Dynamics, Projectile, impulse, impact

I B.Sc., PHYSICS
AVVM Sri Pushpam College, Poondi, Thanjavur (Dt)
1
DYNAMICS
PPRROOJJEECCTTIILLEE
“A projectile is any object that is given an initial velocity and then follows a path
determined entirely by gravity” (Or)
“Any object moving through air without its own source of power and only under the
influence of gravity” eg. - golf ball after being hit
Angle of Projection
The angle between the direction of projection and the horizontal plane through the
point of projection is called angle of projection.
The path described by a projectile is called its trajectory
RANGE
The distance from the point of projection to the point, where the particle reaches the
horizontal plane is called the range on the horizontal plane
Time of flight
The interval of time from the instant of projection to the instant the particle reaches
the horizontal plane is called the time of flight
Unit –I Dynamics
Projectile-range on horizontal and inclined plane – Impulse and impact – Impulsive
force – Laws of impact – Impact of a smooth sphere on a smooth horizontal plane – Direct and
oblique impacts – Loss in kinetic energy – motion of two interacting bodies – reduced mass.

I B.Sc., PHYSICS
2
RRAANNGGEE OONN HHOORRIIZZOONNTTAALL AANNDD IINNCCLLIINNEEDD PPLLAANNEE
OObbttaaiinn aann eexxpprreessssiioonn ffoorr mmaaxxiimmuumm rraannggee ooff aa pprroojjeeccttiillee oonn aann iinncclliinneedd ppllaannee
Let a particle be projected from a point A with velocity u in a direction making an
angle ”” with a horizontal plane through A. it is required to find the range AB on a plane
inclined at an angle  with the horizontal.
The direction of projection lies in a vertical plane through AB. Let BC be the
perpendicular from B to the horizontal plane AC.
The initial velocity of projection u can be resolved into a component )(cosu  
along the plane and a component )-(sinu  perpendicular to the plane.
The acceleration due to gravity g, which acts vertically down can be resolved into a
component sing- in the plane and cosg- perpendicular to the plane. Let T be the time
which the particle takes to go from A to B.
Then in this time, the distance traversed by the projectile perpendicular to the plane is zero.
Hence,
2
.cos
2
1
).sin(0 TgTu   





 2
2
1
atuts
2
.cos).sin(2 TgTu  


cos
)sin(2
g
u
T

 )1(
During this time ‘T’ the horizontal velocity of the projectile ( cosu ) remains constant.
Hence, the horizontal distance described is given by,
u
CA
B


(Velocity=distance/time)
(Distance =velocity X time)

I B.Sc., PHYSICS
3
TuAC .cos 



cos
)sin(2
.cos
g
u
uAC




cos
cos)sin(2 2
g
u
AC

 ---- (2)
The range on the inclined plane


 2
2
cos
cos)sin(2
cos g
uAC
AB




2
2
cos
cos)sin(2
g
u
AB

 ---------- (3)
MMaaxxiimmuumm rraannggee oonn tthhee iinncclliinneedd ppllaannee
To find the direction of projection for the maximum range on the inclined
plane,


2
2
cos
cos)sin(2
g
u
R


 

sin)2sin(
cos2
2

g
u
R
For given values of u and , R is maximum,
When 1,=)-(2sin 
That is 90=-2 
Or 





 
2
1
45=
Therefore, the maximum range on the inclined plane,
hyads /cos 

I B.Sc., PHYSICS
4
)sin1(
cos 2
2



g
u
Rm
)sin1(
)sin1( 2
2





g
u
Rm
)sin1)(sin1(
)sin1(2





g
u
Rm
)sin1(
2


g
u
Rm
IIMMPPUULLSSEE OOFF AA FFOORRCCEE
If a constant force acts on a body for a given interval of time, the product of the force
and the time during which it acts measures the impulse of the force.
S.I unit of impulse is Newton-seconds (Ns)
Another unit for impulse is Kgms-1
Dimensional formula is MLT-1
For example, when a collision takes place between two billiard balls (or) hitting a
cricket ball with a bat, the force acting on body is for very short interval of time. During the
period of collision, the force acting on a body initially increases to a maximum and then
gradually decreases.
The curve in the figure is called a force-time curve and its area is equal to the impulse.
That is,
F
t

