Linear Momentum and Collisions

Contributed by:

Jonathan James Thu, Jan 20, 2022 07:42 PM UTC

Conservation of energy, Linear momentum, Momentum and energy, Impulse momentum theorem, Conservation of momentum, Elastic collisions

1. Physics 111: Mechanics
Lecture 12
Dale Gary
NJIT Physics Department

2. Linear Momentum
and Collisions
 Conservation
of Energy
 Momentum
 Impulse
 Conservation
of Momentum
 1-D Collisions
 2-D Collisions
 The Center of Mass
January 20, 2022

3. Conservation of Energy
 E = K + 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 2 1 2 1 2
mv f  mgy f  kx f  mvi  mgyi  kxi
2 2 2 2
 E = K + 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
1 2 energy”
1 2  or
 1 heat.
2 1 2
 f k d  mv f  mgy f  kx f    mvi  mgyi  kxi 
2 2  2 2 
January 20, 2022

4. 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
January 20, 2022

5. Momentum and Energy
 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
January 20, 2022

6. 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
January 20, 2022

7. 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
January 20, 2022

8. Impulse
 When a single, constant force acts on the
object, there is an impulse delivered to
the object  
 I Ft


I is defined as the impulse
 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
January 20, 2022

9. 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
January 20, 2022

10. 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
January 20, 2022

11. How Good Are the Bumpers?
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
January 20, 2022

12. How Good Are the Bumpers?
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 and3 direction of the average 4
force exerted on the car
pi mvi (1.5 10 kg )( 15m / s )  2.25 10 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
p I 2.64 10 4 kg m / s
Fav    1.76 105 N
t t 0.15s
January 20, 2022

13. Impulse-Momentum
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
January 20, 2022

14. Impulse-Momentum
Theorem 2
 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
January 20, 2022

15. 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.
January 20, 2022

16. Conservation of Momentum
 Start from impulse-
momentum theorem
  
F21t m1v1 f  m1v1i
  
F12 t m2 v2 f  m2 v2i
 
 Since F21t  F12 t
   
 Then m1v1 f  m1v1i  (m2 v2 f  m2 v2i )
   
 So m1v1i  m2 v2i m1v1 f  m2 v2 f
January 20, 2022

17. 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
 
Fnet 0 p 0
 When then
 
 For an pisolated
f  pi
system
 Specifically, the total momentum before the
collision will equal
 the  total momentum after the
collision m1v1i  m2 v2i m1v1 f  m2 v2 f
January 20, 2022

18. 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  m2v2i 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
January 20, 2022

19. Conservation of Momentum
 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
January 20, 2022

20. Types of Collisions
 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
January 20, 2022

21. 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
January 20, 2022

22. 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 v2i (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
 Kinetic energy is NOT
conserved
January 20, 2022

23. An SUV Versus a Compact
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.
(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
January 20, 2022
cars.

24. An SUV Versus a Compact
(a) Find the speed of the m1 1.80 103 kg , v1i 15m / s
entangled cars after the
m2 9.00 10 2 kg , v2i  15m / s
collision.
pi  p f
m1v1i  m2 v2i (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
v f 5.00m / s
January 20, 2022

25. An SUV Versus a Compact
(b) Find the change in the m1 1.80 103 kg , v1i 15m / s
velocity 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
m1v1 m1 (v f  v1i )  1.8 10 4 kg m / s
m2 v2 m2 (v f  v2i ) 1.8 10 4 kg m / s
m1v1  m2 v2 0
January 20, 2022

26. An SUV Versus a Compact
(c) Find the change in the 3
m
kinetic energy of the system 1 1.80 10 kg , v1i 15m / s
consisting 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
January 20, 2022

27. 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
January 20, 2022

28. 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  v22i )(f v2 f  v2i )
m1v1i  m2 v2i m1v1 f  m2 v2 f m1 (v1i  v1 f ) m2 (v2 f  v2i )
v1i  v1 f v2 f  v2i
m1v1i  m2 v2i m1v1 f  m2 v2 f
January 20, 2022

29. Summary of Types of
Collisions
 In an elastic collision, both momentum and
kinetic energy are conserved
v1i  v1 f v2 f  v2i m1v1i  m2 v2 i m1v1 f  m2 v2 f
 In an inelastic collision, momentum is conserved
but kinetic energy is not
m1v1i  m2 v2i m1v1 f  m2 v2 f
 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
m v are
 mthe
v same
( m  m )v
1 1i 2 2i 1 2 f
January 20, 2022

30. Conservation of Momentum
 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
January 20, 2022

31. 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
January 20, 2022

32. 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 v2i m1v1 f  m2 v2 f
January 20, 2022

33. 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  v2i
 Solve: the resulting
equations simultaneously
January 20, 2022

34. One-Dimension vs Two-
Dimension
January 20, 2022

35. 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  m2v2ix m1v1 fx  m2v2 fx
m1v1iy  m2 v2iy m1v1 fy  m2v2 fy
January 20, 2022

36. 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  m2v2 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 be
v1i  v1 f v2 f  v2i
used for one-dimensional situations
January 20, 2022

