
momentum is changing).
⃗I =( ∑ F
⃗) Δt
avg
If Δt is getting bigger Favg is getting
smaller. This means that the seatbelt and
airbag combination could protect
passengers from series injuries and death
during the collision.
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Problem 5: A baseball approaches home plate at a speed of
45m/s, moving horizontally just before being hit by a bat. The
batter hits a pop-up such that after hitting the bat, the baseball is
moving at 55 m/s straight up. The ball has a mass of 145 g and is
in contact with the bat for 2ms. What is the average vector force
the ball exerts on the bat during their interaction?
m=145 g
∆t=2ms
vi=45 m/s
vf=55 m/s
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Problem 16: In a slow-pitch softball game, a 0.2-kg softball crosses
the plate at 15 m/s at an angle of 45° below the horizontal. The batter
hits the ball toward center field, giving it a velocity of 40 m/s at 30°
above the horizontal. (a) Determine the impulse delivered to the ball.
(b) If the force on the ball increases linearly for 4 ms, holds constant
for 20 ms, and then decreases linearly to zero in another 4 ms, what is
the maximum force on the ball?
Solution: a) We use the impulse theorem. ∆p=I
=mvxf-mvxi
Vf=40 m/s
Vi=15 m/s
m=0.2 kg
=mv yf-mv yi
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Problem:
Solution: vi=vf=v
a) Δ p x = p xf − p xi =m(v xf −v xi )
=m(vsin( θ )−vsin( θ ))=0
Δ
p
=
p
−
p
=m(v
−v
)
y
yf
yi
yf
yi
b)
=m(vcos( θ )−(−vcos( θ )))=2 mvcos( θ )
c) Δ p=Δ p x ^i +Δ p y ^j=2 mvcos( θ ) ^j
in + y direction
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Collisions in One Dimension
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Collision: When two or more particles come together for a short time
they produce impulsive force on each other.
●
External forces are negligibly small during the collisions.
●
Collisions can be elastic or inelastic.
●
Linear momentum is conserved during the collisions.
●
We use law of conservation of linear momentum to describe
collisions.
Perfectly inelastic collisions
●
●
●
Two objects collide and then stick together after the collision, then
move with same velocity.
We treat the two objects as a single object after the collision.
Linear momentum are also conserved during this
collision.
Elastic Collisions: Both kinetic energy and linear momentum are
conserved during the collision.
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Collisions in One Dimension
(Continued)
●
In the elastic head-on collisions, if the masses and initial velocities of
objects are known, below equations are used to obtain their final
velocities .
(9.3)
Inelastic Collision
●
●
Kinetic energy is not conserved during the collision, BUT Momentum
is conserved during collision.
Kinetic energy ( KE) is converted to heat, sound, or light.
Types of
Collision
Summary of collisions
Is momentum Is Kinetic energy
conserved?
conserved?
elastic
Yes
yes
inelastic
yes
no
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Problem 25: A railroad car of mass 2.5X104 kg is moving with a
speed of 4 m/s. It collides and couples with three other coupled
railroad cars, each of the same mass as the single car and moving
in the same direction with an initial speed of 2 m/s. (a) What is
the speed of the four cars after the collision? (b) How much
mechanical energy is lost in the collision?
Solution: A railroad car sticks to other 3 coupled cars after the
collision, so that the collision is perfectly inelastic. Let’s take
their final speed vf.
V =4 m/s
1i
V2i= 2m/s
Mass of each car is m=2.5X104 kg
pi=pf ,
p1i+p3i=pf
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Kinetic energy could be transferred to sound, heat energy
Problem 30: As shown in Figure below, a bullet of mass m and speed
v passes completely through a pendulum bob of mass M. The bullet
emerges with a speed of v/2. The pendulum bob is suspended by a
stiff rod (not a string) of length l, and negligible mass. What is the
minimum value of v such that the pendulum bob will barely swing
through a complete vertical circle?