I B.Sc., PHYSICS
5
tFI  ----- (1)
By the second law of motion,
maF  ------ (2)
where,
m- mass of the body and
a- the acceleration produced
Therefore, tFI 
tmaI  ----- (3)
If u be the initial velocity of the body and v the velocity after time t,
t
uv
a

 ------ (4)
t
t
uv
mI 




 

)( uvmI  ------ (5)
The impulse of force acting on a body for an interval of time is measured by
the change of momentum it produces.
IImmppuullssiivvee ffoorrccee
An impulsive force is infinitely great force acting for a very short
interval of time, such that their product is finite.
Is a vector quantities
Problem A force 100 Newton acts on a body 0.3 seconds, calculate the impulse
of the force.
Impulse = Force x time
= 100 x 0.3
= 30 NS [Newton- seconds]

I B.Sc., PHYSICS
6
IIMMPPAACCTT
A ball is dropped from a height ‘h’. It reaches the ground with a velocity
ghu 2 and makes an impact.
These are classified into three types
(i) Perfectly elastic impact
(ii) Inelastic impact
(iii) Imperfectly elastic impact.
(i) Perfectly elastic impact
If the ball is elastic, it rebounds and moves vertically upwards with the
velocity v . If v=u it is said to be perfectly elastic.
(ii) Inelastic impact
If the ball does not rebound at all (i.e.,) v=0, it is said to be inelastic impact.
(iii) Imperfectly elastic impact
If v<u, this case is said to be imperfectly elastic impact.
LLAAWWSS OOFF IIMMPPAACCTT
When two bodies impacts directly, the only force acting at their point of
contact is directed along the normal at the point of contact. The forces between the
two bodies by Newton’s third law of motion are equal in magnitude but in opposite
direction.
Consequently the gain of momentum along the common normal for one
smooth body must be equal to the loss of momentum for the other in the same
direction. Hence the total momentum of the two bodies before and after the impact are
the same.
(i) The law of conservation of momentum hold good, momentum being
measured along the line of impact, the total momentum before impact
is equal to the total momentum after impact.

I B.Sc., PHYSICS
7
(ii) The relative velocity of the two bodies along the line of impact, before
and after is a constant ratio. This constant ratio depends only on the
material of the bodies and not on their masses (or) velocities. It is
called the co-efficient of restitution and denoted by a letter ‘e’. If ‘u1’,
‘u2’ be the velocities of two bodies before impact and ‘v1’, ‘v2’ the
velocities after impact.
e
uu
vv



21
21
Or )u--e(u=v-v 2121
Where, )u-(u 21 - The relative velocities before impact
)v-(v 21 - The relative velocities after impact
(iii) There is no tangential action between the two bodies at the point of
contact. That is due to the impact; there is no change in the velocity of
each body in a direction perpendicular to the common normal at their
point of contact.
CCoo--eeffffiicciieenntt ooff rreessttiittuuttiioonn
The ratio with a negative sign, of the relative velocity of two bodies after impact to
their relative velocity before impact is called the co-efficient of restitution.
e
uu
vv



21
21
DDiirreecctt iimmppaacctt bbeettwweeeenn ttwwoo ssmmooootthh sspphheerreess
Let a smooth sphere ‘A’ of mass m1 moving with velocity u1, impact directly
on another smooth sphere ‘B’ of mass m2 moving with velocity u2 in the same
direction. Let ‘e’ be the coefficient of restitution between them. Since the impact is
direct, there is no force along the common tangent between the two spheres at the
point of contact.
A B
m1
1
v1
m2
u1 u2
v2