37. Glancing Collisions
 The “after” velocities have x and y
components
 Momentum is conserved in the x direction and
in the y direction
 Apply conservation ofv momentum
m1v1ix  m m v  m v separately
2 2 ix 1 1 fx 2 2 fx
to each direction
m1v1iy  m2 v2iy m1v1 fy  m2 v2 fy
January 20, 2022

38. 2-D Collision, example
 Particle1 is moving
v1i
at velocity and
particle 2 is at rest
 In the x-direction,
the initial
momentum is m1v1i
 In the y-direction,
the initial
momentum is 0
January 20, 2022

39. 2-D Collision, example cont
 After the collision, the
momentum in the x-direction
is m1v1f cos  m2v2f cos 
 After the collision, the
momentum in the y-direction
is m1v1f sin  m2v2f sin 
m1v1i  0 m1v1 f cos   m2v2 f cos 
0  0 m1v1 f sin   m2 v2 f sin 
 If the collision is elastic, apply1 1 1
2 2 2
the kinetic energy equation 2 1 1i 2 1 1 f 2 m2v2 f
m v  m v 
January 20, 2022

40. 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
andm the
1.5road
103 kgcan
, m be neglected.
2.5 103 kg
c v
vcix 25m / s, vviy 20m / s, v f ? ?
January 20, 2022

41. 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 104 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 104 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 
January 20, 2022

42. 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   4
1.33
3.75 10 kg m / s
 tan  1 (1.33) 53.1
5.00 10 4 kg m/s
vf  3
15.6 m/s
(4.00 10 kg) sin 53.1
January 20, 2022

43. The Center of Mass
 How should we
define the position
of the moving
body ?
 What is y for Ug =
mgy ?
 Take the average
position of mass.
Call “Center of
Mass” (COM or CM)
January 20, 2022

44. The Center of Mass
 There is a special point in a system
or object, called the center of
mass, that moves as if all of the
mass of the system is concentrated
at that point
 The CM of an object or a system is
the point, where the object or the
system can be balanced in the
uniform gravitational field
January 20, 2022

45. The Center of Mass
 The center of mass of any symmetric object lies
on an axis of symmetry and on any plane of
symmetry
 If the object has uniform density
 The CM may reside inside the body, or outside
the body
January 20, 2022

46. Where is the Center of Mass
?
 The center of mass of particles
 Two bodies in 1 dimension
m1 x1  m2 x2
xCM 
m1  m2
January 20, 2022

47. Center of Mass for
many particles in
3D?
January 20, 2022

48. Where is the Center of Mass
?
 Assume m1 = 1 kg, m2 = 3 kg, and
x1 = 1 m, x2 = 5 m, where is the
center of mass of these twox objects?
m1 x1  m2 x2
CM 
m1  m2
A) xCM = 1 m
B) xCM = 2 m
C) xCM = 3 m
D) xCM = 4 m
E) xCM = 5 m
January 20, 2022

49. Center of Mass
for a System of Particles
 Two bodies and one dimension
 General case: n bodies and three dimension
 where M = m1 + m2 + m3 +…
January 20, 2022

50. Sample Problem : Three particles of masses m1 =
1.2 kg, m2 = 2.5 kg, and m3 = 3.4 kg form an
equilateral triangle of edge length a = 140 cm.
Where is the center of mass of this system? (Hint: m1
is at (0,0), m2 is at (140 cm,0), and m3 is at (70 cm,
120 cm), as n
shown in the figure below.)
1 m1 x1  m2 x2  m3 x3
xCM   mi xi 
M i 1 m1  m2  m3
n
1 m1 y1  m2 y2  m3 y3
yCM   mi yi 
M i 1 m1  m2  m3
xCM  82.8 cm and yCM 57.5 cm
January 20, 2022

51. Motion of a System of
Particles
 Assume the total mass, M, of the
system remains constant
 We can describe the motion of the
system in terms of the velocity and
acceleration of the center of mass of
the system
 We can also describe the momentum
of the system and Newton’s Second
Law for the system
January 20, 2022

52. Velocity and Momentum of a
System of Particles
 The velocity of the center of mass of a
system of particles
 is
 d rCM 1 
vCM    mi v i
dt M i
 The momentum can be expressed as
   
MvCM  mi v i  p i p tot
i i
 The total linear momentum of the system
equals the total mass multiplied by the
velocity of the center of mass
January 20, 2022

53. Acceleration and Force of the
Center of Mass
 The acceleration of the center of mass can
be found by differentiating the velocity with
respect totimedv 1 
CM
aCM    mai i
dt M i
 The acceleration can be related to a force
 
MaCM  Fi
i
 If we sum over all the internal forces, they
cancel in pairs and the net force on the
system is caused only by the external
forces
January 20, 2022

54. Newton’s Second Law
for a System of Particles
 Since the only forces are external, the net
external force equals the total mass of the
system multiplied by the acceleration of the
center of mass:
  
 Fext MaCM
 The center of mass of a system of particles of
combined mass M moves like an equivalent
particle of mass M would move under the
influence of the net external force on the
system
January 20, 2022

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