The document discusses linear momentum, which is defined as the product of an object's mass and its velocity. It provides examples of calculating linear momentum for various objects with given masses and velocities. It also discusses how linear momentum is a vector quantity and how impulse (force over time) results in changes to an object's momentum according to the impulse-momentum theorem. Conservation of momentum is discussed, where the total initial momentum of objects in a closed system equals the total final momentum. Examples demonstrate calculating unknown velocities or momenta using conservation of momentum.

1. MASS
AND
MOMENTUM

2. Momentum, product of the mass of a
particle and its velocity. Momentum is
a vector quantity; i.e., it has both
magnitude and direction. Isaac Newton’s
second law of motion states that the time
rate of change of momentum is equal to
the force acting on the particle.

3. Linear Momentum
Linear momentum is defined as the
product of a system’s mass
multiplied by its velocity:
p = mv

4. The scientific definition of linear momentum
is consistent with most people’s intuitive
understanding of momentum: a large, fast-
moving object has greater momentum than
a smaller, slower object. Linear
momentum is defined as the product of a
system’s mass multiplied by its velocity. In
symbols, linear momentum is expressed
as p = mv.

5. Momentum is directly proportional to the
object’s mass and also its velocity. Thus the
greater an object’s mass or the greater its
velocity, the greater its momentum.
Momentum p is a vector having the same
direction as the velocity v. The SI unit for
momentum is kg · m/s.

6. Example 1:
Calculate the momentum of a 100-kg
football player running at 8.00 m/s.
Given:
Mass: 100 kg
Velocity: 8.00m/s
p = mv
p =( 100kg) (8.00m/s)
p = 800kg ∙ 𝑚/𝑠

7. Example 2:
What is the momentum
of a bowling ball with
mass 5kg and
velocity 10m/s?

8. Momentum describes an
object’s motion.
Linear momentum of an object of mass
(m) moving with a velocity (v) is defined
as the product of the mass and velocity.

9. Momentum is denoted by the symbol
𝒑 . 𝑝 is the tendency of a moving
particle/system to continue moving and
the difficulty it encounters in slowing
down to rest. A common unit of this
vector quantity is kilogram meter per
second (kg•m/s).

10. 𝑝 = m x 𝑣
momentum = mass x velocity

11. Example 3. 0.5-kg ball moves at 2 m/s. Then the ball is hit with
force F opposite to the ball direction, so the ball speed is changed
to 6 m/s. The ball in contact with a hitter for 0.01 second, what is
the change in momentum of the ball.
Mass of ball (m) = 0.5 kg
Initial velocity (𝑣𝑖) = 2 m/s
Final velocity (𝑣𝑓) = -6 m/s
Time interval (t) = 0.01 second
Unknown: The change in
momentum
Solution :
∆p = m(𝑣𝑓) – m(𝑣𝑖) = m (𝑣𝑓–𝑣𝑖 )
∆p = (0.5 kg)(-6 m/s – 2 m/s)
∆p = (0.5 kg)(-8m/s)
∆p = -4 kg ▪m/s

12. Example 4.
The parking brake on a 1200kg
automobile has broken, and the
vehicle has reached a momentum
of 7800 kg▪m/s. what is the velocity
of the vehicle?

13. Linear momentum is a vector quantity
whose direction is the same as that of
velocity (𝑣). It is sometimes referred to
as inertia of a body in motion.

14. A change in momentum takes
force and time.
Now, consider a system acted upon by a force F .
Momentum is closely related to force, in fact, when
Newton expressed mathematically his second law of
motion, the force F will produce an acceleration
(a), he wrote it not as F = ma, but in this form:
𝑭𝒐𝒓𝒄𝒆 =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎
𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍

15. Impulse momentum Theorem
𝐹∆t =∆ 𝑝 or
𝐹∆t =∆ 𝑝 = m∆𝑣𝑓 - m∆𝑣𝑖

16. The expression 𝐹∆t =∆ 𝑝 is called the
impulse-momentum theorem. The term on
the left side of the equation, 𝐹∆t is called
impulse of the force F for the time interval t.
We can see that impulse equals the change
in momentum. Impulse is an impelling force
that acts suddenly and produces motion. A
common unit is Newton second (Ns).

17. Example 5:
A 0.25kg ball was strucked by a baseball bat
from rest up to a speed of 35 m/s the ball was
in contact the bat for 0.02 seconds.
a.What is the change in momentum of the
ball?
b.What was the impulse exerted on the ball?
c.Calculate the average speed?

18. Solution:
a.∆ 𝑝 = m∆v
∆ 𝑝 = m[𝑣𝑓 − 𝑣𝑖]
∆ 𝑝 = (0.25kg)(35m/s - 0)
∆ 𝑝 = 8.75kg ∙ m/s
b. I = ∆ 𝑝
= 8.75N ∙ 𝑠
c. I = F∆t
8.75N.s = F (0.02s)
F = 437.5N

19. Example 6:
A 0.25kg tennis ball moves east at a
speed of 50m/s and strikes a wall. The
ball bounces at speed of (-50m/s). the
contact time between the wall and the ball
was 0.015 seconds. What average force
was exerted by the wall on the ball?

20. Identify the given:
Mass: 0.25kg
Final velocity:-50m/s
Initial velocity: 50m/s
Time: 0.015s

21. Conservation of Momentum
In a closed and isolated system, the total momentum of
the objects before a collision is equal to the total
momentum of the objects after the collision.
𝑃𝑖 =𝑃𝑓
𝑃𝐴 + 𝑃𝐵= 𝑃𝐴
′
+ 𝑃𝐵
′
𝑚𝐴𝑣𝐴 + 𝑚𝐵𝑣𝐵=𝑚𝐴𝑣𝐴′ + 𝑚𝐵𝑣𝐵′
0 = 𝑚𝐴𝑣𝐴′ + 𝑚𝐵𝑣𝐵′
-𝑚𝐴𝑣𝐴′= 𝑚𝐵𝑣𝐵′

22. Example 7: Two ice skaters, Alice and Bob,
are initially at rest on an ice rink. Alice has
a mass of 60 kg, and Bob has a mass of 80
kg. Alice pushes Bob, giving him a velocity
of 4 m/s to the right. What is Alice's
velocity after the push, assuming there is
no external friction?

23. Initial Momentum = Final Momentum
(0 kg) * (Alice's initial velocity) + (0 kg) * (Bob's initial velocity) =
(60 kg) * (Alice's final velocity) + (80 kg) * (Bob's final velocity)
(0) + (0) = (60 kg) * (Alice's final velocity) + (80 kg) * (4 m/s)
0 = 60 kg * (Alice's final velocity) + 320 kg*m/s
60 kg * (Alice's final velocity) = -320 kg*m/s

24. Alice's final velocity = -320 kg*m/s / 60
kg = -5.33 m/s (rounded to two decimal
places) So, Alice's velocity after the push
is approximately -5.33 m/s to the left.

25. Example 8: A 50 kg mass
initially at rest explodes into
two fragments. The 30 kg
fragment moves west at
40m/s. what is the velocity
of the second fragment?