Kinematics 2-D | Physics Optional Notes for UPSC PDF Download

 Page 1


KINEMATICS 2-D       
1-D motion 
v
x
=
dx
dt
 
?? ?? =
?? ?? ?? ????
 
? 2D = 1D + 1D 
Example. Initial velocity = 2??ˆ + 3??ˆ; ?? ? = 4??ˆ + 2??ˆ. 
Find:(a) ?? ? at ?? = 2sec . 
(b) s ? from 0 to 2sec . 
Solution:  
?? ?? = 2 ?? ?? = 3
?? ?? = 4 ?? ?? = 2
?? = 2 ?? = 2
 
?? x
= u
y
+ a
x
t 
v
x
= 2 + 4 × 2 = 10 
s
x
= (
2 + 10
2
)2 = 12 
v ? ? = 10i ˆ + 7j ˆ |v ? ?| = v10
2
+ 7
2
= v 149 
s ? = 12i ˆ + 10j ˆ |s ?| = v12
2
+ 10
2
= v 244 
Example. ?? ? = 6?? ??ˆ + ( 8?? - 4?? 2
) ??ˆ 
Find:(a) v ? ? 
(c) Speed of projection 
Solution: (a) ?? 
x = 6t 
v
x
= 6 
a
x
= 0 
Page 2


KINEMATICS 2-D       
1-D motion 
v
x
=
dx
dt
 
?? ?? =
?? ?? ?? ????
 
? 2D = 1D + 1D 
Example. Initial velocity = 2??ˆ + 3??ˆ; ?? ? = 4??ˆ + 2??ˆ. 
Find:(a) ?? ? at ?? = 2sec . 
(b) s ? from 0 to 2sec . 
Solution:  
?? ?? = 2 ?? ?? = 3
?? ?? = 4 ?? ?? = 2
?? = 2 ?? = 2
 
?? x
= u
y
+ a
x
t 
v
x
= 2 + 4 × 2 = 10 
s
x
= (
2 + 10
2
)2 = 12 
v ? ? = 10i ˆ + 7j ˆ |v ? ?| = v10
2
+ 7
2
= v 149 
s ? = 12i ˆ + 10j ˆ |s ?| = v12
2
+ 10
2
= v 244 
Example. ?? ? = 6?? ??ˆ + ( 8?? - 4?? 2
) ??ˆ 
Find:(a) v ? ? 
(c) Speed of projection 
Solution: (a) ?? 
x = 6t 
v
x
= 6 
a
x
= 0 
v ? ? = 6i ˆ + ( 8 - 8t) j ˆ 
?? ? = -8??ˆ 
When projected, ?? = 0 
PROJECTILE MOTION GENERAL IDEA 
In simpler terms, let's talk about a special kind of motion called projectile motion. 
Imagine a particle moving in a flat vertical plane with an initial speed ??0v0, but its 
acceleration always points downward due to gravity, which we represent as ??g. This 
type of motion is what we call projectile motion. 
Assumptions:- 
Particles remain close to the earth's surface, so acceleration due to gravity remains 
constant. 
Air resistance is neglected. 
The distance that projectile travels is small so that the earth can be treated as a plane 
surface. 
Two straight-line motions:- 
Our goal here is to analyse projectile motion using the tools for two-dimensional motion. 
Th i s feature allows us to break up a problem involving two-dimensional motion into 
two separate and easier one-dimensional problems, 
(a) The horizontal motion is motion with uniform velocity (no effect of gravity) 
(b) The vertical motion is motion of uniform acceleration or freely falling bodies. 
Note: In projectile motion, the horizontal motion and the vertical motion are 
independent of each other, that is either motion does not affect the other. 
 
Treating as two straight line motions:- 
The horizontal Motion(x axis): 
Page 3


KINEMATICS 2-D       
1-D motion 
v
x
=
dx
dt
 
?? ?? =
?? ?? ?? ????
 
