Linear Momentum and

its Conservation

By: Victor Rea Oribe

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

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

Velocity is a vector quantity.

Vector - has magnitude and

direction.

Therefore, Momentum is also

a vector quantity

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

x1 h

3,600 s

x1000 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

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

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

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

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

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

Impulse

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.

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.

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

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

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)

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

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

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

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

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

Find the Impulse

I = ∆p

I = -2373.6 kg . m/s

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

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 – p

0)

t

F =

0 – 112500 kg.m/s

18 s

F = -6250 kg.m/s2

or N

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

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 =

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?

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 = ?

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 = ?

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

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 = ?

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.

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

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

Collision may be classified as

a) elastic collision

– collision in which both momentum

and kinetic energy are conserved.

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

An analysis of the kinetic energy of the two objects revealsthat the total system kinetic energy before the collision is800,000 Joules (200,000 J for the car plus 600,000 J for thetruck).

After the collision, the total system kinetic energy is800000 Joules (800,000 J for the car and 0 J for the truck).

The total kinetic energy before the collision is equal to thetotal kinetic energy after the collision.

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 v

2‘ = 2.0 m/s

Total Momentum Before the

Collision

Ptotal

= p1

+ p2

= m1v

1+ m

2v

2

= (1.2kg)(2.0m/s) + (1.2kg)(0)

= 2.4 kg.m/s

Total Momentum After the

Collision

Ptotal

= p1‘ + p

2‘

= m1v

1‘ + m

2v

2’

= (1.2kg)(0m/s) + (1.2kg)(0)

= 2.4 kg.m/s

Total Kinetic Energy Before the

Collision

KEtotal

= m1

v1

2/ 2 + m

2v

2

2/ 2

= (1.2kg) (2.0m/s)2

/ 2 + (1.2kg)(0)/2

= 2.4 kg.m2/s

2+ 0

= 2.4 N

Total Kinetic Energy After the

Collision

KEtotal

= m1

v1‘

2/ 2 + m

2v

2‘

2/ 2

= (1.2kg) (0m/s)2/2 + (1.2kg)(2.0m/s)2

= 0 + 2.4kg.m2

s2

= 2.4 N

b) Perfectly inelastic collision

– occurs when the two bodies join together

during collision and move together as one

body after collision.

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.

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.

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

= m1v

1+ m

2v

2

= (460.0kg)(10.0m/s) + (460.0kg)(0)

= 4,600.0 kg.m/s

Total Momentum After the Collision

Ptotal

= p1‘ + p

2‘

= (m1+ m

2) 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

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 + m

2v

2

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‘) (v

2‘)

2/ 2

= (460.0kg + 460.0kg) (5.0m/s)2

/2

= (920.0kg) (25m2

/s2

) 2

= 11,500 N.m or J

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

Total Momentum Before the Collision

Ptotal

= p1

+ p2

= m1v

1+ m

2v

2

= (460.0kg)(10.0m/s) + (460.0kg)(0)

= 4,600.0 kg.m/s

Total Momentum After the Collision

Ptotal

= p1‘ + p

2‘

= (m1+ m

2) 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

Total Kinetic Energy Before the Collision

KEtotal

= m1

v1

2/ 2 + m

2v

2

2/ 2

= (460.0kg) (10.0m/s)2

/ 2 + (460.0kg)(0) /2

= (460kg)(100.0m2/s

2/2 + 0

= 23,000 N.m or J

Total Kinetic Energy After the Collision

KEtotal

= m1

v1‘

2/ 2 + m

2v

2‘

2/ 2

= (460.0kg) (2.5m/s)2/2 + (460.0kg)(7.5.0m/s)

2/ 2

= (460.0kg)(6.25m2/s

2/2 + (460.0kg)(56.26m

2/s

2/2

= 2,875.0kg.m2/s

2/ 2 + 25,879.6kg.m

2/s

2/2

= 1,437.5kg.m2/s

2+ 12,939.8kg.m

2/s

2

= 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)

460.0kg460.0kg

10.0m/s V2

= 0m/s

7.5 0 m/s

2.5 0 m /s

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.

Sample Problem #1:

Consider the perfectly elastic collision between

two balls of masses 120g (m1) and 240g (m

2). 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

= m1v

2’+ m

2v

2’

= 120g(v1’) + 240g (28.9cm/s)

= 120g (v1’) + 6936 g.cm/s

After collision

m1

= 120g m2

= 240g

v1’

= 32.0cm/s v2’’

= 28.9cm/s

P’total

= m1v

2’+ m

2v

2’

= 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

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?

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

= ?

Solutions:

Given:

m1

= 70.0 kg m2

= 50.9 kg v‘1

= 1.30 m/s

v1

= 4.62m/s v‘2

= ? ptotal

= ?

ptotal

= m1v

1+ m

2v

2

= (70.0kg)(4.62m/s) + (50.0kg)(0m/s)

= 323.4kg.m/s + 0

= 323.4 kg . m/s

p‘total

= m1v‘

1+ m

2v‘

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’2

= 4.65 m/s

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?

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/sv ‘ = ?

Solutions:

Given:

m1

= 1250 kg v1

= 20 m/s

m1

= 1610 kg v2

= 8.0 m/s v’ = ?

ptotal

= m1v

1+ m

2v

2

= (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

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

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

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

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

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

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’ = ?

Solution

Given:

m1

= 660-kg v1

= 20 m/s

m2

= 1017-kg v2

= 16 m/s

v’ = ?

ptotal

= p1

+ p2

= m1v

1+ m

2v

2

= (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