Lessons about: MOMENTUM, IMPULSE and LAW OF CONSERVATION OF MOMENTUM AND COLLISION

1. Linear Momentum and
its Conservation
By: Victor Rea Oribe

2. Complete the Matrix below by
providing the needed information
Newton’s
Laws of
Motion
Statement
of the law
Illustration or
example or
Diagram
Misconceptions or
difficulties in
understanding the
Laws

3. Momentum
Momentum describes an object’s
resistance to stopping.
The magnitude of momentum (p) is
measured as the product of the
object’s mass (m in kg) and velocity
(v in m/s).
Momentum = mass x velocity
p = m x v
kg.m/s = kg x m/s

4. Velocity is a vector quantity.
Vector - has magnitude and
direction.
Therefore, Momentum is also
a vector quantity

5. Sample Problem # 1
A 295-kg motorcycle moves at a velocity
of 80 km/h. What is its momentum?
Given:
m= 295-kg
v= 80 km/h
80 km
1 h
x 1 h
3,600 s
x 1000 m
1 km
=
80000 m
3600 s
v = 22.22 m/s
p = ?
p = m v
p = (295 kg) ( 22.22 m/s)
p = 6, 554.9 kg.m/s

6. Sample Problem # 2
What is the velocity of a 1,200kg car that
has a momentum of 30,000 kg.m/s?
Given:
m = 1,200 kg
p = 30,000 kg.m/s
v = ?
p = mv
v =
p
m
v =
30,000 kg . m/s
1,200 kg
= 25 m/s

7. Sample Problem # 2
If a car has a momentum of 30,000 kg.m/s
and its velocity is 25m/s, what is its
mass?
Given:
p = 30,000 kg.m/s
v = 25 m/s
m = ?
m =
p
v
m =
30,000
kg
m/s
25 m/s
= 1,200 kg

8. Sample Exercises:
1. A 45.9-g golf-ball is initially at rest on
the tee. As the club hits the ball, the
ball acquires a speed of 76.0 m/s. What
is the change in momentum of the golf
balls?
Given:
m = 45.9 g
m = kg
m = 45.9 g=
1 kg
1000g
m =
45.9 g . kg
1000 g
m = 0.0459 kg
v = 76.0 m/s
p = ?
p = mv
p = (0.0459 kg)(76.0 m/s)
p = 3.5 kg . m/s

9. Sample Exercises:
2. What should be the velocity of a 2600 kg
delivery truck to have the same
momentum as the race car with a mass
of 1,100 kg moving with a velocity of
110 km/h?
For the Race Car
m = 1,100 kg
v = 110 km/h
= 30.55 m/s
p = mv
= (1,100kg)(30.55 m/s)
= 33,605 kg . m/s
For the Delivery Truck
m = 2,600 kg
p = 33, 605 kg.m/s
v = ?
v = p/m
v = 33,605 kg.m/s/2,600 kg
v = 12.925 m/s

10. Research Work:
a. Research on the maximum speeds and
masses of the objects below
b. Arrange the following according to the
increasing magnitude of momentum
c. Assume that all of them are moving in
their maximum speed.
1. Car 6. ant
2. Bus 7. horse
3. Bullet 8. motorcycle
4. Airplane 9. cheetah
5. Turtle 10. ostrich

11. Impulse

12. If a soccer ball is
kicked and attains
a great momentum,
it will be difficult to
stop it
To stop the ball, it is necessary to use
force against its momentum for a given
period of time.
The greater the momentum of an object,
the greater the force, or the longer the
time or both is needed to stop the ball.

13. What is the effect of the force and time in
the motion of the object?
According to Newton’s Second Law of
Motion, an UNBALANCED FORCE causes
an object to accelerate (speed up) or
decelerate (slow down).
If an unbalanced forced acts on an object
against its motion, speed – the object
slows down
If the force acts in the same direction of
the motion – the objects speeds up.