I B.Sc., PHYSICS
8
Hence the velocities of two spheres v1 , v2 after the impact will be along the
common normal at the point of contact.
By the principle of conservation of momentum,
Total momentum after impact = total momentum before impact
i.e., 22112211 umumvmvm  ------ (1)
By Newton’s law, the relative velocity between the spheres along the common
normal after impact is equal to e times the relative velocity between them, along the
same direction but opposite sign.
)( 2121 uuevv  ---------- (2)
Multiplying equ (2) by m2,
)()( 212212 uuemvvm 
22122212 uemuemvmvm  ------ (3)
Adding equ (3) to equ (1)
2212221122122211 uemuemumumvmvmvmvm 
)()1()( 21122211 emmueummmv 
)(
)()1(
21
21122
1
mm
emmueum
V


 ------------- (4)
Multiplying equ (2) by m1,
)()( 211211 uuemvvm 
21112111 uemuemvmvm  ---------- (5)
Subtract equ (5) from equ(1) i.e., (1)-(5);

I B.Sc., PHYSICS
9
2111221121112211 uemuemumumvmvmvmvm 
)()1()( 12211122 emmueummmv 
)(
)()1(
12
12211
2
mm
emmueum
v


 ----------- (6)
An equation (4) and (6) gives the velocities of two spheres after impact along the
common normal.
Case 1:
If m1=m2 & e=1
Substitute in equ (4) and (6)
21 u=v
12 u=v
Case 2:
If the two spheres are inelastic e = 0, and therefore,
21 v=v
Case 3:
Impulse at the impact
The impulse I at the action of B on A is measured by the change of momentum
of A.
)( 111 vumI 
1111 vmumI 









)(
)()1(
21
21122
111
mm
emmueum
mumI









)(
)
21
12112222
111
mm
uemumeumum
mumI

I B.Sc., PHYSICS
10









)(
)(
21
2122211
111
mm
uuemumum
mumI
)(
)(
21
21212211
2
11211
2
1
mm
uumemummumummum
I



)(
)()(
21
21212121
mm
uumemuumm
I



The impulse of the action of A on B is equal and opposite to this.
2222 vmumI 









)(
)()1(
21
12211
222
mm
emmueum
mumI









)(
)
21
21221111
222
mm
eumumeumum
mumI









)(
)(
21
2112211
222
mm
uuemumum
mum
)(
)(
21
21212
2
21212
2
2221
mm
uumemumummumumm



)(
)()(
21
21212121
mm
uumemuumm














)(
)()(
21
21212121
mm
uumemuumm
II 

I B.Sc., PHYSICS
11
Loss of kinetic energy due to direct impact between two smooth spheres
When two bodies make an impact, loss of kinetic energy takes place that can
be calculated as follows,
K.E before impact =
2
22
2
11
2
1
2
1
umum  ------ (1)
K.E after impact =
2
22
2
11
2
1
2
1
vmvm  ------- (2)
Then,
Total loss of K.E = K.E before impact - K.E after impact
i.e., Equ (1)- Equ (2),






 2
22
2
11
2
22
2
11
2
1
2
1
2
1
2
1
vmvmumum
2
22
2
11
2
22
2
11
2
1
2
1
2
1
2
1
vmvmumum 
   2
2
2
22
2
1
2
11
2
1
2
1
vumvum 
     2222211111
2
1
2
1
vuvumvuvum  ----- (3)
But, )( 111 vumI 
i.e., Change in K.E of m1= )(
2
1
11 vuI  and
Change in K.E of m2=
)(
2
1
22 vuI 
Therefore, equ (3) becomes,
   2211
2
1
2
1
vuIvuI 

I B.Sc., PHYSICS
12
  2211
2
1
vuvuI 
  2121 ()
2
1
vvuuI 
By Newton’s experimental law
  2121 ()
2
1
uueuuI 
)( 2121 uuevv 
Or   euuI  1)(
2
1
21
But, )1)(( 21
21
21
euu
mm
mm
I 


Loss of K.E  euueuu
mm
mm


 1)()1)((
2
1
2121
21
21
Loss of K.E )1()(
2
1 22
21
21
21
euu
mm
mm



Case 1:
If the spheres are perfectly elastic e=1, the loss in K.E is zero.
Case 2:
If the spheres are inelastic e=0;
Loss in K.E =
2
21
21
21
)(
2
1
uu
mm
mm


