Horizontal and vertical motion are independent in projectile motion. The horizontal range is maximized when the launch angle is 45 degrees. Uniform circular motion describes objects moving in a circle at constant speed. Centripetal acceleration points toward the center and is required to maintain circular motion. Newton's laws relate forces to motion: the first law states an object in motion stays in motion unless acted on by a net external force, the second law defines acceleration as proportional to net force, and the third law states forces between objects act in equal magnitude but opposite directions.

1. Topic 03
Two Dimensional Motion-2
And
Force and Motion -1
Physics

2. Two Dimensional Motion-2

3. Topics
1. Projectile motion
2. Uniform Circular Motion

4. 1. Projectile motion
A particle moves in a vertical plane, with the only acceleration equal to the
free fall acceleration, g .
In projectile motion, the horizontal motion and the vertical motion are
independent of each other, that is, neither motion affects the other.
The initial velocity of the projectile is:
j
i yo
x v
v
v ˆ
ˆ
0 



5. Horizontal Range, assuming no external forces:
The horizontal range of a projectile is the horizontal distance when it
returns to its launching height.
The distance equations in the x- and y- directions respectively:
Eliminating t:

6. Horizontal Motion: no
acceleration
Vertical Motion;
acceleration = g
Eliminate time, t:
2
2
)
cos
(
2
)
0
0
0
tan


(
v
gx
x
y 


7. Example, projectile motion:
A rescue plane flies at 198 km/h and constant height h=500 m toward a point directly
over a victim, where a rescue capsule is to land. What should be the angle φ of the
pilot’s line of sight to the victim when the capsule release is made. (b) As the capsule
reaches the water, what is its velocity V in unit-vector notation and in magnitude-angle
notation.

8. The speed of
the particle is
constant
A particle
travels around a
circle/circular
arc
Uniform
circular
motion
+
2. Uniform Circular Motion

9. As the direction of the velocity of the particle changes, there is an
acceleration !!!
CENTRIPETAL (center-seeking) ACCELERATION
Here v is the speed of the particle and r is the radius of the circle.

10. The aceleration vector
always points toward the
center
The velocity vector is
always tangent to
the path.

11. Sample problem, top gun pilots

12. We assume the turn is made with uniform circular motion.
Then the pilot’s acceleration is centripetal and has
magnitude a given by a = v2/R.
Also, the time required to complete a full circle is the period given by T
=2πR/v.
Because we do not know radius R, let’s solve for R from the period
equation for R and substitute into the acceleration eqn.
Speed v here is the (constant) magnitude of the velocity during the
turning.

13. To find the period T of the motion, first note that the final velocity
is the reverse of the initial velocity. This means the aircraft leaves
on the opposite side of the circle from the initial point and must
have completed half a circle in the given 24.0 s. Thus a full circle
would have taken T 48.0 s.
Substituting these values into our equation for a, we find

14. Reference.
Halliday D.; Resniick R. and Walker J. (2010). Principles Of Physics , ninth
Edition , John Wiley & SONS Inc, New York, ISBN: 978-0-470-55653-5

15. Force and Motion -1

16. Topics
1. Newton’s Laws
2. Mass and weight
3. Applying Newton’s Laws

17.  Study of relation between force and acceleration
of a body: Newtonian Mechanics.
 Newtonian Mechanics does not hold good for all
situations.
Examples:
1.Relativistic or near-relativistic motion
2.Motion of atomic-scale particles

18. If the body is at rest, it stays at rest.
If it is moving, it continues to move with the same velocity (same
magnitude and same direction).
(1) Newton’s First Law:
If no force acts on a body, the body’s velocity cannot change; that
is, the body cannot accelerate.
1. Newton’s Law

19. a
m
Fnet



The net force on a body is equal to the product of the
body’s mass and its acceleration.
In component form,
z
net,z
y
y
net,
x
x
net, ma
F
;
ma
F
;
ma
F 


The acceleration component along a given axis is caused
only by the sum of the force components along that
same axis, and not by force components along any other
axis.
(2) Newton’s second law

20. reaction
action F
F




(3) Newton’s Third Law
When two bodies interact, the forces on the bodies from each
other are always equal in magnitude and opposite in
direction.