14. Therefore: Unbalanced force causes
change in momentum of an object.
To illustrate the effect, recall the
Newton’s second Law of Motion
Force = mass (m) x acceleration (a)
= kg x m/s2
= Newton
N = kg.m/s2

15. Acceleration (a) is mathematically
defined as
a =
v – v0
t
Relating the two equations:
F = m a a =
v – v0
t
F =
m ( v – v0 )
t
Ft = mv – mv0
Since ∆v = v – v0 Then: Ft = m ∆ v

16. Ft = m ∆ v
The right side of the equation
represents the change in
momentum caused by the
unbalanced force acting at a
certain period of time.
It can be read as the “force (F)
times the time (t) equals the
mass (m) times the change
(delta) in velocity (∆v)

17. In the expression: Ft = mv – mv0
mv represents the final momentum p and
mv0 represents the initial momentum p0 of
an object
The equation can now be written as:
Ft = p – p0 or Ft = ∆p
Again, this equation tells that the product
of the net or unbalanced force, F and time,
t when the force act is equal to change in
momentum, p . This is known as the
IMPULSE- MOMENTUM THEOREM

18. In Physics, Ft is equal to the concept
called IMPULSE (I )
It is describes as the effect of the
unbalanced force on the object.
It is represented by the equation:
I , Impulse = Ft
Since : Ft = ∆ p
Then : I = ∆ p

19. Example # 1
A 430-kg tricycle travelling at 30 km/h
slows down to 10 km/h in 5 s as it
approaches a road intersection. What is
the force applied on the brakes of the
tricycle? What is the impulse exerted on
the vehicle?
Given:
m = 430 kg v = 10 km/h
= 2.78 m/s
v0 = 30 km/h
= 8.3 m/s t = 5 s
Find: (a) F (b) I

20. Solutions:
Find the initial momentum
po = mv0
p0 = (430kg) (8.3 m/s)
= 3,569 kg . m/s
Find the final momentum
p = mv
p = (430 kg) ( 2.78 m/s)
= 1195.4 kg . m/s

21. Find the Force ( F )
Ft = ∆p
∆p = p –p0
= 1195.4 kg . m/s – 3569 kg .m/s
= - 2373.6 kg . m/s
Ft = ∆p
F (5s) = -2373.6 kg . m/s
F =
- 2373.6 kg . m/s
5 s
F = -474.72 kg . m/s2
or - 475N
Note: The
negative sign
indicates F is
a retarding
force

22. Find the Impulse
I = ∆p
I = -2373.6 kg . m/s

23. Example # 1
What force is needed to stop a
4,500 kg tourist bus travelling at 90
km/h in 18 s?
Given:
m = 4,500 kg v = 90 km /h
= 25 m/s
v = 0 t = 18 s
Find: F

24. Solutions:
A. Compute the initial momentum
p0 = mv0
= (4500 kg) ( 25 m/s)
= 112, 500 kg . m/s
B. Compute the final momentum
p = mv
= (4500 kg) ( 0 m/s)
= 0
C. Find the Force (F)
Ft = ∆p
F = ∆p (p – p0 )
t
F =
0 – 112500 kg.m/s
18 s
F = -6250 kg.m/s2 or N

25. Practice Exercise # 1:
A 12.0-kg hammer hits a nail at a velocity of 13.2
m/s in 0.005 s. What is the average force exerted
on the nail?
Given:
m = 12.0kg
v = o
v0 = 13.2 m/s
t = 0.005s
Solution:
A. Compute the initial momentum
p0 = mv0
= (12.0 kg) (13.2 m/s)
= 158.4 kg.m/s
B. Compute the final momentum
p0 = mv
= (12.0 kg) (0)
= 0
C. Compute the Force
Ft = ∆p
F = ∆p / t
F = (0- 158.4 kg.m/s) / 0.005 S
F = - 31,680 kg. m/s2

26. Practice Exercise # 2:
What is the velocity of a 348-kg
motorcylce which has a momentum
of 4833.33 kg.m/s? How much force
is needed to stop it in 5 s. What is
the impulse.
Given:
v0 = ?
v = ?
m = 348 kg
p0 = 4833.33 kg. m/s
F =?
t = 5 s
I = ?
v0 = ?
p = mv0
4833.33 kg.m/s = (348 kg) (v)
v0 = 4833.33 kg.m/s / 348 kg
v0 = 13.9 m/s
Final momentum
p = mv
= (348kg) (0)
p = 0
Ft = ∆p
F = (4833.33kg.m/s – 0) / 5s
F =

27. Quiz (understanding 30%)
1. What is the momentum of an object with m=2.00 kg
and v= 40.0 m/s?
2. A car weighing 15,680 N and moving at 20 m/s is acted
upon by a force of 640 N until it stops.
Find:(a) the mass of the car
3. A car weighing 15,680 N and moving at 20 m/s is acted
upon by a force of 640 N until it stops.
Find: (b) the car’s initial momentum
4. A force of 30000 N j is exerted for 4.00 s, on a 95,000 kg
mass. What is the impulse of the force for this 4.00 s?