Note:
In general e<1, so that (1-e2
) is positive. (u1-u2)2
is always positive. Hence
there is always a loss of K.E due to impact. The K.E lost during impact is converted
into sound, heat etc.,

I B.Sc., PHYSICS
13
Oblique impact between two smooth spheres
Let a smooth sphere of mass m1 moving with velocity u1 impacts obliquely on
a smooth sphere of mass m2 moving with velocity u2.
Let the directions of motion of the spheres before impact make angles  and 
with the common normal at their point of contact, and the velocities of the spheres be
v1 and v2 making angles  and  with the common normal after impact.
By the principle of conservation of momentum,
The total momentum after the impact = The total momentum before the impact
 coscoscoscos 22112211 umumvmvm  ---- (1)
By Newton’s experimental law,
)( 2121 uuevv 
But, in oblique impact
)coscos(coscos 2121  uuevv  ---- (2)
Multiplying equ(2) with m2,
)coscos()coscos( 212212  uuemvvm 

I B.Sc., PHYSICS
14
 coscoscoscos 22122212 uemuemvmvm  ---- (3)
Adding equ (1) to equ (3)
  coscoscoscos 22122211 vmvmvmvm
 coscoscoscos 22122211 uemuemumum  ------ (4)
)1(cos)(cos)(cos 22211211 eumemmummv  
)(
)1(cos)(cos
cos
21
22211
1
mm
eumemmu
v




 ------------- (5)
Multiplying equ (2) with m1, we get,
)coscos()coscos( 211211  uuemvvm 
 coscoscoscos 21112111 uemuemvmvm  ---------- (6)
Subtract equ (6) from equ (1)
 coscoscoscos 21112211 vmvmvmvm 
 coscoscoscos 21112211 uemuemumum 
)(cos)1(cos)(cos 12211212 emmueummmv  
)(
)(cos)1(cos
cos
21
12211
2
mm
emmueum
v




 ------------- (7)
Since, there is no tangential action, so there is no change in the velocity of the sphere
perpendicular to the common normal,
Therefore,
 sinsin 11 uv  ---- (8)
 sinsin 22 uv  ----- (9)
By squaring and adding equation (5) & (8),


 22
1
2
21
2221122
1
22
1
sin
)(
)1(cos)(cos
sincos u
mm
eumemmu
vv 










 22
1
2
21
22211222
1
sin
)(
)1(cos)(cos
)sin(cos u
mm
eumemmu
v 









I B.Sc., PHYSICS
15
2
21
2221122
11
)(
)1(cos)(cos
sin 








mm
eumemmu
uv

 --------- (10)
Similarly
2
21
1221122
22
)(
)(cos)1(cos
sin 








mm
emmueum
uv

 ---------- (11)
Case (i)
If e=1; m1=m2
From equ (5) & (7)
 coscos 21 uv  and ---- (12)
 coscos 12 uv  ---- (13)
Thus, we conclude from the equ (12) & (13) two spheres interchange their velocities
after impact.
------------------------------
Loss of K.E due to oblique impact between two spheres
The velocities of the spheres perpendicular to the common normal remain
unaltered due to the oblique impact between the two spheres so there will be no loss
in K.E perpendicular to the common normal. The only change in K.E will be along
the common normal.
Loss of K.E due to direct impact,
=
2
21
2
21
21
))(1(
)(2
1
uue
mm
mm


--------- (1)
Here, the case is oblique impact, so let take,
cos11 uu 
cos22 uu 
Therefore equ (1) becomes,

I B.Sc., PHYSICS
16
Loss K.E in oblique impact=
2
21
2
21
21
)coscos)(1(
)(2
1
 uue
mm
mm


