PHYSICS 9702
FORCE & MOMENTUM
WE WILL LEARN WHAT IS MOMENTUM IN 1D & 2D
I WILL APPLY LAW OF CONSERVATION & NEWTON’S LAWS
BY: MR. USMAN SHAHZAD
14TH OCT, 2020
10/26/20
LINEAR MOMENTUM
AND COLLISIONS
• Conservation
of Energy
• Momentum
• Impulse
• Conservation
of Momentum
• 1-D Collisions
• 2-D Collisions
• The Center of Mass
10/26/20
CONSERVATION OF ENERGY
• D E = D K + D U = 0 if conservative forces are the only
forces that do work on the system.
• The total amount of energy in the system is constant.
1 2 1 1 1
mv f mgy f kx 2f mvi2 mgyi kxi2
2 2 2 2
• D E = D K + D U = -fkd if friction forces are doing work on
the system.
• The total amount of energy in the system is still constant, but
the change in mechanical energy goes into “internal energy” or
heat. 1 2 1 2 1 2 1 2
f k d mv f mgy f kx f mvi mgyi kxi
2 2 2 2
10/26/20
LINEAR MOMENTUM
• This is a new fundamental quantity, like force, energy. It is a
vector quantity (points in same direction as velocity).
• The linear momentum p of an object of mass m moving with a
velocity v is defined to be the product of the mass and velocity:
p mv
• The terms momentum and linear momentum will be used
interchangeably in the text
• Momentum depend on an object’s mass and velocity
MOMENTUM AND ENERGY
10/26/20
• Two objects with masses m1 and m2 have equal kinetic energy. How
do the magnitudes of their momenta compare?
(A) p1 < p2
(B) p1 = p2
(C) p1 > p2
(D) Not enough information is given
10/26/20
LINEAR MOMENTUM, CONT’D
• Linear momentum is a vector quantity p mv
• Its direction is the same as the direction of the velocity
• The dimensions of momentum are ML/T
• The SI units of momentum are kg m / s
• Momentum can be expressed in component form:
px = mvx py = mvy pz = mvz
10/26/20
NEWTON’S LAW AND MOMENTUM
• Newton’s Second Law can be used to relate the
momentum of an object to the resultant force acting
on it
v (mv )
Fnet ma m
t t
• The change in an object’s momentum divided by the
elapsed time equals the constant net force acting on
the object
p change in momentum
Fnet
t time interval
10/26/20
IMPULSE
• When a single, constant force acts on the object,
there is an impulse delivered to the object
• I Ft
• is defined as the impulse
I
• The equality is true even if the force is not constant
• Vector quantity, the direction is the same as the direction
of the force
p change in momentum
Fnet
t time interval
10/26/20
IMPULSE-MOMENTUM THEOREM
• The theorem states that
the impulse acting on a
system is equal to the
change in momentum of
the system
p Fnet t I
I p mv f mvi
10/26/20
CALCULATING THE CHANGE OF MOMENTUM
p pafter pbefore
mvafter mvbefore
m(vafter vbefore )
For the teddy bear
p m 0 (v) mv
For the bouncing ball
p m v (v) 2mv
How Good Are the Bumpers?
10/26/20
In a crash test, a car of mass 1.5 x 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
How Good Are the Bumpers?
10/26/20
In a crash test, a car of mass 1.5 x 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 10 4 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
p I 2.64 10 4 kg m / s
Fav 1.76 105 N
t t 0.15s
IMPULSE-MOMENTUM
10/26/20
THEOREM
• A child bounces a 100 g superball on the sidewalk.
The velocity of the superball changes from 10 m/s
downward to 10 m/s upward. If the contact time
with the sidewalk is 0.1s, what is the magnitude of
the impulse imparted to the superball?