28. Quiz
1. What is the momentum of an object with
m=2.00 kg and v= 40.0 m/s?
Solution:
ρ = mv
ρ = (2.00 kg)(40.0 m/s)
ρ = 80.0 kg-m/s
Given:
m = 2.00-kg
v = 40.0 m/s
ρ = ?

29. Quiz
2.A car weighing 15,680 N and moving at 20 m/s
is acted upon by a force of 640 N until it stops.
Find:
(a) the mass of the car
Solution:
W = mg
15,680 kg-m/s2 = (m) (9.8 m/s2)
m = 15,680kg-m/s2 ÷ 9.8 m/s2
m = 1,600-kg
Given:
W = 15,680 N
g = 9.8 m/s2
m = ?

30. Quiz
3. A car weighing 15,680 N and moving at 20 m/s
is acted upon by a force of 640 N until it stops.
Find:
(b) the car’s initial momentum
Solution:
ρ = mv
= (15,680 N ÷ 9.8 m/s2) 20 m/s
= 1,600 kg x 20 m/s
= 32,000 kg-m/s
Given:
W = 15,680 N
g = 9.8 m/s2
m = ?
v = 20 m/s
F = 460N

31. Quiz
4. A force of 30000 N is exerted for 4.00 s, on a
95,000 kg mass. What is the impulse of the force
for this 4.00 s?
Solution:
Impulse = Ft
= 30,000 N (4.00 s)
I = 120,000 N-s
Given :
F = 30,000 N
t = 4.00 s
m = 95,000 kg
I = ?

32. Law of Conservation of Momentum
and Collisions
Collision is the interaction that occurs
when two or more objects hit each other.
When two objects collide, each object
exerts a force on another for a short
amount of time.
This force imparts an impulse, or changes
the momentum of each of the colliding
objects.

33. Experiments proved that momentum can
be conserved.
This means that the total momentum of
any set of objects remains the same
unless outside forces act on the object.
Therefore the momentum of each object
involved in the collision changes, the total
momentum of the system remains
constant.
Pbefore = pafter

34. For the collision of two objects
Total momentum before collision = Total
momentum after collision
P1 + p2 = p1’ + p2’
Where:
p1 = momentum of the first object
before collision
p2 = momentum of the second object
before collision
p1’ = momentum of the first object
before collision
p2’ = momentum of the second object
after collision

35. Collision may be classified as
a) elastic collision
– collision in which both momentum
and kinetic energy are conserved.

36. Before collision
m1 v1 + m2 v2
(1,000 kg) (20 m/s) + (3,000 kg) (0 m/s) = 20,000 kg-m/s
After collision
m1 v1’ + m2v2’
(1,000 kg) (-10 m/s) + (3,000 kg) (10 m/s)
-10,000 kg-m/s + 30,000 kg-m/s = 20,000 kg-m/s

37. An analysis of the kinetic energy of the two objects reveals
that the total system kinetic energy before the collision is
800,000 Joules (200,000 J for the car plus 600,000 J for the
truck).
After the collision, the total system kinetic energy is
800000 Joules (800,000 J for the car and 0 J for the truck).
The total kinetic energy before the collision is equal to the
total kinetic energy after the collision.

38. m = 1.2 kg m2 = 1.2 kg
v1 = 2.0 m/s v2 = 0
m = 1.2 kg m2 = 1.2 kg
v’1= 0 m/s v2‘ = 2.0 m/s
Total Momentum Before the
Collision
Ptotal = p1 + p2
= m1v1 + m2v2
= (1.2kg)(2.0m/s) + (1.2kg)(0)
= 2.4 kg.m/s
Total Momentum After the
Collision
Ptotal = p1‘ + p2‘
= m1v1‘ + m2v2’
= (1.2kg)(0m/s) + (1.2kg)(0)
= 2.4 kg.m/s
Total Kinetic Energy Before the
Collision
KEtotal = m1 v1
2 / 2 + m2 v2
2 / 2
= (1.2kg) (2.0m/s)2 / 2 + (1.2kg)(0)/2
= 2.4 kg.m2/s2 + 0
= 2.4 N
Total Kinetic Energy After the
Collision
KEtotal = m1 v1‘ 2 / 2 + m2 v2‘ 2 / 2
= (1.2kg) (0m/s)2/2 + (1.2kg)(2.0m/s)2
= 0 + 2.4kg.m2 s2
= 2.4 N

39. b) Perfectly inelastic collision
– occurs when the two bodies join together
during collision and move together as one
body after collision.