? 2D = 1D + 1D 
Example. Initial velocity = 2??ˆ + 3??ˆ; ?? ? = 4??ˆ + 2??ˆ. 
Find:(a) ?? ? at ?? = 2sec . 
(b) s ? from 0 to 2sec . 
Solution:  
?? ?? = 2 ?? ?? = 3
?? ?? = 4 ?? ?? = 2
?? = 2 ?? = 2
 
?? x
= u
y
+ a
x
t 
v
x
= 2 + 4 × 2 = 10 
s
x
= (
2 + 10
2
)2 = 12 
v ? ? = 10i ˆ + 7j ˆ |v ? ?| = v10
2
+ 7
2
= v 149 
s ? = 12i ˆ + 10j ˆ |s ?| = v12
2
+ 10
2
= v 244 
Example. ?? ? = 6?? ??ˆ + ( 8?? - 4?? 2
) ??ˆ 
Find:(a) v ? ? 
(c) Speed of projection 
Solution: (a) ?? 
x = 6t 
v
x
= 6 
a
x
= 0 
v ? ? = 6i ˆ + ( 8 - 8t) j ˆ 
?? ? = -8??ˆ 
When projected, ?? = 0 
PROJECTILE MOTION GENERAL IDEA 
In simpler terms, let's talk about a special kind of motion called projectile motion. 
Imagine a particle moving in a flat vertical plane with an initial speed ??0v0, but its 
acceleration always points downward due to gravity, which we represent as ??g. This 
type of motion is what we call projectile motion. 
Assumptions:- 
Particles remain close to the earth's surface, so acceleration due to gravity remains 
constant. 
Air resistance is neglected. 
The distance that projectile travels is small so that the earth can be treated as a plane 
surface. 
Two straight-line motions:- 
Our goal here is to analyse projectile motion using the tools for two-dimensional motion. 
Th i s feature allows us to break up a problem involving two-dimensional motion into 
two separate and easier one-dimensional problems, 
(a) The horizontal motion is motion with uniform velocity (no effect of gravity) 
(b) The vertical motion is motion of uniform acceleration or freely falling bodies. 
Note: In projectile motion, the horizontal motion and the vertical motion are 
independent of each other, that is either motion does not affect the other. 
 
Treating as two straight line motions:- 
The horizontal Motion(x axis): 
Because there is no acceleration in the horizontal direction, the horizontal component v
x
 
of the projectile's velocity remains unchanged from its initial value v
( 0x
 throughout the 
motion, 
The vertical motion (y axis): 
The vertical motion is the motion we discussed for a particle in free fall. 
In simple terms, let's look at what happens during projectile motion. In the illustration 
and equation (1.3), the vertical part acts similarly to when you throw a ball straight up. 
Initially, it goes upward with its speed decreasing until it stops, reaching the highest 
point of its path. After that, it starts moving downward, and its speed increases as time 
goes on. 
?? -axis  
Initial velocity ( ?? ?? ) = ?? cos ?? 
acceleration ( a
x
)= 0 
Thus, velocity after time ?? 
v
x
= ucos ?? 
Displacement after time ?? 
x = ucos ?? t 
y-axis 
Initial velocity ( u
y
) = usin ?? 
acceleration ( a
y
) = -g 
Thus, velocity after time ?? 
?? ?? = ?? sin ?? - ???? 
Displacement after time t 
?? = ?? sin ???? - ?? ?? 2
/2 
Resultant velocity 
( ?? ? ?
?? ) = ( ?? cos ?? ) ??ˆ + ( ?? sin ?? - ???? ) ??ˆ
|?? ? ?
?? | = v?? 2
cos
2
 ?? + ( ?? sin ?? - ???? )
2
 
Page 4


KINEMATICS 2-D       
1-D motion 
v
x
=
dx
dt
 
?? ?? =
?? ?? ?? ????
 