(A) 0
(B) 2 kg-m/s
(C) 20 kg-m/s
I p mv f mvi
(D) 200 kg-m/s
(E) 2000 kg-m/s
IMPULSE-MOMENTUM THEOREM 2 10/26/20
• A child bounces a 100 g superball on the
sidewalk. The velocity of the superball changes
from 10 m/s downward to 10 m/s upward. If the
contact time with the sidewalk is 0.1s, what is
the magnitude of the force between the sidewalk
and the superball?
(A) 0
(B) 2 N
I p mv f mvi
(C) 20 N F
t t t
(D) 200 N
(E) 2000 N
10/26/20
CONSERVATION OF MOMENTUM
• In an isolated and closed system, the
total momentum of the system
remains constant in time.
• Isolated system: no external forces
• Closed system: no mass enters or leaves
• The linear momentum of each colliding
body may change
• The total momentum P of the system
cannot change.
10/26/20
CONSERVATION OF MOMENTUM
• Start from impulse-momentum theorem
F21t m1v1 f m1v1i
F12 t m2 v2 f m2 v2i
• Since
F21t F12t
• Then m1v1 f m1v1i (m2 v2 f m2 v2i )
m1v1i m2 v2i m1v1 f m2 v2 f
• So
10/26/20
CONSERVATION OF MOMENTUM
• When no external forces act on a system consisting of two
objects that collide with each other, the total momentum of the
system remains constant in time
Fnet t p p f pi
• When
Fnet 0 p 0
then
• For an isolated
system
p f pi
• Specifically, the total momentum before the collision will equal
the total momentum after the
collision
m1v1i m2 v2i m1v1 f m2 v2 f
10/26/20
The Archer
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 ) 0.417m / s
m1 60.0kg
CONSERVATION OF MOMENTUM 10/26/20
• A 100 kg man and 50 kg woman on ice skates stand facing
each other. If the woman pushes the man backwards so
that his final speed is 1 m/s, at what speed does she recoil?
(A) 0
(B) 0.5 m/s
(C) 1 m/s
(D) 1.414 m/s
(E) 2 m/s
TYPES OF COLLISIONS 10/26/20
• Momentum is conserved in any collision
• Inelastic collisions: rubber ball and hard ball
• Kinetic energy is not conserved
• Perfectly inelastic collisions occur when the objects stick together
• Elastic collisions: billiard ball
• both momentum and kinetic energy are conserved
• Actual collisions
• Most collisions fall between elastic and perfectly inelastic
collisions
10/26/20
COLLISIONS SUMMARY
• In an elastic collision, both momentum and kinetic energy are
conserved
• In a non-perfect inelastic collision, momentum is conserved but
kinetic energy is not. Moreover, the objects do not stick
together
• In a perfectly inelastic collision, momentum is conserved,
kinetic energy is not, and the two objects stick together after the
collision, so their final velocities are the same
• Elastic and perfectly inelastic collisions are limiting cases,
most actual collisions fall in between these two types
• Momentum is conserved in all collisions
10/26/20
MORE ABOUT PERFECTLY INELASTIC COLLISIONS
• When two objects stick together after the
collision, they have undergone a perfectly
inelastic collision
• Conservation of momentum
m1v1i m2 v 2 i ( m1 m2 )v f
m1v1i m2 v2i
vf
m1 m2
• Kinetic energy is NOT conserved
10/26/20
An SUV Versus a Compact
An SUV with mass 1.80 x 103 kg is travelling eastbound
at +15.0 m/s, while a compact car with mass 9.00 x 102
kg is travelling westbound at -15.0 m/s. The cars collide
head-on, becoming entangled.
(a) Find the speed of the entangled
cars after the collision.
(b) Find the change in the velocity
of each car.
(c) Find the change in the kinetic
energy of the system consisting
of both cars.