21. mg
F
or
g
m
F 



Gravitational Force:
A gravitational force on a body is
a certain type of pull that is
directed toward a second body.
Suppose a body of mass m is in
free fall with the free-fall
acceleration of magnitude g.
The force that the body feels
as a result is:
The weight, W, of a body is equal to
the magnitude Fg of the
gravitational force on the body. W
= mg (weight),
2. Mass and weight

22.  In a free-body diagram, the only body shown is the
one for which we are summing forces.
 Each force on the body is drawn as a vector arrow
with its tail on the body.
A coordinate system is usually included, and the
acceleration of the body is sometimes shown with a
vector arrow (labeled as an acceleration).
Newton’s second law; drawing a free-body diagram
The figure here shows two
horizontal forces acting on a
block on a frictionless floor.
3. Applying Newton’s Laws

23. System Force Mass Acceleration
SI Newton ( N ) kilogram (kg) m/s2
Cgs dyne gram (g) cm/s2
British pound (lb) slug ft/s2
Units in Newton’s Second Law
1 dyne = 1 g.m/s2
1 lb = 1 slug.ft/s2
1 N = 105 dyne = 0,2248 lb

24. Sample problem
2/3/2023
A block S (the sliding block) with mass M =3.3 kg.
The block is free to move along a horizontal
frictionless surface and connected, by a cord that
wraps over a frictionless pulley, to a second block H
(the hanging block), with mass m 2.1 kg. The cord
and pulley have negligible masses compared to the
blocks (they are “massless”).
The hanging block H falls as the sliding block S
accelerates to the right.
block S M
m
gH
F

gS
F

N
F

T

T

Find (a) the acceleration of block S, (b) the acceleration of block H, and (c) the
tension in the cord.

25. a
m
F



Key Ideas:
1. Forces, masses, and accelerations are involved, and
they should suggest Newton’s second law of motion:
2. The expression is a vector equation, so we
can write it as three component equations.
3. Identify the forces acting on each of the bodies and
draw free body diagrams
a
m
F




26. Thus, for the block we can write Newton’s second law for a positive-
upward y axis,
(Fnet, y= may), as:
for any vertical acceleration ay of the table and block

27. y y
Hanging
block H
m
a
Sliding
Block S
a
m
(a)
(b)
(a) A free-body diagram for
block S
(b) A free-body diagram for
block H
N
F

gH
F

gS
F

T

T

From the free body diagrams, write Newton’s Second Law
in the vector form, assuming a direction of acceleration for the whole
system. Identify the net forces for the sliding and the hanging blocks:
Fnet,x= max ; Fnet,y= may Fnet,z= maz

28. For the sliding block, S, which does not accelerate vertically.
Also, for S, in the x direction, there is only one force component,
which is T.
gS
N
gS
N
y
y
net, F
F
or
0
F
F
ma
F 




ma
T
ma
F x
x
net, 


For the hanging block, because the acceleration is along the y axis.
We eliminate the pulley from consideration by assuming its mass to
be negligible compared with the masses of the two blocks. With
some algebra,

29. Sample problem
A cord pulls on a box up along a
frictionless plane inclined at q=
300.The box has mass m =5.00
kg, and the force from the cord
has magnitude T =25.0 N. What
is the box’s acceleration
component a along the inclined
plane?

30. For convenience, we draw a coordinate system and a free-body diagram as
shown in Fig. b. The positive direction of the x axis is up the plane. Force
from the cord is up the plane and has magnitude T=25.0 N. The gravitational
force is downward and has magnitude mg =(5.00 kg)(9.8 m/s2) =49.0 N.
Also, the component along the plane is down the plane and has magnitude
mg sinθ as indicated in the following figure.
To indicate the direction, we can write
the down-the-plane component as -mg sin θ.
The positive result indicates that the box
accelerates up the plane.
which gives:
Using Newton’s Second Law, we have :

31. 4. Normal Force:
When a body presses
against a surface, the
surface deforms and pushes
on the body with a normal
force, FN, that is
perpendicular to the surface.
In the figure, forces Fg and
FN and are the only two
forces on the block and they
are both vertical.
(a)A block resting on a table experiences a normal
force perpendicular to the tabletop.
(b) The free-body diagram for the block.

32. Reference.
Halliday D.; Resniick R. and Walker J. (2010). Principles Of Physics ,
ninth Edition , John Wiley & SONS Inc, New York, ISBN: 978-0-
470-55653-5

33. Thank You