40. The total momentum is conserved but the
total kinetic energy before is not equal to
the kinetic energy after collision.
- the kinetic energy is lost in the
deformation caused by the collision or
conversion to other forms of energy like
sound or thermal energy.

41. Before:
m = 460.0 kg m2 = 460.0 kg
v1 = 100.0 m/s v2 = 0
After
m = 460.0 kg m = 460.0 kg
v’ = v1’ = 5.0 m/s
Perfectly inelastic collision between
equal masses occur when two
bodies join together during collision
and move together as one body after
collision.

42. Before:
m = 460.0 kg m2 = 460.0 kg
v1 = 10.0 m/s v2 = 0
After
m = 460.0 kg m = 460.0 kg
v’ = v1’ = 5.0 m/s
Total Momentum Before the Collision
Ptotal = p1 + p2
= m1v1 + m2v2
= (460.0kg)(10.0m/s) + (460.0kg)(0)
= 4,600.0 kg.m/s
Total Momentum After the Collision
Ptotal = p1‘ + p2‘
= (m1+ m2 ) v’
= (460.0kg + 460.0kg)(5.0 m/s)
= 4,600.0 kg.m/s
The total
momentum is
conserved
before and after
the collision

43. Before:
m = 460.0 kg m2 = 460.0 kg
v1 = 100.0 m/s v2 = 0
After
m = 460.0 kg m = 460.0 kg
v’ = v1’ = 5.0 m/s
The total kinetic energy before is not equal
to the kinetic energy after the collision.
Total Kinetic Energy Before the Collision
KEtotal = m1 v1
2 / 2 + m2 v2
2 / 2
= (460.0kg) (10.0m/s)2 / 2 + (460.0kg)(0)
= 32,000 N.m or Joule
Total Kinetic Energy After the Collision
KEtotal =( m1 m2‘) (v2‘)2 / 2
= (460.0kg + 460.0kg) (5.0m/s)2 /2
= (920.0kg) (25m2 /s2 ) 2
= 11,500 N.m or J

44. c) Inelastic collision
Objects of equal masses move off
separately at different speeds in the same
direction after collision.
The figure shows that in an inelastic collision,
object of equal masses move off separately at
different speeds in the same direction after
collision

45. Total Momentum Before the Collision
Ptotal = p1 + p2
= m1v1 + m2v2
= (460.0kg)(10.0m/s) + (460.0kg)(0)
= 4,600.0 kg.m/s
Total Momentum After the Collision
Ptotal = p1‘ + p2‘
= (m1+ m2 ) v’
= (460.0kg + 460.0kg)(5.0 m/s)
= 4,600.0 kg.m/s
460.0kg 460.0kg
10.0m/s
V2 = 0m/s
2.5 0 m /s
7.5 0 m/s

46. Total Kinetic Energy Before the Collision
KEtotal = m1 v1
2 / 2 + m2 v2
2 / 2
= (460.0kg) (10.0m/s)2 / 2 + (460.0kg)(0) /2
= (460kg)(100.0m2/s2 /2 + 0
= 23,000 N.m or J
Total Kinetic Energy After the Collision
KEtotal = m1 v1‘ 2 / 2 + m2 v2‘ 2 / 2
= (460.0kg) (2.5m/s)2/2 + (460.0kg)(7.5.0m/s)2 / 2
= (460.0kg)(6.25m2/s2 /2 + (460.0kg)(56.26m2/s2 /2
= 2,875.0kg.m2/s2 / 2 + 25,879.6kg.m2/s2 /2
= 1,437.5kg.m2/s2 + 12,939.8kg.m2/s2
= 14,377.3 N.m or J.
The total kinetic
energy before is
not equal to the
kinetic energy
after collision.
The kinetic energy is lost
in the deformation caused
by the collision or
conversion to other forms
of energy like sound or
thermal(heat)
460.0kg
460.0kg
10.0m/s V2 = 0m/s
7.5 0 m/s
2.5 0 m /s

47. Sample Problem # 1:
Consider the perfectly elastic collision
between two balls of masses 120g (m1) and
240g (m2). If ball 1 is moving with a velocity
of 32.0 cm/s to the right and ball 2 with a
ve3locity of 25.0 cm/s in the same direction.
Find the velocity of ball 1, if ball 2 has a final
velocity of 28.0 cm/s.