---- (2)
In perfectly elastic body (i.e., e=1) the loss of kinetic energy in oblique impact zero.
Impact of a smooth sphere on a smooth fixed horizontal plane
Let a smooth sphere of mass ‘m’ and whose co-efficient of restitution is ‘e’,
impacts obliquely on a smooth fixed horizontal plane ‘XY’. Let ‘A’ be the point of
contact and ‘AO’ the common normal at the point of contact.
Let ‘u’ be the velocity of the sphere before impact in a direction making an
angle ‘’ with the common normal. Let the velocity of the sphere after the impact be
‘v’ inclined at an angle ‘’ with the common normal.
By newton’s experimental law,
)( 2121 uuevv 
)cos(cos  uev 
 coscos euv  --------- (1)
The sphere and the plane are smooth, there is no change in the velocity of the
sphere in a direction perpendicular to the common normal.
A
X
u sin  v sin 
v cos u cos 
Y
 
O

I B.Sc., PHYSICS
17
 sinsin uv  --------- (2)
Squaring equ (1) & equ (2) and adding these equations,
 22
)2()1(
 222222222
sincossincos uuevv 
)sincos()sin(cos 2222222
  euv
)1sin(cos 22
 
)sincos( 22222
  euv
 2222
sincos(  euv -------- (3)
Dividing equ (2) by (1),




cos
sin
cos
sin
eu
u
v
v

e


tan
tan  ------- (4)
Case 1: If e = 1, that is the sphere is perfectly elastic sphere, sub. in equ (3) and (4)
uv 
 tantan 
 
Case 2: if e = 0, i.e., that is the sphere is inelastic, sub. in equ (3) and (4),
sinuv 
tan

I B.Sc., PHYSICS
18
)(tan 1
 

 90
Case 3: If ;0 Sub. in equ (3) and (4),
uev 
e
0tan
tan 
0tan 
)0(tan 1

0 )0( 
Case 4: The change in K.E of the sphere due to impact on the plane is given by,
)(
2
1 22
uvm 
))((
2
1
uvuvm 
Change in K.E = )(
2
1
uvI 
Since , )( uvmI 

I B.Sc., PHYSICS
19
Motion of two interacting bodies-reduced mass.
Two-body problem
An isolated dynamical system consisting of two freely moving point objects exerting
forces on one another is conventionally termed a two-body problem. Suppose that the first
object is of mass m1 and is located at position vector r1 and the second object is of
mass m2 and is located at position vector r2. Let the first object exert a force f21 on the second.
By Newton's third law, the second object exerts an equal and opposite force, f12=-f21,
on the first. Suppose that there are no other forces in the problem.
The equations of motion of our two objects are thus
f
dt
rd
m 2
1
2
1 ------- (1)
and
f
dt
rd
m 2
2
2
2 ------- (2)
f=f21
Fig. Two-body problem.

I B.Sc., PHYSICS
20
Here, the center of mass of our system is located at
21
2211
mm
rmrm
rcm


 ------- (3)
r
mm
m
rr cm
21
2
1

 ------- (4)
and
r
mm
m
rr cm
21
1
2

 ------- (5)
Hence, we can write
Where r=r2-r1. (From Fig. 1) Substituting the preceding two equations into equ (1)
and equ (2), and making use of the fact that the center of mass of an isolated system does not
accelerate, we find that both equations yield
f
dt
rd
2
2
 ------- (6)
Where
21
21
mm
mm

 ------- (7)
 is called the reduced mass. Hence, we have effectively converted our original two-body
problem into an equivalent one-body problem.
In the equivalent problem, the force ‘f’ is the same as that acting on both objects in
the original problem (except for a minus sign); however, the mass, , is different, and is less
than either of m1 or m2 (which is why it is called the ``reduced'' mass).
We conclude that the dynamics of an isolated system consisting of two interacting
point objects can always be reduced to that of an equivalent system consisting of a single
point object moving in a fixed force field.

Dynamics, Projectile, impulse, impact

More Related Content

What's hot (20)

Similar to Dynamics, Projectile, impulse, impact (20)

Recently uploaded (20)

Dynamics, Projectile, impulse, impact