# Collision by nehalwan

VIEWS: 3 PAGES: 151

AIEEE IIT-JEE CBSE STUDY MATERIAL PHYSICS SAMPLE PAPERS TEST PAPER KEY SOLUTIONS ANSWERS QUESTIONS

• pg 1
```									collision
A process of strong and rapid interaction
between two bodies which brings about a
tremendous (quick) change in momenta of the
interacting bodies is called collision.

During collision, the colliding bodies may or
may not touch physically.

2
For instance, in the collision between ball and
bat physical contact takes place.
On the other hand, in Rutherford’s gold foil
experiment, the -particles do not touch the
gold nuclei.
Explosion can also be called collision because
it is a short timed interaction.
Collision
•In a collision a relatively large force acts on each
colliding particle for a relatively short time.
•The basic idea of a ‘collision’ is that the motion of
the colliding particles (or of at least one of them)
changes rather abruptly .
•We can make a relatively clean separation of
times that are ‘before the collision’ and those that
are ‘after the collision’.

4
Types Of Collision
Collisions are of 3 types :-

1. ELASTIC (ALSO CALLED PERFECTLY ELASTIC)
COLLISION
Symptom: The colliding bodies separate out after
collision.
Properties

(i) Total momentum of the system is conserved.
(ii) Total K.E of the system is conserved.

5
2.   PERFECTLY INELASTIC
COLLISION)
Symptom: Colliding bodies get embedded to each other
and become one body after collision. In this case

(i) Total momentum remains conserved, as in the case
of Perfectly Elastic Collisions.

(ii) Total K.E. of the system generally decreases.
Sometimes, total K.E. of the gross body may
increase also, if some internal energy gets released
during collision.

6
3.   INELASTIC COLLISION
It is a case of lying between perfectly elastic and
perfectly inelastic collisions.

7
ON COLLISION BETWEEN TWO
BODIES

If the initial and final velocities of colliding
masses lie along the same line (along the
common normal), then it is known as head-on
collision.

m1            m2          m1          m2

u1          u2            v1             v2

10

Let the smooth small balls of masses m1 and m2
moving in positive x-direction with speeds u1 and
u2 (u1 > u2) collide.

u1                        u2

m1                      m2

before collision

11
m1       m2

V   During
collision

J

J
m1                 m2

12
m1    m2

V

v1                v2

m1               m2

After collision

13
Just after the collision let they move in positive
x-direction with speeds v1 and v2 respectively.

v1
v2
m1                  m2
Just after collision
In case of collision linear momentum of
system of particles is remains
unchanged.
Since no net impulse acts on the system (m1 + m2),
its momentum remains constant
Or      Pf = Pi
m1v1  m2v2  m1u1  m2u2   ...(i )

15
Problem 33.

Consider a head- on collision between two particles of
masses m1 and m2. The initial speeds of the particles
are u1 and u2 in the same direction. The collision starts
at t = 0 and the particles interact for a time interval
t. During the collision, the speed of the first particle
varies as
t
(t )  u1       (1  u1)
t

Find the speed of the second particle as a function of
time during the collision.

16
Solution.
m1             m2
Before collision :        u1                 u2

At t = 0 collision occure the particles interact
for a time t.
m1             m2

v1(t)          v2(t)

As linear momentum of the system of particles remain
constant.
m1u1  m2u2  m1v1(t )  m2v2 (t )
     t            
m1u1  m2u2  m1 u1     (1  u1 )  m2v2 (t )
     t           
t
m2v2 (t )  m2u2  m1     (v1  u1)
t
m1t
v2 (t )  u2        (v1  u1)
m2 t

m1            m2
Before collision :        u1             u2

m1           m2
During collision :
v1(t)         v2(t)

18
Problem 34.

A bullet of mass m moving at a speed v hits a ball of
mass M kept at rest. A small part having mass m’
breaks from the ball and sticks to the bullet. The
remaining ball is found to move at a speed v1 in the
direction of the bullet.
Find the velocity of the bullet after the collision.

19
Solution.
m       v        m
M
Just Before collision :
bullet            ball
v’            v1
Just After collision :    (m + m’)           (m - m ’)
(M-m)
bullet             ball

Linear momentum in x-direction ‘bullet + ball’
system will be conserved.
mv  0  (m  m ')v ' (M  m ')v1
mv  (M  m ') v1
v' 
(m  m ')
NEWTON’S EXPERIMENT LAW
Newton’s Empirical Formula for Collision
NEWTON’S EXPERIMENT LAW

When two bodies collide directly, the relative
velocity after collision bears a constant ratio to
their relative velocity before collision and is
opposite in direction.

