MEMBERS
CERVANTES, DIPLOMA, DUENA,
JARUYAN, MEDADO, MONTES,
OLVIDO, POBLACION
The Center of Mass of a
system is that one point
where any uniform force is
acted upon the object.
In a system, the average
position of all parts are
weighted according to their
masses.
VISUAL OF COM
Next
stage
MOMENTUM
IMPULSE
a measure of how quickly
force changes in
magnitude AND direction.
COLLISION
NEXT
STAGE
Agenda Page
M = 𝑚1 + 𝑚2 + 𝑚3 + ⋯ + 𝑚𝑁
i = 1, 2, 3, …, N
𝑥𝑐𝑜𝑚 =
𝑚𝑖 𝑥𝑖
𝑀
FOR X-AXIS
𝑦𝑐𝑜𝑚 =
𝑚𝑖 (𝑥, 𝑦, 𝑧)𝑖 = 𝑚1 (𝑥, 𝑦, 𝑧)1 +𝑚2 (𝑥, 𝑦, 𝑧)2 + ⋯ + 𝑚𝑁 (𝑥, 𝑦, 𝑧)𝑁
𝑚𝑖 𝑦𝑖
𝑀
FOR Y-AXIS
𝑧𝑐𝑜𝑚 =
𝑚𝑖 𝑧𝑖
𝑀
FOR POINT
OBJECT
𝑥𝑐𝑜𝑚
𝑙
=
2
FOR GEOMTRIC
SHAPE
Agenda Page
𝑝 = 𝑚𝑣
m – mass (kg) v - velocity (m/s)
𝐽 = 𝐹∆𝑡
F = applied force (N)
∆𝑡 = elapsed time (s)
𝐹∆𝑡 = 𝑚∆𝑣
Impulse = Ft
Impulse = m(𝑣𝑓 − 𝑣𝑖 )
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = (𝑚1 + 𝑚2 )𝑣𝑓
Center of Mass (COM) Unit
(m) - meters
Momentum (p) Unit
(kg-m/s) – kilogram
-meter per second
Impulse (J) Unit
(N-s) – Newton-second
Collision Unit
(kg-m/s) – kilogram
-meter per second
problem
Two point masses 3kg and 5kg are at
4m and 8m from the origin on X-axis.
Locate the position of center of mass
of the two point masses from:
a.) the origin;
b.) 3kg mass.
Agenda Page
solution
Given: 𝑚1 = 3𝑘𝑔 and 𝑚2 = 5𝑘𝑔
a.) from the origin,
So 𝑥1 = 4𝑚 and 𝑥2 = 8𝑚
𝑥𝑐𝑜𝑚
𝑚1 𝑥1 + 𝑚2 𝑥2
=
𝑚1 + 𝑚2
𝑥𝑐𝑜𝑚
3 ∗ 4 + (5 ∗ 8)
=
3+5
𝑥𝑐𝑜𝑚
52
=
= 6.5𝑚
8
b.) from 3kg mass,
So 𝑥1 = 0𝑚 and 𝑥2 = 4𝑚
J = Ft
J = (3000N)(0.05s)
J = 150 Ns
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖
= 𝑚1 + 𝑚2 𝑣𝑓
(4kg*6m/s)+(2kg*0m/s)
= (4kg+2kg)𝑣𝑓
24kg-m/s =(6kg )𝑣𝑓
𝑣𝑓 = 4 m/s
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓
(3000kg*10m/s) +(1000kg*0m/s) =
(3000kg*𝑣1𝑓 ) +(1000kg*15m/s)
30000kg 2 -m/s = 3000kg𝑣1𝑓 + 15000kg-m/s
15000kg 2 -m/s =3000kg𝑣1𝑓
𝑣1𝑓 = 5 𝑚/𝑠
1. It is a measure of how quickly
force changes in magnitude AND
direction.
2. In a system, the average position
of all parts are weighted according
to their?
3. WHAT OCCURS WHEN TWO OBJECTS
COMING INTO EACH OTHER FOR A VERY
SHORT PERIOD?
4. WHAT IS THE QUANTITY OF MOTION OF
AN OBJECT THAT HAS BOTH A MAGNITUDE
AND A DIRECTION.
5. WHAT IS THE FORMULA FOR GETTING
THE CENTER OF THE MASS OF GEOMETRIC
SHAPES?
a 0.150-kg basketball
moving at a speed of
45.0m/s crosses the plate
and strikes the 0.250-kg
catcher’s mitt
(originally at rest). The
catcher’s mitt
immediately recoils
backwards (at the same
speed of the ball) before
the catcher applies an
external force to stop
its momentum. Determine
the post-collision
velocity of the mitt and
ball.
Item 1
IMPULSE
Item 2
PROBLEM SOLVING
Item 5
MASSES
𝑥
𝑙
𝑐𝑜𝑚 =
2
MOMENTUM
Item 4
COLLISION
Item 3
𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 = (𝑚1 +𝑚2 )𝑣𝑓
(0.15kg*45m/s)
+(0.25kg*0m/s) =
(0.15kg+0.25kg)𝑣𝑓
6.75kg-m/s =( 0.4kg)𝑣𝑓
𝑣𝑓 = 16.875 𝑜𝑟 16.9 𝑚/𝑠
The velocity of the mitt and ball
is 16.9 m/s.