? 2D = 1D + 1D 
Example. Initial velocity = 2??ˆ + 3??ˆ; ?? ? = 4??ˆ + 2??ˆ. 
Find:(a) ?? ? at ?? = 2sec . 
(b) s ? from 0 to 2sec . 
Solution:  
?? ?? = 2 ?? ?? = 3
?? ?? = 4 ?? ?? = 2
?? = 2 ?? = 2
 
?? x
= u
y
+ a
x
t 
v
x
= 2 + 4 × 2 = 10 
s
x
= (
2 + 10
2
)2 = 12 
v ? ? = 10i ˆ + 7j ˆ |v ? ?| = v10
2
+ 7
2
= v 149 
s ? = 12i ˆ + 10j ˆ |s ?| = v12
2
+ 10
2
= v 244 
Example. ?? ? = 6?? ??ˆ + ( 8?? - 4?? 2
) ??ˆ 
Find:(a) v ? ? 
(c) Speed of projection 
Solution: (a) ?? 
x = 6t 
v
x
= 6 
a
x
= 0 
v ? ? = 6i ˆ + ( 8 - 8t) j ˆ 
?? ? = -8??ˆ 
When projected, ?? = 0 
PROJECTILE MOTION GENERAL IDEA 
In simpler terms, let's talk about a special kind of motion called projectile motion. 
Imagine a particle moving in a flat vertical plane with an initial speed ??0v0, but its 
acceleration always points downward due to gravity, which we represent as ??g. This 
type of motion is what we call projectile motion. 
Assumptions:- 
Particles remain close to the earth's surface, so acceleration due to gravity remains 
constant. 
Air resistance is neglected. 
The distance that projectile travels is small so that the earth can be treated as a plane 
surface. 
Two straight-line motions:- 
Our goal here is to analyse projectile motion using the tools for two-dimensional motion. 
Th i s feature allows us to break up a problem involving two-dimensional motion into 
two separate and easier one-dimensional problems, 
(a) The horizontal motion is motion with uniform velocity (no effect of gravity) 
(b) The vertical motion is motion of uniform acceleration or freely falling bodies. 
Note: In projectile motion, the horizontal motion and the vertical motion are 
independent of each other, that is either motion does not affect the other. 
 
Treating as two straight line motions:- 
The horizontal Motion(x axis): 
Because there is no acceleration in the horizontal direction, the horizontal component v
x
 
of the projectile's velocity remains unchanged from its initial value v
( 0x
 throughout the 
motion, 
The vertical motion (y axis): 
The vertical motion is the motion we discussed for a particle in free fall. 
In simple terms, let's look at what happens during projectile motion. In the illustration 
and equation (1.3), the vertical part acts similarly to when you throw a ball straight up. 
Initially, it goes upward with its speed decreasing until it stops, reaching the highest 
point of its path. After that, it starts moving downward, and its speed increases as time 
goes on. 
?? -axis  
Initial velocity ( ?? ?? ) = ?? cos ?? 
acceleration ( a
x
)= 0 
Thus, velocity after time ?? 
v
x
= ucos ?? 
Displacement after time ?? 
x = ucos ?? t 
y-axis 
Initial velocity ( u
y
) = usin ?? 
acceleration ( a
y
) = -g 
Thus, velocity after time ?? 
?? ?? = ?? sin ?? - ???? 
Displacement after time t 
?? = ?? sin ???? - ?? ?? 2
/2 
Resultant velocity 
( ?? ? ?
?? ) = ( ?? cos ?? ) ??ˆ + ( ?? sin ?? - ???? ) ??ˆ
|?? ? ?
?? | = v?? 2
cos
2
 ?? + ( ?? sin ?? - ???? )
2
 
& tan ?? =
usin ?? - gt
ucos ?? 
where ?? is angle that velocity vector makes with horizontal. Also known as direction or 
angle of motion 
Vectorial treatment:-(Optional) 
Lets say a particle is projected at an angle ?? from horizontal with a velocity. Now if we 
take the point of projection as the origin and take vertically upward as the positive ?? -axis 
and the direction of projection horizontally as the x-axis. 
u ?? = ucos ?? i ˆ + usin ?? j ˆ
a ? ? = -g ˆ
 