22
If two bodies collide their relative velocity along their
common normal after collision bears a constant ratio to
their relative velocity before collision. This constant
ratio is called
‘e’ = coefficient of restitution.

–(v1 – v2 )
Co-efficient of restitution,e =             ;
(u1 – u2 )

23
where,    v1 = velocity of particle (1) after collision
v2 = velocity of particle (2) after collision
u1 = velocity of particle (1) before
collision
u2 = velocity of particle (2) before
collision

(–) Relative velocity after collision
e
Relative velocity before collision

24
[0  e  1] i.e., the quantity  lies between 0 and 1.
where e = 0, the collision is said to be perfectly
inelastic (because in case of perfectly inelastic collision
v1 = v2). e = 1, the collision is said to be perfectly
elastic.
Velocity Of Approach

Since u1 > u2, m1 approaches m2 with a
velocity u12 = vapp = u1 – u2, just before the
collision.

1                      2
u1 - u2

m1 approaches m2 with
vapp = u1 - u2
Velocity Of Separation

Just after the collision m1 separates from
m2 with a velocity
V12 = vsep = v1 – v2

1              2
v1  v 2

m1 separates from m2 with
vsep = v1 - v2
      

Since vsep and vapp  are oppositely directed and
is generally less than vapp, we can write

- e vapp = vsep
Where e is called the coefficient of restitution
which tells us what fraction (coefficient) of the
relative velocity between the colliding bodies
(vapp) is restituted (restored).

or    e (u1  u2 )  v2  v1    ...(ii )
(b)   For oblique impact:
When two bodies collide obliquely, their relative
velocity resolved along their common normal after
impact is in a constant ratio to their relative velocity
before impact (resolved along common normal) and is in
the opposite direction.
Common normal
f
           b          q
v2
u1       u2              v1
v1 cos q  v2 cos f
 e
u1 cos   u2 cos b

v1 cos q – v2 cos f = – e (u1 cos  – u2 cos b)
Common Tangent (CT) & Common Normal (CN)

CT

CN

These CN & CT direction
have no nothing to do
with the directions of
velocity of  the   two
bodies.
Oblique collision in case of smooth
surfaces
(a)    Common Normal (CN)
Impule is exerted in CN–direction only. Both bodies
exert equal and opposite forces (action and reaction) on
each other. Hence, momentum and velocities change
accordingly in CN–direction.

Momentum changes in CN-direction only
Apply ‘e’ (coefficient of restitution) in CN–direction
only.
(b)   Common Tangent (CT)
No impulse in Common Tangen tdirection (in case of
smooth surfaces)
Neither momentum nor velocities change in CT-
direction.
If a ball strikes with a velocity u1 with
a velocity wall which itself approaching
it with a velocity u2
u1
|Relation velocity of separation|
= |Relative velocity of approach|
v1
v1 – v2  –e(u1 – u2 )

v1 – (–u2 )  –e[u1 – (–u2 )]              u2

v1  –e(u1  u2 ) – u2
v2 = u2
In case of perfectly elastic collision,
u1
v1  –e(u1  u2 ) – u2

v1  –u1 – u2 – u2  –(u1  2u2 )          v1
v1  –(u1  2u2 )
u2

v2 = u2
Illustration 32.

A ball moving with a velocity v strikes a wall moving
toward the wall with a velocity u. An elastic impact
occurs. Determine the velocity of the ball after the
impact. What is the cause of the change in the
kinetic energy of the ball ? Consider the mass of
the wall to be infinitely great.

v    u
Solution.
(v2 – v1)
Co-efficient of restitution,e =            ;
(u1 – u2 )
((u) – v1)
e =1              ;
(v – (u))
v  u  (u  v1)
v1  (v  u)  u  (v  2u)
1
Kinetic energy before impact    mv 2
2
1
Kinetic energy after impact    m (v  2u)2
2

The change in kinetic energy is equal to 2mu(u + v).
Now we calculate the work of reaction forces
acting on the ball during the impact
. Let the collision continue for t seconds.
Assume the reaction force to be constant (the
result does not depend on this assumption).
Since the impact changes the momentum by
2m(v + u), the force of reaction is
2m (v  u)
F 
t
The work of this force is
2m (v  u)ut
W  Fs  F (ut )                  2m (v  u) u
t
So we can see that this work is equal to change in
kinetic energy.
Problem.

Two particle of masses m1 = 1 kg and m2 = 10 kg
are moving in a straight line as shown in the
figure.
10 kg

10 m/s                  5 m/s

1 kg

If particle collide elastically find the velocity of
the particles after collision.

43
Solution.