Solution: We deal with an isolated system, so that ∆U+∆K=0, pi=pf
vf=0
Before collision:
●
Bullet’s initial speed=v
●
Bob’s initial speed=0
After collision:
●
Bullet’s final speed=v/2
● Let’s assume that bob gains a speed of v after the collision. After the
b
collision, bob will use its kinetic energy to reach the top of vertical
circle at height 2l.
●
Let’s apply conservation of mechanic energy to bob between bottom
and top of the circle to find vb.
Δ K +Δ U =0
( K f −K i )+(U f −U i )=0
1
2
2
M (0 −v b )+ Mg(2 l−0)
2
v b =2 √ gl
●
The bob will swing barely through a complete vertical circle
this means that its velocity is going to be zero at the top.
Apply law of conservation of linear momentum
p1i=initial momentum of the bullet
p1f=final momentum of the bullet
pi=pf
p1i+p2i=p1f+p2f
P2i=0 (the bob at rest),
before the collision
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Problem 81: A bullet of mass m =8 g is fired into a block of mass M=
250 g that is initially at rest at the edge of a table of height h=1 m. The
bullet remains in the block, and after the impact the block lands d= 2m
from the bottom of the table. Determine the initial speed of the bullet.
●
There are two steps in this problem:
Step 1: Collision of bullet with the block
Step 2: Projectile motion of the block.
The block is initially at rest so
Its initial momentum is zero.
pi=pf
Δ x=d=v xi t
1 2
Δ y=0−h=v yi t− g t
2
Notice that vertical component of velocity
vyi is zero After the collision.
vxi=vf
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Problem 79: A 0.4-kg blue bead slides on a frictionless, curved wire,
starting from rest at point A in Figure below, where h =1.50 m. At point
B, the blue bead collides elastically with a 0.6-kg green bead at rest.
Find the maximum height the green bead rises as it moves up the wire.
Solution: We use conservation of mechanical
energy to find blue bead’s speed before the
collision.
●
Blue bead’s PE energy is transferred to KE. Assume that its final
speed becomes v1.
Δ K +Δ U =0
●
●
Before the collision, the green bead is at rest (v2i=0), the blue bead’s
initial speed is v1i=v1.
The masses and velocities of the objects are known, we can use Eq.
(9.3) to calculate final speed of v2f the after collision.
Apply law of conservation of mechanical energy for green bead to obtain its maximum height
Δ K +Δ U =0
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Problem 89: A 5-g bullet moving with an initial speed of vi =400 m/s is
fired into and passes through a 1-kg block as shown in Figure. The
block, initially at rest on a frictionless, horizontal surface, is connected
to a spring with force constant 900 N/m.The block moves d= 5cm to
the right after impact before being brought to rest by the spring. Find
(a) the speed at which the bullet emerges from the block and (b) the
amount of initial kinetic energy of the bullet that is converted into
internal energy in the bullet– block system during the collision.
Solution: a) m=5 g, M=1 kg, vi =400 m, k=900 N/m
●
Assume that after the collision the block
moves with speed of Vi.
●
We consider two steps in this problem.
Step 1: Collision of bullet with the block
pi=pf
●
Step 2: Conservation of mechanical energy in the block-spring system.
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Problem 89: (b) the amount of initial kinetic energy of the bullet that is
converted into internal energy in the bullet– block system during the
collision.
Solution: b) m=5 g, M=1 kg, vi =400 m, k=900 N/m
374 J energy is transferred to the internal energy!!
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Collisions in Two Dimensions
If two particles collide and their motion is not along a single axis, the
collision is two-dimensional.
●
Last section we studied head on collisions, in which both objects
move on a line. Two dimensional collision is not head-on collision.
●
If the system is isolated, the law of conservation of linear momentum is
applied to the collision shown in the figure.
●
●
We can rewrite the above equation in component forms
Shorthand notations of subscripts
(initial, final)-->(i,f)
(x-axis, y-axis)-->(x,y)
(particle 1, particle 2)-->(1,2)
●
●
Notice that y components of the initial
momentum of two particles is ZERO.
V1iy,2iy=0
Let’s substitute definitions of v1ix ,v2ix , v1fx , v2fx into the above eqs.
