1. LAW OF MOMENTUM
Ex 1) A 3.0 kg object traveling 6.0 m/s east has a perfectly elastic collision with a 4.0 kg object
traveling 8.0 m/s W.
After the collision, the 3.0 kg object will travel 10. m/s west.
a) Total momentum before the collision?
m1 = 3.0 kg
v1 = 6.0 m/s(East)
m2 = 4.0 kg
v2 = - 8.0 m/s (West)
Total momentum before:
= m1v1 + m2v2
= 3.0 kg(6.0 m/s)+4 kg(-8 m/s)
= 18. kg(m/s) + -32. kgm/s
= - 14. kg(m/s)
OR
14. kg(m/s) West
Ex2) After the collision, the 3.0 kg object will travel 10. m/s west.
b) What velocity will the 4.0 kg object have after the collision?
m1 = 3.0 kg
v1′ = -10. m/s
m2 = 4.0 kg
v2′ = ?
Total Momenta Before Interaction = Total Momenta After Interaction

2. Before
-14. kg(m/s) = m1v1′ + m2v2′
-14. kg(m/s) =
3.0 kg(-10. m/s)+4.0 kg(v2′)
-14. kg(m/s) =
(-30. kgm/s)+4.0 kg(v2′)
16. kg(m/s) = 4.0 kg(v2′)
v2′ = 4.0 m/s
OR
v2′ = 4.0 m/s East
Ex 3) A 10. kg Block A moves with a velocity of 2.0 m/s to the right and collides with a 10. kg
Block B which is at rest. After the collision Block A stops moving and Block B moves to the
right.
a)
b)
c)
d)
e)
f)
Find the total momentum after the collision
Total mom. before = Total mom. after:
= mAvA + mBvB
= 10. kg(2.0m/s)+10. kg(0 m/s)
= 20 kgm/s
FALLING BODIES
1. A basketball dropped from rest 1.00 m above the floor rebounds to a height of 0.67 m.
Assuming the ball is not moving horizontally, calculate its velocity …
a. before it hit the floor on the way down and
b. just after it left the floor on the way up.
If the ball is in contact with the floor for 0.10 s determine its acceleration …
c. on the way down,
d. while it is contact with the floor, and
e. on the way up.
solution
a. The first half of this problem is much like the the skydiving problem we just solved.
v0 = 0 m/s
v2 = v02 + 2aΔs
Δs = 1.00 m v = √(2aΔs)

3. 9.8 m/s2 v = √[2(9.8 m/s2)(1.00 m)]
??
v = 4.4 m/s down
a=
v=
b. The second half of this problem is also like the first only the velocity reduces to zero
instead of starting at zero.
v=
Δs =
a=
v0 =
0 m/s
0.67 m
9.8 m/s2
??
v2 =
v0 =
v0 =
v0 =
v02 + 2aΔs
√(2aΔs)
√[2(9.8 m/s2)(0.67 m)]
3.6 m/s up
b. There is another way to solve the second half of this question using the notion of
proportionality. 0.67 m is approximately 2/3 of 1.00 m. Velocity is proportional to the
square root of distance when acceleration is constant (and the initial velocity is zero).
Thus …
v ∝ √Δs ⇒
v2
⎞½
⎛ Δs2
=
⎝ Δs1
⎠
v1
v2
⎛ 0.67 m ⎞½
=
≈ √⅔
⎝ 1.00 m ⎠
4.4 m/s
v2 =
3.6 m/s up
c. This method is no easier, but it serves as a check on our first calculation. Since both
results are identical, we must have done it right.
d. This question is designed to see if you've been paying attention. The acceleration of a
falling body is 9.8 m/s2 down on the surface of the earth.
e. This part requires computation. Use the definition of acceleration. Let's say that down is
negative. Then …
a=
Δv
v − v0
(3.6 m/s) − (−4.4 m/s)
=
=
= 80 m/s2 up
Δt
Δt
0.01 s
f. There is very little work to do here. Just write the answer. The acceleration due to gravity
is still 9.8 m/s2 down even if the basketball is rising.
2. A diver jumps from a 3.0 m board with an initial upward velocity of 5.5 m/s. Determine …
a. the time the diver was in the air
b. the maximum height to which she ascended
c. her velocity on impact with the water
solution