Linear momentum of colliding particle will be
conserved.
v1              v2

m1                       m2
m1u1  m2u2  mv1  mv 2
1 10  10  5  1 v1  10v 2
60  v 1  10v 2            …(i)

44
Now applying equation of coefficient of restitution.
 v  v1              v 2  v1
e 2                1
 u1  u2 
            u1  u2

v 2  v1  u1  u2  10  5
v 2  v1  5                          …(ii)

Solving equation (i) and (ii) we can get the values of
v1 and v2.
10
v1  m / s
11
65
v2       m/ s
11

45
Problem.

Two identical balls of mass m are moving on a
horizontal surface with velocity u1 and u2 as shown
in figure. The balls collide elastically find the
velocities of the balls after collision.

46
Solution.

* Both balls are moving on horizontal smooth surface
with velocity u1 & u2 (given u1 > u2).
* They collide elastically (e = 1)

* Find the velocities of the balls after collision.

47
m        m         m1   m2
u1       u2

v1         v2

48
Conservation of linear moment
mu1 + mu2 = mv1 + mv2
v1 + v2 = u1 + u2
Now using restitution equation
v 2  v1
e            1
u1  u2
v 2  v1  u1  u2

Solving equation (i) and (ii)

v1 = u2
v2 = u1

49
Inelastic Collision

Let us consider a case of inelastic collision 9e > 1) in
which a projectile of mass m1 collides with a stationary
target of mass m2 with a velocity v1 as shown below .

m1               m2        m1                  m2

u1          Rest           v1             v2
Before                        After

A moving particle m1 hits a stationary particle
m2 inelastically (e > 1).
Applying momentum conservation

m1u1  m1v1  m2v 2

For inelastic collisions, e > 1, we have

eu1  v 2  v1

After solving the above two equations, we get
 m  em2 
v1   1        u1
 m1  m2 
 (1  e )m1 
v2               u1
 m1  m2 
Note :-

• Velocity of the target is always more than that
of the projectile for finite values of m1 and m2
and O < e < 1.

* For completely inelastic collision (e = 0) both the
particles move together.

* When the target is infinitely massive, it remains
stationary after the collision and the projectile
rebounds with a speed eu1.
The loss in kinetic energy of the system is given by

1  m1m2 
K             (1  e 2 ) u1
2

2  m1  m2 
The fraction of kinetic energy loss is given by

K m2 (1  e 2 )

Ki   m1  m2
Note :-

* for an elastic collision (e = 1), there is no kinetic energy
loss.

* for completely inelastic collision (e = 0), the kinetic energy
loss is maximum and it is given by

 K        m2
        
 K i max m1  m2

* greater the mass of the target, larger the loss of kinetic
energy.
Problem.
A particle is dropped from a height h0 on a
horizontal floor. The coefficient of restitution
between the particle and the floor is e.
(i) Determine the first
height of rebound h1.

h0
h1

56
Solution.

(i) The ball strikes the floor with a speed
v 0  2 gh0

And it rebounds from the                      ev0
floor with a speed            v0
v1  ev 0  e 2 gh0
The approach velocity is
Therefore, the first         v0 and the separation
height of rebound is         velocity is ev0.
v1
2
h1      e 2 h0
2g

57
Problem.
A particle is dropped from a height h0 on a
horizontal floor. The coefficient of restitution
between the particle and the floor is e.
(ii) Determine the height
of the particle after
nth rebound

h0
h1

58
u=0   h
e2 h
e4 h
e6 h

u         eu          e2 u          e3 u
(ii) obviously, the particle’s velocity after nth
rebound will be
v n  e nv0

Therefore, the height after nth rebound will be
vn
2
hn      e 2 n h0
2g

60
Problem.
A particle is dropped from a height h0 on a
horizontal floor. The coefficient of restitution
between the particle and the floor is e.

(iii) Determine the
number of rebounds
after which the
rebound height
h0
becomes half of the
initial height h0.                     h1

61
vn
2
hn      e 2 n h0
2g

(iii) Taking log on both the sides, we get
log hn  2n log e  log h0

 h 
log  n 
n      h0 
2log e
hn 1
Since                               (given)
h0 2
log(1/ 2)
n
2log e

62
Problem.
A particle is dropped from a height h0 on a
horizontal floor. The coefficient of restitution
between the particle and the floor is e.

(iv) Compute the total
distance travelled
by the particle
before it stops
bouncing.         h0
h1

63
(iv) The total distance travelled by the particle is
H  h0  2h1  2h2  .....

H  h0  2e2 h0  2e 4 h0  ......

H  h0 [1  2e2 (1  e2  e 4  .....)]
          1                                         1
H  h0 1  2e 2      2       [ 1  e2  e 4  ....... 