Now since acceleration is uniform 
 
Velocity after time t: -V
? ? ?
= u ?? + a?? ? ? ??
? V
? ? ?
= ucos ?? i ˆ + usin ?? j ˆ + ( -g) t 
? ?? ? ?
= ?? cos ?? ??ˆ + ( ?? sin ?? - ???? ) ??ˆ 
Displacement after time ?? : - ?? ? = ??? ??? +
1
2
?? ?? 2
? ?? ?
= ( ?? cos ?? ??ˆ) ?? + ( ?? sin ?? ??ˆ) ?? +
1
2
( -?? ??ˆ) ?? 2
 
? ?? ?
= ( ?? cos ?? ??ˆ) ?? + (?? sin ???? -
1
2
?? ?? 2
)??ˆ 
LEVEL GROUND PROJECTION 
Lets say we project a particle with velocity ?? at an angle ?? from horizontal 
Time of flight ( ?? ) : 
T =
2usin ?? g
 
Considering vertical motion 
s
y
= 0; u
y
= vsin ?? ; a
y
= -g 
0 = usin ?? T - gT
2
/2 ? T =
2usin ?? g
 
Page 5


KINEMATICS 2-D       
1-D motion 
v
x
=
dx
dt
 
?? ?? =
?? ?? ?? ????
 
? 2D = 1D + 1D 
Example. Initial velocity = 2??ˆ + 3??ˆ; ?? ? = 4??ˆ + 2??ˆ. 
Find:(a) ?? ? at ?? = 2sec . 
(b) s ? from 0 to 2sec . 
Solution:  
?? ?? = 2 ?? ?? = 3
?? ?? = 4 ?? ?? = 2
?? = 2 ?? = 2
 
?? x
= u
y
+ a
x
t 
v
x
= 2 + 4 × 2 = 10 
s
x
= (
2 + 10
2
)2 = 12 
v ? ? = 10i ˆ + 7j ˆ |v ? ?| = v10
2
+ 7
2
= v 149 
s ? = 12i ˆ + 10j ˆ |s ?| = v12
2
+ 10
2
= v 244 
Example. ?? ? = 6?? ??ˆ + ( 8?? - 4?? 2
) ??ˆ 
Find:(a) v ? ? 
(c) Speed of projection 
Solution: (a) ?? 
x = 6t 
v
x
= 6 
a
x
= 0 
v ? ? = 6i ˆ + ( 8 - 8t) j ˆ 
?? ? = -8??ˆ 
When projected, ?? = 0 
PROJECTILE MOTION GENERAL IDEA 
In simpler terms, let's talk about a special kind of motion called projectile motion. 
Imagine a particle moving in a flat vertical plane with an initial speed ??0v0, but its 
acceleration always points downward due to gravity, which we represent as ??g. This 
type of motion is what we call projectile motion. 
Assumptions:- 
Particles remain close to the earth's surface, so acceleration due to gravity remains 
constant. 
Air resistance is neglected. 
The distance that projectile travels is small so that the earth can be treated as a plane 
surface. 
Two straight-line motions:- 
Our goal here is to analyse projectile motion using the tools for two-dimensional motion. 
Th i s feature allows us to break up a problem involving two-dimensional motion into 
two separate and easier one-dimensional problems, 
(a) The horizontal motion is motion with uniform velocity (no effect of gravity) 
(b) The vertical motion is motion of uniform acceleration or freely falling bodies. 
Note: In projectile motion, the horizontal motion and the vertical motion are 
independent of each other, that is either motion does not affect the other. 
 