If the collision shown in the figure is elastic,kinetic energy is also
conserved.
v2i=0
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Problem 41: A billiard ball moving at 5 m/s strikes a stationary ball of
the same mass. After the collision, the first ball moves at 4.33 m/s at an
angle of 30 with respect to the original line of motion. Assuming an
elastic collision (and ignoring friction and rotational motion), find the
struck ball’s velocity after the collision.
Solution:
●
Apply law of conservation of linear moment for the two balls in xdirection.
m v 1ix +mv 2ix =m v 1 fx +mv 2 fx
m v 1ix +0=m v 1 fx +mv 2 fx
●
v2ix=0
Apply law of conservation of linear moment for the two balls in ydirection.
m v 1iy +mv 2iy =m v 1 fy +mv 2 fy
0=m v y 1 f +m v y 2 f
v 2 fy =−v 1 fy =−(4.33)sin (30)=−2.16 m/ s
2
2 fy
2
2 fx
v 2 f = √ v +v
−1 v 2 fy
ϕ =tan (
)
v 2 fx
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Problem 36: Two automobiles of equal mass approach an inter- section.
One vehicle is traveling with speed 13 m/s toward the east, and the
other is traveling north with speed v2i. Neither driver sees the other. The
vehicles collide in the intersection and stick together, leaving parallel
skid marks at an angle of 55 north of east. The speed limit for both
roads is 35 mi/h, and the driver of the northward-moving vehicle claims
he was within the speed limit when the collision occurred. Is he telling
vf
the truth? Explain your reasoning.
Solution: Use law of conservation of linear
momentum for 2 vehicles.
The driver of the northbound car is NOT telling the truth, his/ her speed
v2i =41.5 mi/h is bigger than the speed limit of 35 mi/h.
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Problem 42: A 90-kg fullback running east with a speed of 5 m/s is
tackled by a 95-kg opponent running north with a speed of 3 m/s. (a)
Explain why the successful tackle constitutes a perfectly inelastic
collision. (b) Calculate the velocity of the players immediately after the
tackle. (c) Determine the mechanical energy that disappears as a result
of the collision. Account for the missing energy.
Solution: a)When 95 kg opponents grabs the 90kg
Fullback, he will not let him to go thus they move
with together. Their collision is perfectly inelastic
Collision.
b) Use law of conservation of linear momentum for 2 players.
For x-direction:
For y-direction:
c)
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The Center of Mass
●
The center of mass (CM) of a system is the point that moves as
if all the mass of the system were concentrated there and all
external forces acted at that point.
●
Translational motion of a system → governed by CM
●
Internal forces do not affect CM motion
The center of mass of the pair of particles is given by
The above equation is extended to a system of many particles in
three dimensions.
For a system of N particles with masses mi at position vectors r⃗i
Cartesian components:
CM lies closer to the heavier mass
If m1=m2 → CM is at the midpoint
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The Center of Mass (Continue)
●
Center of Mass of continuous mass distributions is given by
If the object is
Newton’s second law for CM is :
F⃗net =m a⃗CM
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Example 9.10: A system consists of three particles located as shown
in Figure below. Find the center of mass of the system. The masses
of the particles are m1=m2 =1 kg and m3 = 2 kg.
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48. A uniform piece of sheet metal is shaped as shown in
Figure below . Compute the x and y coordinates of the
center of mass of the piece.
Solution:
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49. A rod of length 30.0 cm has linear density (mass per length)
given by λ =50+20 x , where x is the distance from one end,
measured in meters, and l is in grams/meter. (a) What is the mass
of the rod? (b) How far from the x= 0 end is its center of mass?
Solution:
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Example: The figure above shows a triangular
lamina ABC. The coordinates of A, B and C are
(0, 4) (9, 0) and (0, –4) respectively. Particles of
mass 4m, 6m and 2m are attached at A, B and C
respectively.
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REFERENCES
1) Physics For Scientist and Engineers, SERWAY
and JEWETT, 9th Edition.
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Recommended problems in chapter 9:
4,18,22,34,33,44, 37,70,71,75, 80,77,91
example 9.6,
These problems are highly recommended. You make a good
practice by solving these problems.
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