7-1 impulse and linear momentum
Impulse (change in momentum)
The rate of change of momentum of
a body is equal to the net force
acting on it :
During a collision, a force F acts on
an object, thus causing a change in
momentum of the object:
The product of Force and time is known
as IMPULSE
units of impulse are Ns
A change in momentum is called “impulse”:
Think of hitting a soccer ball: A force F acting over a time Dt causes a change Dp in the
momentum (velocity) of the ball.
Linear Momentum and Collisions
The linear momentum of a particle of mass m and velocity v is defined
as


p  mv
The linear momentum is a vector quantity. It’s direction
is along v.
The components of the momentum of a particle:
px  m  vx
py  m  vy
pz  m  vz
momentum also depends on the mass.
So changing the mass of an object will also change
the momentum vector.
Therefore to change momentum one must change
the mass or velocity or both.
Regardless of what changes, the momentum vector
is always in the same direction as the velocity
vector.
units of momentum are kg m/s
m1
v1
p2 = -m2v2
p1 = m1v1
v2
m2
Note that particles moving in opposite directions
have momentum which are opposite sign!
example
• A 1200 kg car drives west at 25 m/s for 3
hours. What is the car’s momentum?
• 1200 kg = mass
• 25m/s, west = velocity
• 3 hours = time
P = mv = 1200 x 25 = 30000 kg m/s,west
example
•a car of mass 1 000 kg with a velocity of 8
m/s. The momentum of the car is .
p = m・ v
= (1000)(8) = 8000 kg ・ m / s
•a motor cycle of mass 250 kg travelling at 32 m/s.
The momentum of the motor cycle is
p = m・ v
= (250)(32)= 8000 kg ・ m / s
If the boulder and the boy
have the same momentum,
will the boulder crush the
boy?
Hint: Which would have the
larger speed?
1.
When two objects with different masses collide, the force
on the less massive object is larger than the force on the
more massive object.
Both objects experience the same magnitude of force
but in opposite directions. However, greater damage
is suffered by the less massive object because of the
greater momentum of the more massive one.
example
2 cars are heading east, car A is traveling 30m/s, car B is
traveling 60m/s. Each car weighs 2000Kg.
– What is the momentum of car A?
– What is the momentum of car B?
– If car B crashes into car A, what is the total momentum?
p=mv
Car (A) momentum = 30m/s x 2000Kg
pX = 60,000 mi-lbs/hr east
Car (B)momentum = 60m/s x 2000Kg
pY = 120,000 m-Kg/s west
Total momentum = pA - pB
= 120,000 - 60,000
= 60,000 m-Kg/s west
Impulse (change in momentum)
The rate of change of momentum of
a body is equal to the net force
acting on it :
During a collision, a force F acts on
an object, thus causing a change in
momentum of the object:
The product of Force and time is known
as IMPULSE
units of impulse are Ns
A change in momentum is called “impulse”:
Think of hitting a soccer ball: A force F acting over a time Dt causes a change Dp in the
momentum (velocity) of the ball.
How do the safety features of a car protect
you in a collision?
• In an accident, the Dp stays the same, but you can
change F and t of impact.
• When you increase the time that the collision lasts,
the force of impact decreases.
Ft
• All safety features in a car are designed to reduce the
force by increasing the time you are in contact with
that force.
• Watch the video!
Remember: Ft = Dp
the momentum change of an object is the
mass X velocity change
the impulse equals the momentum change
Impulse is a vector quantity and has
the same direction as the average force.
Impulse
Final
momentum
Initial
momentum
Washing a Car: Momentum Change and
Force.
Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s
and is aimed at the side of a car, which stops it. (That is, we ignore
an splashing back.) What is the force exerted by the water on the
car?
If the water splashes back from the car, the change in
momentum will be greater in magnitude, and so will the
force on the car be greater in magnitude. Note that pfinal
will now point in the negative x direction. So the result
for F will be minus something of magnitude depending
on the water’s rebound speed.). To put it simply, the car
exerts not only a force to stop the water, but also an
additional force to give it momentum in the opposite
direction.
Force & Impulse
• Two boxes, one heavier than the other, are initially at rest on a horizontal
frictionless surface. The same constant force F acts on each box for exactly
1 second.
– Which box has the most momentum after the force acts?
(a) heavier
(b)
lighter
(c) same
Dp
so Dp  Fav Dt
Dt
In this problem F and Dt are the same for both boxes!
We know Fav 
The boxes will have the same final momentum.
F
light
F
heavy
Why does an egg break or not break?
• An egg dropped on a tile floor breaks, but an
egg dropped on a pillow does not. Why?
FΔt= mΔv
In both cases, m and Δv are the same. If Δt goes
up, what happens to F, the force?
Right! Force goes down. When dropped on a
pillow, the egg starts to slow down as soon as
it touches it. A pillow increases the time the
egg takes to stops.
Impulse and contact time:
Impulses and Contact Time
Example
• The baseball has a mass of 0.14kg and an initial speed of 30 ms-1.
It rebounds from the bat with a speed of 40 ms-1 in the opposite
direction and is in contact with the bat for 0.002 s. Determine the
average force on a baseball hit by a bat.
Example :
• The engine of a model rocket is rated with a total impulse of
5.00 Ns and a thrust duration 0f 1.20 s. what is the average
force exerted by the engine?
 FDt
Impulse
F 
Dt
5.00