 1 e 
]
                                                    1 e 2

1  e2 
H  h0 
 1  e2 


64
Problem.
A particle is dropped from a height h0 on a
horizontal floor. The coefficient of restitution
between the particle and the floor is e.
(v) Compute the total
time taken to stop
bouncing.

h0
h1

65
1  e2 
H  h0 
 1  e2 

(v) The total time taken by the particle is
T  t0  2t1  2t2  ......
2h0    2h1    2h2
T         2     2      .....
g      g      g

Since h1 = e2h0, h2 = e4h0 , therefore
2h0
T        1  2e  2e 2  .....
g                       
2h0
T         1  2e(1  e  e 2  .....)
g                             

 1  e  2h0        
1  e  e 2  ...... 
1 
T                    
 1 e  g                                       1 e 


66
Problem.

A particle of mass m moving with a speed v
strikes a smooth horizontal surface at an angle .
The particle rebounds at an angle b with a speed
v. Determine an expression for v and b if the
coefficient of restitution is e.
m
m      v
u
       b

A ball strikes a smooth horizontal
surface at an angle  and rebounds at
an angle b.

67
Solution.

Since no external impulse acts in the horizontal
direction, therefore, momentum of the ball is
conserved in the horizontal direction. That is,

mu cos  = mv cos b
or            u cos  = v cos b                  …(i)

Using the definition of the coefficient of restitution,
we get
e(u sin ) = v sin b                …(ii)

68
Dividing (ii) by (i), we get

tan b = e tan 

or    b = tan-1(e tan )]
On squaring and adding, we get

v 2  u2 (cos2   e2 sin2 )

v  u cos2   e 2 sin2 

69
Problem.
Refer to fig. The pendulum on
the left is pulled aside to the
positions shown. It is then
released and allowed to collide
with the other pendulum, which     h
is at rest.

(i) What is the speed of the ball on the left just
before collision?
(ii) If collision is elastic calculate the maximum
height reach by the right ball.
(iii) How high, in terms of h, does the
combination swing? Assume that the two
balls have equal masses.

h
Solution.
Using conservation of mechanical energy

1
mu2  mgh
2                              1

u  2 gh
h       u
(i) If collision is elastic :-
1   2

The ball (1) will become at rest and ball (2) will
start moving with velocity u and swing to the same
height h same as ball (1).
(ii) If the balls stick together just after
collision:-
Using conservation of linear momentum just before
and just after collision.
mu  0  2mv
u
 v        …(i)
2
For maximum height
reached by
combined body using                           h’
v
conservation of
mechanical energy.
1
2 mg h '  (2m) v 2
2
2
u
v '2  2  u2
h'        
2g    2g   8g

Hence, h '  ( 2 gh )2 h

8g      4
h
h' 
4

h’
v
Illustration 33.

In figure, there are n identical
spheres of mass m suspended
with wires of equal length. The
spheres are almost in contact
with each other. The sphere 1
is pulled aside and released. If
sphere 1 strikes sphere 2 with
velocity v1,,find an expression
for the velocity vn of the nth
sphere immediately after being          v1                     vn
struck by the one adjacent to      '   1  e 
n 1

vn                  v1
it. The coefficient of              2 
restitution for all the impacts
is e.
Solution.
From the definition of coefficient of
restitution, we have,
'    '
v2  v1
e
v1  0

or,            '    '
ev1  v2  v1     … (i)

From law of conservation of momentum, we have
'     '
mv1  mv2  mv1
'    '
v1  v2  v1
'    '
or,            v1  v2  v1
'         '
v1  v1  v2      … (ii)

From equations (i) and (ii), on eliminating v’1, we get

'    e  1          … (iii)
v2         v1
 2 
Similarly for second and third balls,
'    ''
v 3  v2
e '
v2  0
'    '    ''
ev2  v3  v2                      … (iv)

From conservation of momentum, we get,
'     '     ''
mv2  mv3  mv2
''   '     '
v2  v2  v 3                      … (v)
From equations (iv) and (v), we can solve v’3.
'
(e  1)v2
'
v3                             … (vi)
2
On substituting expression for v’2 from equation (iii)
in equation (vi), we get,
2
 e  1
… (viii)
'
v3         v1
 2 

In a similar way, we can generalize the result,
n 1
'   1  e 
vn                  v1
 2 
Problem.
A heavy ball of mass 2M moving with a velocity v0
identical balls each of mass M as shown in figure.
Determine the velocity of each ball after collision.

2M      v0
M M M

A heavy ball collides
identical balls.

81
Solution.