4. a. This is a problem in which close attention to signs is a must. Let's assume that up is
positive.
v0 =
Δs =
a=
v=
+5.5 m/s
−3.0 m
−9.8 m/s2
??
Δs =
v0Δt + ½aΔt2
(−3.0 m) = (+5.5 m/s)Δt + ½(−9.8 m/s2)Δt2
0=
4.9Δt2 − 5.5Δt − 3
+5.5 ± √[(5.5)2 + 4(4.9)(3)]
Δt =
2(4.9)
Δt =
+1.5 s or −0.40 s
b. Quadratics have two solutions. What do they mean? Surely the negative solution is
nonsense. How could a diver dive and yet strike the water before she left the board? The
answer is 1.5 s.
c. At the point of maximum height, the diver's velocity would be zero. Thus …
v0 =
v=
a=
s0 =
Δs =
+5.5 m/s
0 m/s
−9.8 m/s2
+3.0 m
??
v2 =
Δs =
Δs =
Δs =
v02 + 2aΔs
v2 − v02
2a
(0 m/s)2 − (+5.5 m/s)2
2(−9.8 m/s2)
+1.5 m
3. The following passages are excerpts from "The Long, Lonely Leap" by Captain Joseph W.
Kittinger, Jr. USAF as they appeared in National Geographic magazine. It is the story of his
record-setting, high altitude parachute jump from a helium balloon over New Mexico on
16 August 1960.
An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At
ground control the radar altimeters also have stopped on readings of 102,800 feet, the figure that
we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float
altitude.
At zero count I step into space. No wind whistles or billows my clothing. I have absolutely no
sensation of the increasing speed with which I fall.
Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before
hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude. An Air Force
camera on the gondola took this photograph when the cotton clouds still lay 80,000 feet below.
At 21,000 feet they rushed up so chillingly that I had to remind myself they were vapor and not
solid.
Solution
For most skydivers, the acceleration experienced while falling is not constant. As a skydiver's
speed increases, so too does the aerodynamic drag until their speed levels out at a typical
terminal velocity of 55 m/s (120 mph). Air resistance is not negligible in such circumstances.
The story of Captain Kittinger is an exceptional one, however. At the float altitude where his
dive began, the earth's atmosphere has only 1.5% of its density at sea level. It is effectively a
vacuum and offers no resistance to a person falling from rest.

5. The acceleration due to gravity is often said to be constant, with a value of 9.8 m/s2. Over the
entire surface of the earth up to an altitude of 18 km, this is the value accurate to two significant
digits. In actuality, this "constant" varies from 9.81 m/s2 at sea level to 9.75 m/s2 at 18 km. At the
altitude of Captain Kittinger's dive, the acceleration due to gravity was closer to 9.72 m/s2.
Given this data it is possible to calculate the maximum speed of Captain Kittinger during his
descent. First we will need to convert the altitude measurements. To save calculation time we
will only convert the change in altitude and not each altitude. Given that he stepped out of the
gondola at 102,800 feet, fell freely until 96,000 feet, and then continued to accelerate for another
6,000 feet; the distance over which he accelerated uniformly was …
102,800 − 96,00 + 6,000 = 12,800 feet
12,800 feet 1609 m
= 3900 m
1
5280 feet
It's now just a matter of choosing the correct formula and plugging in the numbers.
v=
v0 =
a=
Δs =
??
0 m/s
9.72 m/s2
3900 m
v2 =
v=
v=
v=
v02 + 2aΔs
√(2aΔs)
√(2(9.72 m/s2)(3900 m))
275 m/s
This result is amazingly close to the value recorded in Kittinger's report.
614 mile 1609 m 1 hour
= 274 m/s
1 hour
1 mile 3600 s
As one would expect the actual value is slightly less than the theoretical value. This agrees with
the notion of a small but still non-zero amount of drag. This number is another world record —
the fastest speed attained by a human without the use of an engine.
GRAPHING MOTION
Motion graphs can tell you how far a body has travelled, how fast it is moving and all the speed
changes there have been.
EXAMPLE 1:
EXAMPLE 2:

6. EXAMPLE 3:
SECOND LAW OF MOTION
EXAMPLE 1.
Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas
station, and he makes the car go 0.05 m/s/s. Using Newton's Second Law, you can compute
how much force Mike is applying to the car.
Answer = 50 newtons
However, the Second Law gives us an exact relationship between force, mass, and
acceleration. It can be expressed as a mathematical equation:
or
FORCE = MASS times ACCELERATION
EXAMPLE 2:
A net force of 15 N is exerted on an encyclopedia to cause it to accelerate at a rate of 5 m/s2.
Determine the mass of the encyclopedia.
Use Fnet= m * a with Fnet = 15 N and a = 5 m/s/s.

7. So (15 N) = (m)*(5 m/s/s)
And m = 3.0 kg
EXAMPLE 3:
Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is
doubled, then what is the new acceleration of the sled?
Answer: 3 m/s/s
The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional)
and divided by 2 (since a and m are inversely proportional)
ENERGY
EXAMPLE 1
A second-order reaction was observed. The reaction rate constant at 3 °C was found to be 8.9 x
10-3 L/mol and 7.1 x 10-2 L/mol at 35 °C. What is the activation energy of this reaction?
SOLUTION:
Activation energy is the amount of energy required to initiate a chemical reaction. The activation
energy can be determined from reaction rate constants at different temperatures by the equation
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
where
Ea is the activation energy of the reaction in J/mol
R is the ideal gas constant = 8.3145 J/K·mol
T1 and T2 are absolute temperatures
k1 and k2 are the reaction rate constants at T1 and T2
Step 1 - Convert °C to K for temperatures
T = °C + 273.15
T1 = 3 + 273.15
T1 = 276.15 K
T2 = 35 + 273.15
T2 = 308.15 K
Step 2 - Find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(7.1 x 10-2/8.9 x 10-3) = Ea/8.3145 J/K·mol x (1/276.15 K - 1/308.15 K)
ln(7.98) = Ea/8.3145 J/K·mol x 3.76 x 10-4 K-1
2.077 = Ea(4.52 x 10-5 mol/J)
Ea = 4.59 x 104 J/mol
or in kJ/mol, (divide by 1000)
Ea = 45.9 kJ/mol

8. Answer:
The activation energy for this reaction is 4.59 x 104 J/mol or 45.9 kJ/mol.
WORK:
EXAMPLE 1:
A constant force of 10.0 N, 225 acts on an object as it moves 5.0 m, 120 . Determine the work done on
the object by the force.
ANSWER: −13 J
EXAMPLE 2:
An object moves on the x-axis beginning at x = 2 m and ending at x = – 3 m. During this interval
it is subjected to a force given by F(x) = (4 N/m3)x3. Determine the work done on the object.
ANSWER:
65 J
EXAMPLE 3:
A spring of length 50.0 cm is characterized by the constant k = 80.0 N/m. Determine the work
required to compress it from a length of 40.0 cm to 35.0 cm.
ANSWER: 0.500 J
KINETIC ENERGY:
Kinetic Energy:
Kinetic Energy is defined as the work needed to accelerate a body of a given mass from rest
to its current velocity.
Kinetic Energy Formula :
Kinetic Energy: Ek = ½ mv2
Mass: m = Ek / ½ v2
Velocity:
where,
Ek = Kinetic Energy,
M = Mass of object,
V = Velocity or Speed of object,
Kinetic Energy Example:
Case 1: Determine the Kinetic energy of a 500kg roller coaster train which moves at a speed of