1.20
 4.17 N
Impulse
Example
• A 68 kg soccer player kicks a stationary 0.425kg ball giving it a speed 0f
13.7 ms-1. The player foot is in contact with the ball for 0.097 s. What is
the average force on the ball?
The cart’s change of momentum is
A. 30 kg m/s.
B. 10 kg m/s.
C. –10 kg m/s.
D.–20 kg m/s.
E. –30 kg m/s.
example
A 57 gram tennis ball falls on a tile floor. The ball
changes velocity from -1.2 m/s to +1.2 m/s in
0.02 s. What is the average force on the ball?
Mass = 57 g = 0.057 kg
Δvelocity = +1.2 – (-1.2) = 2.4 m/s
Time = 0.02 s
FΔt= mΔv
F x (0.02 s) = (0.057 kg)(2.4 m/s)
F= 6.8 N
Car Crash
Would you rather be in a
head on collision with an
identical car, traveling at
the same speed as you,
or a brick wall?
Assume in both
situations you come to a
complete stop.
It Does Not Matter!
Look at FΔt= mΔv
In both situations, Δt, m, and Δv are the
same! The time it takes you to stop
depends on your car, m is the mass of
your car, and Δv depends on how fast you
were initially traveling.
example
• If the halfback experienced a force of 800 N
for 0.9 seconds to the north, determine the
impulse
Given: F = 800 N
• F (Δ t ) = m D v
t = 0.9 s
Find :
• 800N ( 0.9s ) = 720 N . s
• the impulse was 720 N . s
Impulse (F Δt)
or
• a momentum change of 720 kg . m/s
example
• A 0.10 Kg model rocket’s engine is designed to
deliver an impulse of 6.0 N*s. If the rocket
engine burns for 0.75 s, what is the average
force does the engine produce?
• F (Δ t ) = m D v
Given: F = 800 N
• 6.0 N . s = F ( 0.75s )
t = 0.9 s
• 6.0 N . s / 0.75s = F
Find :
• 8.0 N = F
Average Force
example
• A Bullet traveling at 500 m/s is brought to rest
by an impulse of 50 N*s. What is the mass of
the bullet?
Given: v = 500 m/s
• F ( Δt ) = m D v
F Δ t= 50 N . s
• 50 N . s = m ( 500 m/s – 0 m/s ) Find :
• 50 kg-m/s 2 . s / 500 m/s = m
m=?
• 0.1 kg = m
Example
A baseball (m=0.14kg) has initial velocity of v0=38m/s as it approaches a bat. The bat applies an
average force F that is much larger than the
weight of the ball, and the ball departs from the
bat with a final velocity of vf=+58m.
(a)Determine the impulse applied to the ball by the
bat.
(b) Assuming time of contact is Dt =1.6*10-3s, find the
average force exerted on the ball by the bat.
(a)
 (0.14kg)(58m / s)  (0.14kg)( 38m / s)
= +13.4 kg.m/s
(b)
example
2 cars are heading east, car A is traveling 30m/s, car B is
traveling 60m/s. Each car weighs 2000Kg.
– What is the momentum of car A?
– What is the momentum of car B?
– If car B crashes into car A, what is the total momentum?
p=mv
Car (A) momentum = 30m/s x 2000Kg
pX = 60,000 mi-lbs/hr east
Car (B)momentum = 60m/s x 2000Kg
pY = 120,000 m-Kg/s west
Total momentum = pA - pB
= 120,000 - 60,000
= 60,000 m-Kg/s west
In a crash test, a car of mass 1.5103 kg collides with a wall and
rebounds as in figure. The initial and final velocities of the car are vi=-15
m/s and vf = 2.6 m/s, respectively. If the collision lasts for 0.15 s, find
(a) the impulse delivered to the car due to the collision
(b) the size and direction of the average force exerted on the car