Conserving linear momentum
2Mu0  2Mv1  Mv2
2u0  2v1  v2
v 2  v1 v 2  v1
Now using restitution equation      e           
u1  u2 v 0  0
1

v 2  v1  v 0

Heavy ball 1,                      2M    v0
v
M     M M
 2M  M 
v1           v0  0
 2M  M       3             1             2     3  4

 2(2M )      4
ball 2.      v2           v0  v0
 2M  M      3

82
As the ball 2 gains momentum, immediately it
transfers it to the ball 3, which in turn transfer to
the ball 4.
Thus, the ball 4 comes out with a velocity v4 = 4/3
v0 leaving the balls 1 and 2 stationary.

1   u0/3    2         4u0/3

3    4

83
Now, the heavy ball collides the ball 2 with a
velocity u0/3 and their final velocities are:
v0       4
v1 '       v2  v0
'

9        9
The ball 2 transfer, its momentum to the ball 3,
which starts moving with a velocity v2 = 4/9 u0,
leaving the ball 2 stationary.

1                      4v0/9   4v0/3
v0/9
2     3       4

84
Now, the heavy ball collides with the ball 2 with a
velocity u0/9 and their final velocities are:
u0               4
v'      ;     v1 
'
u0
27              27

2M
u0/27           4u0/27           4u0/9       4u0/3
M                M             M
1             2                 3            4
Final

85
u        4u      4u 
pi = 2Mu0    Pf  2M  0   M  0   M  0 
 27      9       3 
 1   2 2 2
Pf  2Mu0       
 27 27 9 3 

Pf  2Mu0

Note: that the collision does not occur actually in
the sequence described in the problem. In fact all
the balls attain their final velocities simultaneously.

86
Problem 56.

Two balls having masses m and 2m are fastened to two
light strings of some length l. The other ends of the
strings are fixed at O. The strings are kept in the
same horizontal line and the system is released from
rest. The collision between the balls is elastic.
(a) Find the velocities of the balls rise after the
collision ?
(b) How high will the balls rise after the collision ?
1     0       1

m                          2m

87
Solution.
Applying
m                                  2m
conservation of                               O            L

linear momentum               1                                  2
just before and                                O
just after collision
mu – 2mu = mv1 + 2mv2
       
v1 + 2v2 = -u           … (i)

u = 2g           u = 2g

v1               v2

88
Applying restitution equation

v2 – v1 = e (u1 - u2)
(v2 – v1) = 1 (u - (-u))

v2 – v1 = 2u                  … (ii)
Solving (i) & (ii), we get,

50 g
v1                 toward left
3
2g
v2                  toward right
3
Now applying conservation of mechanical energy for
ball 1 & 2
2
1  50 g 
For ball (1) :     m        mgh
2   3   
1 50g
mgh  m
2    9
25
h
9
But the ball can not go beyond 2 hence maximum height
reached by ball 1 is 2.

2
1  2g                        h' 
m      mgh '
2  3                                 3

91
Particle & Wedge Collision
Problem.

Determine the velocity of approach in the following
situations.

Rest                          m
u1
u1
M
u2
m            M
q                                  q
Solution.

Velocity of approach
VOA  (u1  u2 )n

VOA  u1 sin q  0

VOA  u1 sin q
t
n           +
Common tangent
m    u1                       direction
2
q
1                             Common normal
q        M       direction
m
1
t
u1

+
2          q
M

q               u2

VOA  (u1  u2 )n  u1 cos q  ( u2 sin q)

VOA  u1 cos q  u2 sin q
Problem.

Determine the velocity of separation in the
following situations.

m
q
v1                  
b   m   v
M
V
q
v2
Solution.

VOS  (v 2  v 1 )n
n
v1
 [(v 2 )  ( v 1 cos q)]            q
1
VOS  v 2  v 1 cos q         +
2

v2
Solution.

VOS  (v 2 sin q)  ( v 1 cos b )

VOS  v 2 sin q  v1 cos b
m
n

            +
b   m       v1
M
V2
q
Illustration 28.
A smooth ball is dropped from a height h on a
smooth incline, as shown in figure. After collision
the velocity of the ball is directed horizontally
(a) Find the coefficient of restitution.
(b) If the collision is elastic, what is the
impulse on the ball?

h
v’

q
Solution.

(a) Normal axis and tangential axis are shown in
the figure. Reaction of the incline is along
n-axis and in the absence of friction there
is no force along t-axis; therefore velocity
along t-axis remains unchanged,
n-axis
 cos q  u sin q   …(i)                                     j
v sin q

v
q
v cos q     i
u cos q     u        u sin q
q

(a)
From the definition of coefficient of
restitution,
 sin q
e                  …(ii)
 u cos q

  eu cot q         …(iii)

From equations (i) and (iii), (eu cot q)cos q  u sin q

e  tan q    2
(b) When the collision is elastic, the
component of velocity along n-axis is
reversed in direction. Therefore the
change in velocity                    n-axis
  2 cos q                            q
There is no change in velocity                     q
v
along the t-axis, therefore no
impulse along t-axis.
u
q
 m 
(b)
 2m cos q
Velocity of the ball when it strikes the plane
 2gh

Thus impulse      2m 2gh cos q
Problem.