9. 20 m/s.
M = 500 kg, V = 20 m/s
Step 1: Substitute the values in the below kinectic energy formula:
Kinetic Energy: Ek = ½ mv2
= ½ x 500 x 202
= 0.5 x 500 x 400
Kinetic Energy: Ek = 100000 Joules or 1 x 105 Joules
Case 2: What is the speed of a 55 kg woman running with a kinetic energy of 412.7 J?
Ek = 412.7, M = 55 kg
Step 1: Substitute the values in the below Velocity formula:
Example 3.
What is the kinetic energy of a 45 kg object moving at 13 m/sec?
1. First we identify the information we are given in the problem:
mass = 45 kg
velocity = 13 m/sec
2. Next, we place this information into the kinetic energy formula:
KE = 1/2 mv2
KE = 1/2 (45 kg)(13 m/sec)2
3. Solving the equation gives a kinetic energy value of 3802.5 J
4. Note: the unit for energy is the same as for work: the Joule (J)
POTENTIAL ENERGY
Potential energy, on the other hand, is energy of position, not of motion. The amount of potential
energy possessed by an object is proportional to how far it was displaced from its original
position. If the displacement occurs vertically, raising an object off of the ground let's say, we
term this Gravitational Potential Energy. We can calculate the gravitational potential energy of
an object with this formula:
GPE = weight x height

10. An increase in the weight of an object or the height to which it is raised will result in an increase
in the potential energy the object possesses. Once the object is dropped, the potential energy
begins to decrease due to reduced height, but we also now see an increase in kinetic energy
because the velocity is also increasing.
Sample Problem
Case 1: A 37 N object is lifted to a height of 3 meters. What is the potential energy of this
object?
Identify the information given to you in the problem:
weight = 37 N
height = 3 meters
Insert the information into the gravitational potential energy formula:
GPE = weight x height
GPE = 37 N x 3 meters
Solving the problem gives a potential energy value of 111 J.
Case2: A cat had climbed at the top of the tree. The Tree is 20 meters high and the cat weighs
6kg. How much potential energy does the cat have?
m = 6 kg, h = 20 m, g = 9.8 m/s2(Gravitational Acceleration of the earth)
Step 1: Substitute the values in the below potential energy formula:
Potential Energy: PE = m x g x h
= 6 x 9.8 x 20
Potential Energy: PE = 1176 Joules
This example will guide you to calculate the potential energy manually.
Case 3: On a 3m ledge, a rock is laying at the potential energy of 120 J. What will be the mass of
the rock.
PE = 120 J, h = 3m, g = 9.8 m/s2(Gravitational Acceleration of the earth)
Step 1: Substitute the values in the below Velocity formula:
POWER ENERGY:
1. Applied force vs. position graph of an object is given below. Find the kinetic energy gained by
the object at distance 12m.

11. By using work and energy theorem we say that; area under the graph gives us work done by the
force.
ΔEK=W=area under the graph=(8+4)/2.8-8(12-8)
ΔEK=12.4-8.4=16 joule
2. Box having mass 3kg thrown with an initial velocity 10 m/s on an inclined plane. If the box
passes from the point B with 4m/s velocity, find the work done by friction force.
We use conservation of energy theorem.
EA=EB+Wfriction
Wfriction=1/2.m.V2-(mgh+1/2mVL2)
Wfriction=1/2.3.102-(3.10.2+1/2.3.42)
Wfriction=66 joule
3. Three different forces are applied to a box in different intervals. Graph, given below, shows
kinetic energy gained by the box in three intervals. Find the relation between applied forces.
Slope of the EK vs. position graph gives applied force
I. interval: F1=(20-0)/(5-0)=4N
II. interval: F2=(30-20)/(10-5)=2N

12. III. interval: F3= (0-30)/(15-10)=-6N
FIII>FI>FII
Department of Education
Region V
Division of Camarines Sur
Sipocot North District
ANIB NATIONAL HIGH SCHOOL

13. Submitted by:
ARWIN VILLAMONTE IV
Submitted to:
MS. RONALYN BORRAS
Department of Education
Region V

14. Division of Camarines Sur
Sipocot North District
ANIB NATIONAL HIGH SCHOOL
Submitted by:
RICHARD ROYALES
Submitted to:
MS. RONALYN BORRAS