pi  mvi  (1.5 103 kg)( 15m / s)  2.25 10 4 kg  m / s
p f  mv f  (1.5 103 kg)(2.6m / s)  0.39 104 kg  m / s
I  p f  pi  mv f  mvi
 (0.39 10 4 kg  m / s )  (2.25  10 4 kg  m / s )
 2.64 10 4 kg  m / s
Dp I
2.64 10 4 kg  m / s
Fav 


 1.76 105 N
Dt Dt
0.15s
How do the safety features of a car protect
you in a collision?
• In an accident, the Dp stays the same, but you can
change F and t of impact.
• When you increase the time that the collision lasts,
the force of impact decreases.
Ft
• All safety features in a car are designed to reduce the
force by increasing the time you are in contact with
that force.
• Watch the video!
Remember: Ft = Dp
A 10 g rubber ball and a 10 g clay ball are thrown at a wall with
equal speeds. The rubber ball bounces, the clay ball sticks. Which
ball exerts a larger impulse on the wall?
A. They exert equal impulses because they have equal
momentum.
B. The clay ball exerts a larger impulse because it sticks.
C. Neither exerts an impulse on the wall because the
wall doesn’t move.
D. The rubber ball exerts a larger impulse because it
bounces.
The two particles are both moving to the right. Particle 1 catches up with
particle 2 and collides with it. The particles stick together and continue on
with velocity vf. Which of these statements is true?
A.
B.
C.
D.
E.
vf = v2.
vf is less than v2.
vf is greater than v2, but less than v1.
vf = v1.
vf is greater than v1.
example
• A 2000 kg truck traveling west at 12 m/s collides with a 1200
kg car traveling east at 16 m/s. They collide and remain stuck
together. What is their final velocity?
1.5 m/s west
Example
Rain comes straight down with velocity of v0=15m/s and hits the roof of a car perpendicularly.
Mass of rain per second that strikes the car roof
is 0.06kg/s.
Assuming the rain
comes to rest upon
striking the car
(vf=0m/s), find the
average force exerted
by the raindrop.
36
-(0.06kg/s)(-15m/s) = 0.9 N
According to action-reaction law, the force
exerted on the roof also has a magnitude
of 0.9 N points downward: - 0.9N
example
A 10 g rubber ball and a 10 g clay ball are thrown at
a wall with equal speeds. The rubber ball bounces,
the clay ball sticks. Which ball exerts a larger
impulse on the wall?
A. They exert equal impulses because they have
equal momenta.
B. The clay ball exerts a larger impulse because it
sticks.
C. Neither exerts an impulse on the wall because
the wall doesn’t move.
D. The rubber ball exerts a larger impulse
because it bounces.
In which case is the impulse greater when You drop an egg onto
a) the floor
b) a thick piece of foam rubber
In both cases, the egg does not bounce.
A) case 1
B) case 2
C) the same
In which case is the average force greater
A) case 1
B) case 2
C) the same
T. Norah Ali Almoneef
Two identical balls are dropped from the same height onto the
floor.In case 1 the ball bounces back up, and in case 2 the ball sticks to
the floor without bouncing. In which case is theimpulse given to the
ball by the floor larger?
1. Case 1
2. Case 2
3. The same
the impulse is greater for case 1 because: the change in momentum
of the object is proportional to the change in velocity which is greater
in case 1 because it has a greater final velocity (down then up) than
case 2 (which is only from down to zero). Impulse must be
greater for case 1.
Example: suppose m=1 kg, v(initial)=-1 m/s
mv(initial)= -1 kg-m/s
Case 1 mv(final)= +1 kg-m/s Impulse = 1- (-1)=2
