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1. UNIVERSITY OF HARGEISA
2014
CABDIRISAAQ
AFGAAB
2/23/2014
PHYSICS

2. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB2
Name: ___________________________________________
Class: ___________________________________________
ID: _____________________________________________
Deadline: ___________________________________________

3. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB3
CONTENTS
1. Definition and over view of the momentum
and collision.
2.Formulae
3.Drawing and figures
4.In collision of a cars how injuries of
passenger can be reduced

4. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB4
Introduction of momentum
Figure (1)
Momentum
Momentum is product of objects, mass and velocity of particle.
It is a vector quantity directed through the particle in the directed of motion.
The linear momentum of a body or of a system of particles is the vector sum of
the linear moment of the individual particles.
If a body of mass M is translated with a velocity V, its momentum is MV, which
is the momentum of particle of mass M at the centre of gravity of body.
The momentum which depends on the mass and the velocity.
Momentum= mass × velocity
(kg) (m/s)
PPPP = m ×××× v

5. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB5
Example
Figure (2)
Calculate the momentum of a bus of mass 2000kg travelling 5ms-1?
Solution
Given
M=2000kg
V=5ms-1
PPPP =?
PPPP = m ×××× v
2000kg×5ms-1
=10000kgms-1 or 10000Ns

6. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
6
Momentum and Newton’s Second Law
The importance of momentum, unlike the importance of energy, was recognized early
in the development of classical physics.
Momentum was deemed so important that it was called the “
Newton actually stated his
The net external force equals the change in momentum of a system divided by the time
over which it changes. Using symbols, this law is
Where Fnet is the net external force,
change in time.
Law of conservation of momentum
The law of conservation states that when two or more bodies act upon one another,
their total momentum remains constant, provided no external force are acting.
Applying the law of conservation of momentum to a collision of mass
travelling at u1 and u2 before collision, and
Momentum before collision=Momentum after collision
If the two bodies travel on coupled to gather with a common velocity
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB
Momentum and Newton’s Second Law
The importance of momentum, unlike the importance of energy, was recognized early
in the development of classical physics.
deemed so important that it was called the “quantity of motion
Newton actually stated his second law of motion in terms of momentum:
external force equals the change in momentum of a system divided by the time
over which it changes. Using symbols, this law is
Fnet = ∆p
∆t,
is the net external force, ∆p is the change in momentum, and
Law of conservation of momentum
The law of conservation states that when two or more bodies act upon one another,
their total momentum remains constant, provided no external force are acting.
Applying the law of conservation of momentum to a collision of mass
before collision, and v1 and v2 after collision:
Momentum before collision=Momentum after collision
If the two bodies travel on coupled to gather with a common velocity
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
AFGAAB
Momentum and Newton’s Second Law
The importance of momentum, unlike the importance of energy, was recognized early
quantity of motion.”
terms of momentum:-
external force equals the change in momentum of a system divided by the time
is the change in momentum, and ∆t is the
Law of conservation of momentum
The law of conservation states that when two or more bodies act upon one another,
their total momentum remains constant, provided no external force are acting.
Applying the law of conservation of momentum to a collision of mass m1 and m2,
after collision:
Momentum before collision=Momentum after collision
If the two bodies travel on coupled to gather with a common velocity v.

7. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB7
Conservation of Momentum
Momentum is an important quantity because it is conserved. Yet it was not conserved
in the examples in Impulse and Linear Momentum and Force, where large
changes in momentum were produced by forces acting on the system of interest. Under
what circumstances is momentum conserved?
The answer to this question entails considering a sufficiently large system. It is always
possible to find a larger system in which total momentum is constant, even if
momentum changes for components of the system. If a football player runs into the
goalpost in the end zone, there will be a force on him that causes him to bounce
backward. However, the Earth also recoils —conserving momentum—because of the
force applied to it through the goalpost. Because Earth is many orders of magnitude
more massive than the player, its recoil is immeasurably small and can be neglected in
any practical sense, but it is real nevertheless.
Consider what happens if the masses of two colliding objects are more similar than the
masses of a football player and Earth—for example, one car bumping into another, as
shown in Figure (3) on next page . Both cars are coasting in the same direction
when the lead car (labeled m2) is bumped by the trailing car (labeled m1). The only
unbalanced force on each car is the force of the collision. (Assume that the effects due
to friction are negligible.) Car 1 slows down as a result of the collision, losing some
momentum, while car 2 speeds up and gains some momentum. We shall now show
that the total momentum of the two-car system remains constant.

8. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB8
Figure (3)
Impulse
Impulse is product of force and time interval over which it act.
Impulse of constant force F, of the force and the time t for which it act. If the force
varies with time, the impulse is the integral of the force with respect to the time during
which the force acts. In either case, impulse of force equals the change of
momentum produced by it. An impulsive force is one that is very large but acts
only for a very short time; it can be represented by a DIRAC FUNCTOIN.

9. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
9
1. Two stationary rocks of mass 100kg and 200kg respectively are held together.
An explosion between the rocks pushes then apart with no less of mass.
velocity of the 100kg
rock after the explosion.
Given
Before
m1=100kg m
m2=200kg
u1=0
u2=0
0=100(4) + 200(v2)
200v2=-400
V2=-2m/s
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB
Example
Two stationary rocks of mass 100kg and 200kg respectively are held together.
An explosion between the rocks pushes then apart with no less of mass.
velocity of the 100kg rock after the explosion is 4m/s. Find the velocity
rock after the explosion.
Solution
After
100kg m1=100kg
=200kg m2=200kg
=0 v1=4m/s
v2=?
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
AFGAAB
Two stationary rocks of mass 100kg and 200kg respectively are held together.
An explosion between the rocks pushes then apart with no less of mass. The
rock after the explosion is 4m/s. Find the velocity of 200kg

10. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
10
Introduction of
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB
Introduction of collision
Collision
Figure (4)
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
AFGAAB
)
Figure (5)

11. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
11
A collision is thought of as being one of three kinds
1. Elastic collision
unchanged after the collision, none bein
nuclear
particle is scattered without exiting or breaking up the struck nuclear.
This formula:
2. Inelastic collision (or Inelastic collision of the first kind):
the total kinetic energy
some other form of energy increased.
Figure
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB
Figure
A collision is thought of as being one of three kinds
Elastic collision: - One in which the total kinetic energy of transl
unchanged after the collision, none being translated into other forms. In
nuclear physics, an elastic collision is the one in which the incoming
particle is scattered without exiting or breaking up the struck nuclear.
This formula:
Inelastic collision (or Inelastic collision of the first kind):
the total kinetic energy of translation is decreased by
some other form of energy increased.
Figure (7)
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
AFGAAB
Figure (6)
One in which the total kinetic energy of translation is
g translated into other forms. In
physics, an elastic collision is the one in which the incoming
particle is scattered without exiting or breaking up the struck nuclear.
Inelastic collision (or Inelastic collision of the first kind):- One in which
of translation is decreased by the collision while

12. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB12
3. Superelastic collision (or inelastic collision of the second kind):- One in
which the total kinetic energy of translation is increased by the collision
while some other form of energy decreased.
In all kind of collision total energy, mass, momentum, and angular momentum are conserved.
Figure (8)
Figure (9) Figure (10)

13. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB13
Figure (11)
Figure (12)
Figure (13)

14. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
14
1. A 3kg mass with velocity 4m/s collides with stationary any 2kg mass. After
impact, the 3kg mass continue indirection at 2m/s. The 1kg mass moves at 6m/s
in the same direction as the 3kg mass. Is this an
collision?
given
KE before
m1=3kg
m2=1kg
u1=4m/s
u2=0
1/2(3) (4)2 + 1/2(1) (0)2 =
24j + 0j = 6j + 18j
24j = 24j
It was the elas
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB
Example
A 3kg mass with velocity 4m/s collides with stationary any 2kg mass. After
impact, the 3kg mass continue indirection at 2m/s. The 1kg mass moves at 6m/s
in the same direction as the 3kg mass. Is this an example of elastic or inelastic
Solution
KE after
m1=3kg
m2=1kg
v1=2m/s
v2=6m/s
= 1/2 (3) (2)2 + 1/2 (1) (6)2
24j + 0j = 6j + 18j
UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Title: momentum and collision
AFGAAB
A 3kg mass with velocity 4m/s collides with stationary any 2kg mass. After
impact, the 3kg mass continue indirection at 2m/s. The 1kg mass moves at 6m/s
example of elastic or inelastic

15. UNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISAUNIVERSITY OF HARGEISA
Faculty of engineering
Assignment
Title: momentum and collision
By: AFGAAB15