A ball of mass ‘m’ moving horizontally which
velocity ‘u’ hits a wedge of mass ‘M’. The wedge
is situated on a smooth horizontal source. If after
striking with wedge the ball starts moving in
vertical direction and the wedge starts moving in
v
horizontal plane. Calculate
(i) The velocity (v) at
which the ball moves
in vertical direction.      V            u

(ii)     The   velocity   of                    m
q
wedge V.
Problem.
A ball of mass ‘m’ moving horizontally which
velocity ‘u’ hits a wedge of mass ‘M’. The wedge
is situated on a smooth horizontal source. If after
striking with wedge the ball starts moving in
vertical direction and the wedge starts moving in
v
horizontal plane. Calculate
(iii) The impulse
imparted by the ball
on the wedge.       V               u
m
(iv) The impulse imparted
q
by ground on the
wedge.
Problem.31

We can solve this problem by one of two methods
* Using conservation of linear momentum

* Impulse approach.                             1

Using conservation of                  q
2                   vn
linear momentum         V
As no external force are         M
acting on the system in                 q
horizontal direction the
t
linear momentum should
be constant conserved in
horizontal direction.
Solution.

For ball :
J sin q
mu  J sin q  0   …(i)
J cos q
J sin q  mu
…(ii)
J cos q  mv               M
J sin q
q
From (i) and (ii)
J’
v  u cot q
From (i)    J  mu cos ec q        J sin q

For wedge:    J '  J cos q
J cos q
J '  mu cot q
M
J sin q  MV        J sin q                  J sin q
V
M               q
mu
V
M              J’
Problem.
A ball of mass ‘m’ moving horizontally which
velocity ‘u’ hits a wedge of mass ‘M’. The wedge
is situated on a smooth horizontal source. If after
striking with wedge the ball starts moving in
vertical direction and the wedge starts moving in
v
horizontal plane. Calculate

(v) The coefficient of
restitution e = ?                     u
V
(vi) The heat developed                        m
during collision .               q
(v) The coefficient of restitution is given as

(v 2  v1 )n  e (u1  u2 )n                 v
n
V sin q  (v cos q)  e u sin q  0
V sin q  v cos q
e                                  2           q
u sin q            V                         u   1
V v          m
e   cot q        (cot q) cot q
u u          M                              q
m
e    cot 2 q
M                                       +
(vi) The heat developed
1     1    1  2
H  mu   mv  MV 
2      2

2     2    2   
1      1              1  mu  
2

H  mu2   m (u cot q)2  M     
2      
 2             2  M   
1                   m
H  mu 2 1  cot 2 q 
2                   M

A Wedge of mass M Is kept at rest on sooth surface , a
particle of mass m
hits the wedge normally .Find the velocity of wedge and
particle just after collision. Take coefficient of
restitution as e.

v0
vn

V
mv 0 sin q  MV  mv n sin q         …(i)

Using Newton’s law of collision.
(v 2  v1 )n  e (u1  u2 )
(v sin q  v n )  e (v 0  0)                 v0
…(ii)
vn
Solving (i) and (ii) we get
V
(1  e ) mv 0 sin q
V
M  m sin q 2
Using impulse method :

The impulse acting on the ball and wedge can be
represented as

v            v0
J                         vn
J   vn
V               J sin q          V

q
J
v
V                     J sin q
vn
J
q

J sin q  MV                              …(i)

mv  J  mv n                             …(ii)

and Newton’s law of restitution

(v sin q  v n )  e (v 0  0)            …(iii)
Solving (i), (ii) and (iii) we get

(1  e ) mv 0 sin q
V
M  m sin q 2
Collision of two Discs
Oblique Collision

R2           U2
O2                   d
O1   U1
R1
t

n
O2                             d
q              sin q 
R1  R2
d
q
O1         A
At the time of collision let the velocity of Disc ‘1’
and Disc ‘2’ just after collision in the direction of
normal and tangent are (v1,n, v1, t) and (v2,n, v2, t)
respectively then
t

n
O2        V2,n
q

d
q
O1         A
v1,n                        v2t

v1t
The velocity of Discs will be unchanged in the
direction of tangent
v1,t  u1 sin q     and    v2,t  u2 sin q …(i)
t

n
O2        V2,n
q

d
q
O1         A
v1,n                       v2t

v1t
Conservation of linear momentum is applicable in any
direction as there is no external impulse on the
system.