10/26/20
An SUV Versus a Compact
(a) Find the speed of the entangled m 1.80 103 kg , v 15m / s
1 1i
cars after the collision.
m2 9.00 10 kg , v2i 15m / s
2
pi p f
m1v1i m2 v2i (m1 m2 )v f
m1v1i m2 v2i
vf
m1 m2
v f 5.00m / s
10/26/20
An SUV Versus a Compact
(b) Find the change in the velocity m1 1.80 103 kg , v1i 15m / s
of each car.
m2 9.00 10 2 kg , v2i 15m / s
v f 5.00m / s
v1 v f v1i 10.0m / s
v2 v f v2i 20.0m / s
m1v1 m1 (v f v1i ) 1.8 10 4 kg m / s
m2 v2 m2 (v f v2i ) 1.8 10 4 kg m / s
m1v1 m2 v2 0
10/26/20
An SUV Versus a Compact
(c) Find the change in the kinetic
m
energy of the system consisting 1 1 .80 10 3
kg , v1i 15m / s
of both cars. m2 9.00 10 2 kg , v2i 15m / s
v f 5.00m / s
1 1
KEi m1v1i m2 v22i 3.04 105 J
2
2 2
1 1
KE f m1v1 f m2 v22 f 3.38 10 4 J
2
2 2
KE KE f KEi 2.70 105 J
10/26/20
MORE ABOUT ELASTIC COLLISIONS
• Both momentum and kinetic energy are
conserved
m1v1i m2 v2i m1v1 f m2 v2 f
1 1 1 1
m1v1i m2 v2i m1v1 f m2 v22 f
2 2 2
2 2 2 2
• Typically have two unknowns
• Momentum is a vector quantity
• Direction is important
• Be sure to have the correct signs
• Solve the equations simultaneously
10/26/20
ELASTIC COLLISIONS
• A simpler equation can be used in place of the KE equation
1 1 1 1
m1v1i m2 v2i m1v1 f m2 v22 f
2 2 2
2 2 2 2
m1 (v12i v12f ) m2 (v22 f v22i )
v v ( v v )
m1 (v11i i v1 f )( v21ii v1 f ) m21(fv2 f v2 2 f v2 f v2 i )
i )(
m1v1i m2 v2 i m1v1 f m2 v2 f m1 (v1i v1 f ) m2 (v2 f v2 i )
v1i v1 f v2 f v2 i
m1v1i m 2 v2 i m1v1 f m 2 v 2 f
10/26/20
SUMMARY OF TYPES OF COLLISIONS
• In an elastic collision, both momentum and kinetic energy are
conserved
v1i v1 f v 2 f v2 i m1v1i m2 v2 i m1v1 f m 2 v 2 f
• In an inelastic collision, momentum is conserved but kinetic energy is
not m1v1i m 2 v2 i m1v1 f m 2 v 2 f
• In a perfectly inelastic collision, momentum is conserved, kinetic
energy is not, and the two objects stick together after the collision, so
1i the
m1vare
their final velocities v 2 i ( m1 m2 )v f
m2same
CONSERVATION OF MOMENTUM 10/26/20
• An object of mass m moves to the right with a speed
v. It collides head-on with an object of mass 3m
moving with speed v/3 in the opposite direction. If the
two objects stick together, what is the speed of the
combined object, of mass 4m, after the collision?