Case 2 mv(final)= 0 Impulse = 1 - 0 = 1
Note: the direction (upward) important.
T. Norah Ali Almoneef
Is it possible for a system of two objects to have zero total
momentum while having a non-zero total kinetic energy?
1. YES
2. NO
in an isolated system, two ice skaters starting at rest and pushing
on one another will move in opposite directions thus the
momentum of the two are equal and opposite and total momentum
is zero. but they are moving apart after the push and therefore the
KE is non-zero.
two hockey pucks moving towards each other with the same speed
on a collision course have zero total momentum, but a non zero
total kinetic energy
T. Norah Ali Almoneef
The diagram depicts Before and After velocities of an 800-kg car
in two different collisions with a wall. In case A, the car rebounds
upon collision. In case B, the car hits the wall, crumples up and
stops. Assume that the collision time for each collision are the
same.
1. In which case does the car experience the greatest momentum change?
a. Case A
b. Case B
c. Both the same
d. Insufficient
information
2. In which case does the car experience the greatest impulse?
a. Case A
b. Case B
c. Both the same
d. Insufficient information
3. The impulse encountered by the 800-kg car in case A has a magnitude of
___ N•s.
a. 0
b. 800
c. 3200
d. 4000
e. 7200
f. Not enough information to determine.
T. Norah Ali Almoneef
A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle θ of
33o and then rebounds with the same speed and angle (Fig). It is in
contact with the wall for 10.4 m s.
(a) The impulse was experienced by the wall is.
(b) The average force exerted by the ball on the wall is
According to the impulse-momentum theorem:
The impulse was experienced by the wall is
Δp = m v sin − (−m v sin ) = 2m v sin
= 2× 0.32× 6.22× sin 33o = 2.17 N⋅ s
The average force exerted by the ball on the wall is
T. Norah Ali Almoneef
Δp =F Δt
Example
A baseball (m=0.14kg) has initial velocity of v0=-38m/s as it
approaches a bat. The bat applies an average force F that is
much larger than the weight of the ball, and the ball departs
from the bat with a final velocity of vf=+58m.
(a) Determine the impulse applied to the ball by the bat.
(b) Assuming time of contact is Dt =1.6*10-3s, find the average
force exerted on the ball by the bat.
T. Norah Ali Almoneef
Conservation of momentum 2 particle system
m
m
1
2
Before colliding
N1
W1
m
F12
W1
N1
m
1
W2
N2
F21
m
2
during collision
W2
N2
m
1
2
After collision
N1
W1
N2
W2
From Newton’s 3rd Law
F12 = - F21
F12
OR
F12 + F21 = 0
is force of 1 on 2
F21
F12 Δ t = p 1 f - p 1i
F21 Δ t = p2f - p 2i
P 1 f - P 1i
+
P2f - P 2i = 0




P1i  P2i  P1f  P2 f
is force of 2 on 1
Conservation of linear momentum
• If Fext = 0 then the total momentum remains constant.
• The total momentum of an isolated system remains
constant
•momentum before = momentum after
Example
Starting from rest, two
skaters push off against
each other on smooth level
ice (friction is negligible).
One is a woman (m1=54kg),
and one is a man(m2=88kg).
The woman moves away
with a velocity of
vf1=2.5m/s. Find the recoil
velocity vf2 of the man.
For the two skater system, what are the forces?
In horizontal direction
internal forces
F12 F21
No external forces.
system taken together
0
isolated system
conservation of momentum
m1vf1 + m2vf2 = 0
after pushing
vf 2 
 m1v f 1
m2
before pushing
 (54kg)( 2.5m / s)