48. Sample Problem #1:
Consider the perfectly elastic collision between
two balls of masses 120g (m1) and 240g (m2). If
ball 1 is moving with a velocity of 32.0 cm/s to
the right and ball 2 with a ve3locity of 25.0 cm/s
in the same direction. Find the velocity of ball 1,
if ball 2 has a final velocity of 28.0 cm/s.
Before collision
m1 = 120g m2 = 240g
v1 = 32.0cm/s v2 = 25.0cm/s
ptotal = m1 v1 + m2 v2
= 120g(32.0cm/s) + 240g (25.0 cm/s)
= 9840 g.cm/s
After collision
m1 = 120g m2 = 240g
v1’ = 32.0cm/s v2’’ = 28.9cm/s
P’total = m1v2’ + m2 v2’
= 120g(v1’) + 240g (28.9cm/s)
= 120g (v1’) + 6936 g.cm/s

49. After collision
m1 = 120g m2 = 240g
v1’ = 32.0cm/s v2’’ = 28.9cm/s
P’total = m1v2’ + m2 v2’
= 120g(v1’) + 240g (28.9cm/s
= 120g (v1’) + 6936 g.cm/s
9840 g. cm/s = 120g (v‘1) + 6936 cm/s
9840 g.cm/s – 6936 cm/s = 120g (v’1)
120g (v’1) = 2,904 g.cm/s
v’1 = 2,904 g.cm/s / 120g
v’1 = 24.2 cm/s
Before collision
m1 = 120g m2 = 240g
v1 = 32.0cm/s v2 = 25.0cm/s
ptotal = m1 v1 + m2 v2
= 120g(32.0cm/s) + 240g (25.0 cm/s)
= 9840 g.cm/s

50. Sample Problem #2:
A 70.0-kg ice skater moving on ice at 4.62 m/s
hits from behind another 70.9 kg skater, who
is initially at rest.
After collision, the skaters move in the same
direction at different speeds.
The first skater moves at 1.30 m/s.
What is the speed of the second skater?
What is the total momentum of the skaters
before collision?
Is the collision elastic or inelastic?

51. Sample Problem # 2:
A 70.0-kg ice skater moving on ice at 4.62 m/s hits from
behind another 50.9 kg skater, who is initially at rest.
After collision, the skaters move in the same direction at
different speeds. The first skater moves at 1.30 m/s.
What is the speed of the second skater?
What is the total momentum of the skaters before
collision?
Is the collision elastic or inelastic?
m1 = 70.0kg
v1 = 4.62m/s
m2 = 50.9kg
v2 = 0 m/s
m1 = 70.0kg
v‘1 = 1.30 m/s
m2 = 50.9kg
v ‘2 = ?

52. Solutions:
Given:
m1 = 70.0 kg m2 = 50.9 kg v‘1 = 1.30 m/s
v1 = 4.62m/s v‘2 = ? ptotal = ?
ptotal = m1v1 + m2v2
= (70.0kg)(4.62m/s) + (50.0kg)(0m/s)
= 323.4kg.m/s + 0
= 323.4 kg . m/s
p‘total = m1v‘1 + m2v‘2
= (70.0 kg) (1.30 m/s) + (50.9 kg) (v’2)
= 91 kg . m/s + (50.9 kg) (v’2)
323.4 kg.m/s = 91 kg.m/s + (50.9-kg) (v’2)
323.4 kg.m/s – 91 kg.m/s = (50.0-kg) (v’2)
232.4 kg.m/s = (50.0-kg) (v’2)
(50.0-kg)(v’2) = 232.4 kg.m/s
v’2 = 232.3 kg.m/s ÷ 50.0 kg
v’ = 4.65 m/s

53. Sample Problem # 3:
One rainy day, a car with a mass of 1250-kg
moving at 20.0 m/s hits the rear end of
another car with a mass of 1610-kg moving at
8.0 m/s in the same direction.
What is the final velocity of the two cars if
they stick together.
What is the change in kinetic energy of the
system?
What type of collision occurred in the
system?

54. Sample Problem # 3:
One rainy day, a car with a mass of 1250-kg moving at 20.0 m/s
hits the rear end of another car with a mass of 1610-kg moving at
8.0 m/s in the same direction.
What is the final velocity of the two cars if they stick together.
What is the change in kinetic energy of the system?
What type of collision occurred in the system?
m1 = 1250-kg
m2 = 1610-kg
v1 = 20.0 m/s
v2 = 8.0 m/s v ‘ = ?