(A) 0
(B) v/2
(C) v
(D) 2v
(E) 4v
10/26/20
PROBLEM SOLVING FOR 1D COLLISIONS, 1
• Coordinates: Set up a
coordinate axis and define the
velocities with respect to this
axis
• It is convenient to make your
axis coincide with one of the
initial velocities
• Diagram: In your sketch,
draw all the velocity vectors
and label the velocities and the
masses
10/26/20
PROBLEM SOLVING FOR 1D COLLISIONS, 2
• Conservation of Momentum:
Write a general expression for
the total momentum of the
system before and after the
collision
• Equate the two total momentum
expressions
• Fill in the known values
m1v1i m2 v2 i m1v1 f m 2 v 2 f
10/26/20
PROBLEM SOLVING FOR 1D COLLISIONS, 3
• Conservation of Energy: If
the collision is elastic, write a
second equation for
conservation of KE, or the
alternative equation
• This only applies to perfectly
elastic collisions
v1i v1 f v2 f v2 i
• Solve: the resulting equations
simultaneously
10/26/20
ONE-DIMENSION VS TWO-DIMENSION
10/26/20
TWO-DIMENSIONAL COLLISIONS
• For a general collision of two objects in two-dimensional
space, the conservation of momentum principle implies
that the total momentum of the system in each direction is
conserved
m1v1ix m2 v2ix m1v1 fx m2 v2 fx
m1v1iy m2 v2iy m1v1 fy m2 v2 fy
10/26/20
TWO-DIMENSIONAL COLLISIONS
• The momentum is conserved in all directions
m1v1ix m2 v2ix m1v1 fx m2 v2 fx
• Use subscripts for
m1v1iy m2 v2iy m1v1 fy m2 v2 fy
• Identifying the object
• Indicating initial or final values
• The velocity components
• If the collision is elastic, use conservation of kinetic
energy as a second equation
• Remember, the simpler equation can only
v1i bev1used
vfor one-
v2 i
f 2f
dimensional situations
10/26/20
GLANCING COLLISIONS
• The “after” velocities have x and y components
• Momentum is conserved in the x direction and in the y
direction
• Apply conservation of momentum separately to each
direction m1v1ix m2 v2ix m1v1 fx m2 v2 fx
m1v1iy m2 v2iy m1v1 fy m2 v2 fy
10/26/20
2-D COLLISION, EXAMPLE
• Particle 1
is moving at
v
velocity 1iand particle 2
is at rest
• In the x-direction, the
initial momentum is
m1v1i
• In the y-direction, the
initial momentum is 0
10/26/20
2-D COLLISION, EXAMPLE CONT
• After the collision, the momentum in the
x-direction is m1v1f cos q + m2v2f cos f
• After the collision, the momentum in the
y-direction is m1v1f sin q + m2v2f sin f
m1v1i 0 m1v1 f cos m2 v2 f cos
0 0 m1v1 f sin m2 v2 f sin
• If the collision is elastic, apply the
1 1 1
kinetic energy equation m1v12i m1v12f m2 v22 f
2 2 2
10/26/20
Collision at an Intersection
A car with mass 1.5×103 kg traveling
east at a speed of 25 m/s collides at an
intersection with a 2.5×103 kg van
traveling north at a speed of 20 m/s.
Find the magnitude and direction of
the velocity of the wreckage after the
collision, assuming that the vehicles
undergo a perfectly inelastic collision
and assuming that friction between the
vehicles and the road can be
neglected.
mc 1.5 103 kg , mv 2.5 103 kg
vcix 25m / s, vviy 20m / s, v f ? ?
10/26/20
Collision at an Intersection
mc 1.5 103 kg, mv 2.5 103 kg
vcix 25 m/s , vviy 20 m/s , v f ? ?
p xi mc vcix mv vvix mc vcix 3.75 10 4 kg m/s
p xf mc vcfx mv vvfx (mc mv )v f cos
3.75 10 4 kg m/s (4.00 103 kg)v f cos
p yi mc vciy mv vviy mv vviy 5.00 10 4 kg m/s
p yf mc vcfy mv vvfy (mc mv )v f sin
5.00 10 4 kg m/s (4.00 103 kg)v f sin
10/26/20
Collision at an Intersection
mc 1.5 103 kg , mv 2.5 103 kg
vcix 25m / s, vviy 20m / s, v f ? ?
5.00 10 4 kg m/s (4.00 103 kg)v f sin
3.75 10 4 kg m/s (4.00 103 kg )v f cos
5.00 10 4 kg m / s
tan 1.33
3.75 10 kg m / s
4
tan 1 (1.33) 53.1
5.00 10 4 kg m/s
vf 15.6 m/s
(4.00 103 kg) sin 53.1