 1.5m / s
88kg
It is important to realize that the total linear
momentum may be conserved even when the
kinetic energies of the individual parts of a system
change.
example
• Assume that an 8kg mass m1 moving to the right at 4ms-1
collides with a 6kg mass m2 moving to the left at 5ms-1.
what is the total momentum before and after the collision?
(Ans: 2kgms-1)
example
• A 1400kg cannon mounted on wheels fires a 60kg ball in a horizontal
direction with a velocity of 50ms-1. Assume that the cannon can move
freely, what will be its recoil velocity? (Ans: vc=-2.14ms-1)
example
What is the recoil velocity of a 4.0-kg rifle that shoots a 0.050-kg
bullet at a speed of 280 m/s?
the total momentum is conserved. The initial momentum of
the rifle-bullet system is 0 because the riffle and the bullet
were at rest at the beginning. By the Law of Conservation of
Momentum,
example
A 3-kg mass is moving with an initial velocity v1. The mass collides
with a 5-kg mass m2, which is initially at rest. Find the final
velocity of the masses after the collision if it is perfectly inelastic.
According to the Law of Conservation of Momentum
example
An archer stands at rest on frictionless ice and fires a 0.5-kg arrow
horizontally at 50.0 m/s. The combined mass of the archer and bow is
60.0 kg. With what velocity does the archer move across the ice after
firing the arrow?

pi  p f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1  60.0kg, m2  0.5kg, v1i  v2i  0, v2 f  50m / s, v1 f  ?
0  m1v1 f  m2 v2 f
m2
0.5kg
v1 f  
v2 f  
(50.0m / s )  0.417 m / s
m1
60.0kg
In an inelastic collision
1. both kinetic energy and momentum are conserved.
2. only kinetic energy is conserved.
3. only momentum is conserved.
4. neither kinetic energy nor momentum are conserved.
•1: False by definition of inelastic collision
• 2: False by definition of inelastic collision
• 4: False by definition of collision
T. Norah Ali Almoneef
Collisions
• Term collision is used to represent an event during which 2
particles come close to each other and interact by means of forces.
• Time interval during which the velocities of the particles change
from initial to final values is assumed to be short.
• The interaction forces are assumed to be much greater than any
external forces present.
• When two particles of masses m1 and m2 collide, the impulsive
forces may vary in time in complicated ways.
• The two particles form an isolated system, and the momentum
must be conserved.
• The total momentum of an isolated system just before a collision
equals to the momentum of the system just after the collision.
• The total kinetic energy of the system, may or may not be
conserved, depending on the type of the collision.
T. Norah Ali Almoneef
Elastic and inelastic collisions in one dimension
Momentum is conserved in any collision, elastic and
inelastic.
Mechanical Energy is only conserved in elastic collisions.
Perfectly inelastic collision: After colliding, particles stick together. There is a loss of
energy (deformation).
Elastic collision: Particles bounce off each other without loss of energy.
Inelastic collision: Particles collide with some loss of energy, but don’t stick
together.
Perfectly inelastic collision of two particles
OR completely inelastic collision
(Particles stick together) thy moves as one object


pi  p f



m1v1i  m2v2i  (m1  m2 )v f
Notice that p and v are
vectors and, thus have a
direction (+/-)
( totally in elastic collision)
There is a loss in energy Eloss
Ki  Eloss  K f
1
1
1
2
2
2
m1v1i  m2v2i  (m1  m2 )v f  Eloss
2
2
2
There is a loss in energy Eloss
Ki  Eloss  K f
1
1
1
2
2
2
m1v1i  m2v2i  (m1  m2 )v f  Eloss
2
2
2
Totally Inelastic collision:
All masses are equal.
After
Before
m
2m
m
vf
vi
Pi  mvi
Pf  2mv f
1
v f  vi
2
T. Norah Ali Almoneef
Calculate the final velocity
T. Norah Ali Almoneef
Example
Car 1 has a mass of m1=65*103kg and moves at a
velocity of v01= vi1=+0.8m/s. Car 2 has a mass of
m2=92*103kg and a velocity of v02= vi2= +1.3m/s.
Neglecting friction, find the common velocity vf of
the cars after they become coupled.
64
(m1+m2) vf = m1v01 + m2v02
After collision
Before collision
m1v01  m2 v02
vf 
m1  m2
(65  10 3 kg)(0.8m / s)  (92 10 3 kg)(1.3m / s)