Treating as two straight line motions:- 
The horizontal Motion(x axis): 
Because there is no acceleration in the horizontal direction, the horizontal component v
x
 
of the projectile's velocity remains unchanged from its initial value v
( 0x
 throughout the 
motion, 
The vertical motion (y axis): 
The vertical motion is the motion we discussed for a particle in free fall. 
In simple terms, let's look at what happens during projectile motion. In the illustration 
and equation (1.3), the vertical part acts similarly to when you throw a ball straight up. 
Initially, it goes upward with its speed decreasing until it stops, reaching the highest 
point of its path. After that, it starts moving downward, and its speed increases as time 
goes on. 
?? -axis  
Initial velocity ( ?? ?? ) = ?? cos ?? 
acceleration ( a
x
)= 0 
Thus, velocity after time ?? 
v
x
= ucos ?? 
Displacement after time ?? 
x = ucos ?? t 
y-axis 
Initial velocity ( u
y
) = usin ?? 
acceleration ( a
y
) = -g 
Thus, velocity after time ?? 
?? ?? = ?? sin ?? - ???? 
Displacement after time t 
?? = ?? sin ???? - ?? ?? 2
/2 
Resultant velocity 
( ?? ? ?
?? ) = ( ?? cos ?? ) ??ˆ + ( ?? sin ?? - ???? ) ??ˆ
|?? ? ?
?? | = v?? 2
cos
2
 ?? + ( ?? sin ?? - ???? )
2
 
& tan ?? =
usin ?? - gt
ucos ?? 
where ?? is angle that velocity vector makes with horizontal. Also known as direction or 
angle of motion 
Vectorial treatment:-(Optional) 
Lets say a particle is projected at an angle ?? from horizontal with a velocity. Now if we 
take the point of projection as the origin and take vertically upward as the positive ?? -axis 
and the direction of projection horizontally as the x-axis. 
u ?? = ucos ?? i ˆ + usin ?? j ˆ
a ? ? = -g ˆ
 
Now since acceleration is uniform 
 
Velocity after time t: -V
? ? ?
= u ?? + a?? ? ? ??
? V
? ? ?
= ucos ?? i ˆ + usin ?? j ˆ + ( -g) t 
? ?? ? ?
= ?? cos ?? ??ˆ + ( ?? sin ?? - ???? ) ??ˆ 
Displacement after time ?? : - ?? ? = ??? ??? +
1
2
?? ?? 2
? ?? ?
= ( ?? cos ?? ??ˆ) ?? + ( ?? sin ?? ??ˆ) ?? +
1
2
( -?? ??ˆ) ?? 2
 
? ?? ?
= ( ?? cos ?? ??ˆ) ?? + (?? sin ???? -
1
2
?? ?? 2
)??ˆ 
LEVEL GROUND PROJECTION 
Lets say we project a particle with velocity ?? at an angle ?? from horizontal 
Time of flight ( ?? ) : 
T =
2usin ?? g
 
Considering vertical motion 
s
y
= 0; u
y
= vsin ?? ; a
y
= -g 
0 = usin ?? T - gT
2
/2 ? T =
2usin ?? g
 
Maximum Height(H) : 
H =
u
2
sin
2
 ?? 2 g
 
Vertical velocity at maximum height v
y
= 0 
0 = ?? 2
sin
2
 ?? - 2???? ? ?? =
?? 2
sin
2
 ?? 2?? 
Horizontal Range ( ?? ) : 
?? =
?? 2
sin 2?? 2?? =
2?? ?? ?? ?? ?? 
Total time T =
2usin ?? g
 
Velocity in horizontal direction u
x
= ucos ?? 
Total displacement in horizontal direction R = ucos ?? T 
?? =
?? 2
sin 2?? 2?? 
Example. A body is thrown with initial velocity 10 m /sec . at an angle 37
°
 from 
horizontal. 
Find 
(i) Time of flight 
(ii) Maximum height. 
(iii) Range 
(iv) Position vector after ?? = 1sec . 
 Ans. (i) 1.2sec (ii) 1.8 m (iii) 8.6 m (iv) ( 16i ˆ - 8j ˆ) 
Example. 50 m /s /37
°
 
g = 10 m /s
2
. Find T and H and R ? 
Solution: T =
2usin ?? g
=
2×50×sin 37
°
10
= 10 ×
3
5
= 6sec . 
?? =
( ?? sin ?? )
2
?? =
( 50 × sin 37
°
)
2
2 × 10
=
(50 ×
3
5
)
2
2 × 10
=
900
2 × 10
= 45 m