We have already calculated the velocity in tangent
direction.
We need to calculate velocity in normal direction.
Hence conserving linear momentum in normal
direction.
…(ii)
m1u1 cos q  m2u2 cos q  m1v1,n  m2v2,n
Now using Newton law of restitution we get

(v2,t  v1,n )  e (u1 cos q  u2 cos q)       …(iii)
t

n
O2          V2,n
q

d
q
O1          A
v1,n                            v2t

v1t
From equation (ii) and (iii) we can calculate the
value of v1,n and v2,n.

t

n
O2        V2,n
q

d
q
O1         A
v1,n                        v2t

v1t
Illustration 35.

In figure ball 1 with an initial speed of 10 m/s collides
elastically with the stationary balls 2 and 3, whose
centres are on a line perpendicular to the initial velocity
of ball 1 and that are initially in contact with each
other. The three balls are identical. Ball 1 is aimed
directly towards contact point of 2 and 3. Assume all
motion is frictionless. Determine the velocity of each ball
after the collision.

2
v0               x

1                  3
Solution.

In the absence of friction each impulse is directed along
the line joining centres of the colliding balls, normal to
the colliding surfaces.
After collision the target balls leave the collision along
normals, while the incident ball leaves along x-axis.
Let v0 be the velocity of the incident ball before the
collision,
V be its velocity afterwards and v be the speed of each
of the target balls after the collision.
Each ball has mass m.
From conservation of momentum along x-axis, we have,

mv0  mV  2mv cos q

2           v
R           Sin q = R/2R =1/2

v     1   R         R
x
q
q

R   R

v
3
From conservation of momentum along x-axis, we have,

mv0  mV  2mv cos q                            … (i)

 2.0 m/s

Thus the incident ball rebounds in the negative x-direction.
Particle & Wedge Collision
Illustration 37.
A right angled wedge ABC of mass M = 4 kg and
base angle a = 53o is resting over a smooth
horizontal plane. A shell of mass m = 0.5 kg moving
horizontal with velocity v0 = 40 m/s collides with
the wedge, just above point A. As a consequence,
wedge starts to move towards left with velocity v =
5 m/s. Calculate
C     (a)   maximum height reached by the shell.
(b)   heat generated during
v0            collision,

B           53°      A
Illustration 31.
Figure shows a smooth
spherical ball of mass m
striking two identical
equilateral triangular               m
wedges of mass M. At
the instant of impact
v0
velocity of the ball is v0.
Taking coefficient of      M          M
restitution e, determine
the velocities of the      60°      60°
sphere and the wedges
just after collision.
Solution.
Let J be impulse between ball and wedges and v1 and v2
be the velocities of the ball and the wedge.
From impulse-momentum equation on the ball,
2J sin 30° = mv1 – (-mv0)
v1
J = mv1 + mv0         … (i)
For the wedge,
J
J cos 30° = Mv2         … (ii) v        J
2                              v2
M                J
60°                 60°
On eliminating J from equations (i) and (ii), we have,
2
Mv2  mv1  mv 0                 … (iii)
3
From the definition of coefficient of restitution,
v1 cos 60  v2 cos 30 v1  3v2
e                        
v 0 cos 60           v0

or,       ev 0  v1  3v2                        … (iv)
On solving equations (iii) and (iv) for v1 and v2, we get,

(2eM  3m) v0
v1 
2M  3m

3 (1  e)mv0
v2 
2M  3m
Illustration 36.
A small particle of mass m is
released from a height h on a
large smooth sphere kept on a
perfectly smooth surface, as                h
shown in figure. Collision
between particle and sphere is
perfectly inelastic. Determine
O
the velocities of particle and
R/2
sphere after collision.

2    2           6m2 gh     gh
v  v n  vt                 2

4 (4M  m)    2
Solution.
Collision takes place between smooth bodies; therefore
tangential velocity of particle is unchanged.
vt  u sin q
R /2 1
Where   sin q      and u  2gh
R   2
or,     q  30
y
n-axis
vn
x                        u
R                    vn
V                  q
q          vt
R/2         t-axis

vt
u
vt                             … (i)
2

From conservation of momentum along x-axis,
Pi  Pf
0  0  MV  mv n sin q  mvt cos q       … (ii)
y
n-axis
vn
x                                u
R                            vn
V                  q
q          vt
R/2         t-axis

vt
From the definition of coefficient of restitution,

V sin q  (v n )
e                   0
u cos q  0

[e = 0, for a perfectly inelastic collision]

or,        V sin q  v n
V
vn                                … (iii)
2
On substituting expressions for vn and vt in
equation (ii), we obtain

 V 1    u 3
MV  m     m  
 2 2    2 2
    m      u 3      where u  2 gh
M   V  m
    4       4

or,          m 6 gh
V 
(4M  m)

2    2             6m2 gh     gh
v  v n  vt                   2

4 (4M  m)    2
Illustration 34.