3
3
(65  10 kg  92 10 kg)
=+1.1 m/s
v = 10
v=0
M
M
Before Collision
p = Mv
vf = 5
M
M
Mv = 2Mvf
vf = ½ v
After Collision
p = 2Mvf
example

An SUV with mass 1.80103 kg is travelling eastbound at +15.0 m/s, while
a compact car with mass 9.00102 kg is travelling westbound at -15.0 m/s.
The cars collide head-on, becoming entangled.
•
Find the speed of the entangled cars
after the collision.
•
Find the change in the velocity of
each car.
•
Find the change in the kinetic energy
of the system consisting of both cars.
(a)
Find the speed of the entangled cars after
the collision.
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
pi  p f
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
v f  5.00m / s
(b)
Find the change in the velocity of each car.
v f  5.00m / s
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
Dv1  v f  v1i  10.0m / s
Dv2  v f  v2i  20.0m / s
m1Dv1  m1 (v f  v1i )  1.8 104 kg  m / s
m2 Dv2  m2 (v f  v2i )  1.8 104 kg  m / s
m1Dv1  m2 Dv2  0
March 24, 2009
(c)
Find the change in the kinetic energy of the
system consisting of both cars.
m1  1.80 103 kg, v1i  15m / s
m2  9.00 102 kg, v2i  15m / s
v f  5.00m / s
1
1
2
KEi  m1v1i  m2 v22i  3.04 105 J
2
2
1
1
2
KE f  m1v1 f  m2 v22 f  3.38 10 4 J
2
2
DKE  KE f  KEi  2.70 105 J
March 24, 2009
example
Two blocks are traveling toward each other. The first has a speed of
10cm/sec and the second a speed of 60 cm/sec. After the collision, the
second block is moving with a speed of 20 cm/sec in a direction
opposite to its initial velocity. If the mass of the first block is twice that
of the second, determine the velocity of the first block after the
collision.
Since the surface is frictionless, the net force acting on the system is
0, and from the Law of Conservation of Momentum
P1i + P2i = P1f + P2f
Adding the x-components we have m1v1i - m2v2i = m1v1f + m2v2f
Since m1 = 2m2, we find that 2v1i - v2i = 2v1f + v2f
(2)(10) - (60) = 2v1f + 20
Thus, 2v1f = - 60 and v1f = - 30 cm/sec ('-' means to left).
The two particles are both moving to the right. Particle
1 catches up with particle 2 and collides with it. The
particles stick together and continue on with velocity
vf. Which of these statements is true?
A.
B.
C.
D.
E.
vf = v2.
vf is less than v2.
vf is greater than v2, but less than v1.
vf = v1.
vf is greater than v1.
A 1000 kg car moving north at (13.4m/s) collides with a 2000 kg moving east at (22.4m/s).
The cars get entangled and moved together. What is their final speed? How much energy is
converted to other forms?
Answer Say we define our coordinates so that north is the y direction and east is the x direction. Then the initial momentum of the first car is
p1 = ( (1000 kg)(13.4 m/s)) = (1.34 X 104 kg m/s)
T. Norah Ali Almoneef
T. Norah Ali Almoneef
example
T. Norah Ali Almoneef
Example
Car 1 has a mass of m1=65X103kg and moves at a velocity of
v01=+0.8m/s. Car 2 has a mass of m2=92X103kg and a velocity
of v02=+1.3m/s. Neglecting friction, find the common velocity
vf of the cars after they become coupled.
(m1+m2) vf = m1v01 + m2v02
After collision
Before collision
m1v01  m2 v02
vf 
m1  m2
(65  10 3 kg)(0.8m / s )  (92  10 3 kg)(1.3m / s )