55. Solutions:
Given:
m1 = 1250 kg v1 = 20 m/s
m1 = 1610 kg v2 = 8.0 m/s v’ = ?
ptotal = m1v1 + m2v2
= (1250kg)(20m/s) + (1610kg)(8.0m/s)
= 25,000 kg.m/s + 12,880 kg.m/s
= 37,880 kg . m/s
p‘total = p’1 + p’2
= (m1 + m2) v’
= (1250-kg + 1610-kg) (v’)
37,880 kg.m/s = (1250-kg + 1610-kg) (v’)
37,880 kg.m/s = (2,860-kg) (v’)
(2,860-kg) (v’) = 37,880 kg . m/s
v’ = 37,880 kg . m/s ÷ 2,860-kg
v’ = 13.2 m/s

56. Practice Exercise #
1. A 4.0x103 kg.m/s2 force acts on a
1,200-kg car to stop it in 8.0s.
What is the initial velocity of the
car in km/h?
Given:
F = 4.O X 103 kg.m/s
m = 1,200-kg
v = ?
t = 8.0s
F = m (v/t)
4.0 x 103 kg.m/s = 1,200-kg (v/8.0s)
4.0 x 103 kg.m/s = (1,200-kg) (v) / 8.0s
4.0 x 103 kg.m/s = (150-kg/s)(v)
150-kg/s (v) = 4.0 x103 kg.m/s
v = 4.0 x103 kg.m/s ÷ 150 kg/s
v = 27 m/s2
27m/s x 1km/1000m x 3600s/1h = 97.2km/h

57. Practice Exercise # 2
What force is needed to stop a 500-
kg tricycle in 6.0s if it is travelling at
60 km/h?
Given:
F = ?
m = 500 kg
t = 6.0 s
v = 60 km/h x 1000m/1km x 1h/3600s
= 17m/s
F = m(v/t)
= (500-kg) (17m/s ÷ 6.0s)
= (500-kg) (2.8 m/s2)
= 1,400 kg . m/s2

58. Practice Exercise # 3.a
A player hits a football which is initially at
rest and moves 22 m/s. Find the
momentum of the football if it has mass
of 0.48kg.
Given:
v0= 0
v = 22 m/s
p =?
m = 0.48kg
p = m(v – v0)
= (0.48kg) (22 m/s – 0)
= (0.48kg)(22m/s)
= 10.56 kg . m/s

59. Practice Exercise # 3.b
A player hits a football (0.48-kg) which is
initially at rest and moves 22 m/s. Find
the force exerted on it when the time of
contact is 0.03 s.
Given:
m = 0.48-kg
v0= 0
v = 22 m/s
F =?
t = 0.03 s
F = m (v –v0÷ t)
= (0.48kg) (22m/s -0 ÷ 0.03s)
= (0.48kg) (733 m/s2)
= 351.8 kg . m/s2

60. Practice Exercise # 3.c
A player hits a football (0.48-kg) which is
initially at rest and moves 22 m/s. Find the
Impulse in the football.
Given:
m = 0.48-kg
v0 = o
v = 22 m/s
p = ?
p0 =?
p0 = mv0
= (0.48kg)(0)
= 0
p = mv
= (0.48kg)(22m/s)
= 10.56 kg . m/s
I = ∆p
= p – p0
= 10.56 kg.m/s – 0
= 10.56 kg.m/s

61. Practice Exercise # 4
A 660-kg motorcycle moving 20m/s rear-
ends a 1017-kg car travelling at 16 m/s on
ca concrete highway in the same direction.
a) Sketch the situations showing the before
and after collision conditions for both
vehicles.
b) If the two vehicles stick together, how
fast did they move immediately after
collision?
Given:
m1 = 660-kg v1 = 20 m/s
m2 = 1017-kg v2 = 16 m/s
v’ = ?

62. Solution
Given:
m1 = 660-kg v1 = 20 m/s
m2 = 1017-kg v2 = 16 m/s
v’ = ?
ptotal = p1 + p2
= m1v1 + m2v2
= (660kg)(20m/s) + (1017kg)(16 m/s)
= 13,200kg.m/s + 16272kg.m/s
= 29,472 kg . m/s
p’total = p1 + p2
= (m1 + m2 ) (v’)
= (660kg + 1017kg) (v’)
= (1,677 kg) (v’)
ptotal = p’total
29,472 kg . m/s = (1,677 kg) (v’)
(1,677 kg)(v’) = 29,472 kg . m/s
v’ = 29,472 kg . m/s ÷ 1,677 kg
v’ = 17.6 m/s