A small particle of mass m, moving with velocity
u horizontally, strikes a triangular wedge of
mass M. After collision the particle starts
moving along the incline. Determine the velocity
of the wedge immediately after collision and the
maximum height to which it can ascend on the
wedge. Assume that friction is negligible at all
the surfaces.

M
m
q
u2       M2 cos2 
h
2g (M  m)(M  m cos2 )

mu sin2                Mu cos 
V           2
and v rel 
M  m sin              M  m sin2 
Solution.
Let vrel be the velocity of the particle
relative to the wedge and V the velocity of
the wedge after collision. After collision x-
and y-components of resultant velocity of
particle are
v x  v rel cos q  V       v y  v rel sin q

Vrel sin q             vrel
N    vrel

v
N
q                                         Vrel cos q

V
Momentum of system is conserved along x-direction
only.

mu  m (v rel cos q  V )  MV    … (i)

At the instant of collision the force between the
particles is normal to the surface, hence the
velocity of the particle along the wedge remains
constant.
u cos   v rel  V cos          … (ii)
On solving equations (i) and (ii), we get,

mu sin2                Mu cos 
V           2
and v rel           2  … (iii)
M  m sin              M  m sin 

The particle and wedge moves on a single body till
the block reaches maximum height. At this position
the particle has only x-component of velocity, let it
be u.
From law of conservation of energy, we have,
mu
mu  (m  M)v '   or, v '             … (iv)
(m  M)
From law of conservation of energy, we have,
1     1                1
MV  m (v rel  V )  (M  m)v '2    … (v)
2       2      2

2     2                2

On substituting V, vrel and v’ in equation (v) and
solving for h, we get,
u2       M2 cos2 
h
2g (M  m)(M  m cos2 )
Problem.
A wedge of mass M = 3.6 kg and having base angle
q= 370 is resting over a smooth horizontal surface. A
ball of mass m = 1 kg is thrown vertically downwards
such that it strikes the wedge with velocity v0 = 11
ms-1 at height h = 48 cm from the base of the wedge
as shown in figure.
m

v0

M                h = 48cm
q
Coefficient of restitution between ball and wedge
is e = 0.5. Assuming all the surfaces to be
smooth. calculate
(i) Velocity of wedge just after collision.

(ii) Vertical component of velocity of ball just after collision.
(iii) Time of flight of ball from the instant of collision with
wedge to the instant when ball strikes the floor.
(iv) Distance between ball and right edge of the wedge when
ball strikes the floor.
m

v0

M                  h = 48cm
q
Solution.
When the ball with strike the
wedge the impulsive reaction will
develop normal to the inclined                      n
surface of the wedge. The
horizontal component of this
q
impulse will give motion to the    V            J
wedge in horizontal direction.
Which vertical component of                q
the impulsive reaction will be
J’           t
supported by impulse given by
ground in vertical direction, as
shown in the diagram given
below,
vy
n
q            y
vx

q                            x
J
vS
If we consider (wedge + ball) a system there exits a
external impulse in vertical direction. Hence we can conserve
the momentum of the system only in horizontal direction as
there will be no component of external impulse (from ground)
in the direction. The momentum of the ball can be conserved
in the direction of common tangent as there will be no
impulse on the ball in the common tangent direction.
Conserving momentum of the system in horizontal
direction.
ˆ        ˆ
  MV i  mv x i                       …(i)
Where v = velocity of the wedge just after collision
Vx = horizontal component of velocity of ball just after
collision.
For the ball velocity of it should be same as tangent
direction just before and just after collision.
v0 sin q  v x cos q  v y sin q      …(ii)
Now we are left with three unknowns v, vx and vy hence we
need another equation which can be achieved by Newton’s
restitution equation.
(v 2  v1 )n       v sin q  (v x sin q  v y cos q)
e                
(u1  u2 )n                   v 0 cos q

From equations (i), (ii) and (iii) we get
V =2ms-1, vx = 7.2 ms-1 and vy = 1.4 ms-1
(downward)
Let the time when ball strikes the ground be T
1
h  vtT  gT 2
2
Substituting the values we get

T  0.2 sec

Distance between ball and right edge of wedge when ball
strikes the ground l = R – x + s

V = vx, T – h/tan q = V.T    = 1.20 m

V
h
q
x           x     R
S
l

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