(65  10 3 kg  92  10 3 kg)
=+1.1 m/s
T. Norah Ali Almoneef
T. Norah Ali Almoneef
T. Norah Ali Almoneef
example
Two cars travelling, one from west to east and the other from
south to north got stuck and move together of the first one is 45
km /h and that of the second is 30 km/ h . The masses of the
first and second cars are 1000 g and 1500 respectively, What is
the velocity of the wreck after collision ?
T. Norah Ali Almoneef
In elastic collision after the collision they
moves as tow objects
m1v f 1  m 2 v f 2  m1v 01  0
Total momentum
after collision
Total momentum
before collision
1
1
1
2
2
2
m1v f 1  m2 v f 2  m1v01  0
2
2
2
Total kinetic energy
after collision
v f1 
m1v01  m2 v f 2
m1
Total kinetic energy
before collision
Elastic collision of two particles
(Particles bounce off each other without loss of energy.
Momentum is conserved:




m1v1i  m2 v2i  m1v1 f  m2 v2 f
Energy is conserved:
1
1
1
1
2
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v2 f
2
2
2
2
If V2 not equal zero
Elastic collision:
Collision of Two Large Balls:
All masses are equal.
Before
m
m
After
m
m
v
V1 = V
Pi  mv
Pf  mv
1 2
KE i  mv
2
1
KE f  mv 2
2
T. Norah Ali Almoneef
v
V2 = 0
V1=0
V 2= V
v f  vi
T. Norah Ali Almoneef
Elastic Collisions
• Case 1
Objects A and B are of the same mass, the objects exchange velocities
– If m1=m2, then v1=u2 and v2 = u1
If 2nd object is initially at rest
– If u2=0, then v1=u2=0 and v2 = u1
• Case 2
If m2>>m1, u2=0
A small object collides with a big object which is at rest.
– then v1
– u1 and v2 0
• When a small/light object such as a ball collides with a very
large/heavy object such as bus, the ball rebounds with the same
speed as that before collision while the bus remains at rest.
• Case 3
If m1>>m2, u2=0
A heavy object collides with a light object which is at rest.
– If u2=0, then v1 u1 and v2 2u1
• After collision, the heavy object continues its motion with the
same velocity. The light object moves off with a velocity which is
twice the initial velocity of the heavy object.
example
• A 2000 kg truck traveling west at 12 m/s
collides with a 1200 kg car traveling east at 16
m/s. They collide and remain stuck together.
What is their final velocity (include direction)?
1.5 m/s west
In an inelastic collision
1. both kinetic energy and momentum are conserved.
2. only kinetic energy is conserved.
3. only momentum is conserved.
4. neither kinetic energy nor momentum are conserved.
•1: False by definition of inelastic collision
• 2: False by definition of inelastic collision
• 4: False by definition of collision
T. Norah Ali Almoneef
example
The version shown in Fig. consists of
a large block of wood of mass M =
5.4 kg, hanging from two long cords.
A bullet of mass m = 9.5 g is fired
into the block, coming quickly to rest.
The block + bullet then swing
upward, their center of mass rising a
vertical distance h = 6.3 cm before
the block comes momentarily to rest
end. What is the speed of the bullet
just before the collision?
Example
• In a game of billiard, a player wishes to sink
the target ball in the corner pocket as shown
in the figure. If the angle of the corner pocket
is 350, at what angel θ is the cue ball
deflected? Assume collision elastic.
solution
Example
A 1500 kg car traveling east with a speed of 25.0 ms-1
collides at an intersection with a 2500 kg van traveling north
at a speed of 20.0 ms-1 as shown in the figure. Find the
direction and magnitude of the velocity of the wreckage
after the collision, assuming that the vehicles undergo a
perfectly inelastic collision (that is they stick together).
solution
Simple Examples of Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.