Transcript System of particles

Physics
Session Opener
A boxer wisely moves his head
backward just before receiving
a punch. How does this tactics
help reduce the force of
impact?
Session Objectives
Session Objective
1. Impulse – Momentum Theorem
2. Elastic Collisions
3. Inelastic Collisions
Definition of Momentum
Consider
All of them have MASS (m)
All of them have VELOCITY (v)
We say, all of them have MOMENTUM (P)
Where,
MOMENTUM = MASS x VELOCITY
Or,
P = mv
Linear Momentum of a System
of Particles
mi
vi
Linear Momentum of a System
of Particles
Pi=mivi
Ptotal 
n
n
P  mv
i1
i
i1
i
i
n
VCM
mv
 n


  mivi    mi  VCM
i1
 i1

 mi
i1
n
i
i
n
i1
Ptotal
 n

   mi  VCM
 i1

Hence,
If total mass is M, then
Psystem  MVCM
Ptotal  MVCM
Questions
Illustrative Example
v
u
v
Newton’s Second Law and
Linear Momentum
“Rate of change of momentum is equal
to the net force acting on the particle.”
Since,
n
  Fi,j  0
i1 i j
and
a
Fext  MaCM
Fext
Also,
P system  MV CM
dpsystem
dt
dPsystem
d
Hence,
F

M
VCM  MaCM
ext
dt
dt
 
Kinetic Energy and Linear
Momentum
P  mv
[Linear Momentum]
  
Then, P.P  mv . mv
 P2  m2 v2
P2 m2 v2

2
2
P2
1
 mv2  KE
2m 2
P2
KE 
2m
P  2m KE
[Kinetic Energy]
[Linear Momentum]
Conservation of Momentum
If no net external force acts on a
system,the linear momentum of the
system remains constant.
Fext  MaCM
a
If Fext=0 then,
Since,
F ext
Fext
dP
dP


0
dt
dt
P  cons tant
[Conservation of Linear Momentum]
Question
Illustrative problem
A projectile moving with velocity V
in space bursts into two parts of
mass in ratio 1:2. The smaller part
becomes stationary. What is the
velocity of the other part?
3
(a) 4v
(b) v
2
4v
2v
(c)
(d)
3
3
Solution
Let the masses be m and 2m after
explosion.
In an explosion the momentum
remains constant.
So. 3m x V=m x 0+2mV2
 V2 
3
V
2
Newton’s Third Law and Conservation of
Linear momentum
Fext
P1  P2  cons tant
dP1 dP2

0
dt
dt
m1
dv1
dv2
 m2
0
dt
dt
m1 a1  m2 a2  0
F1  F2  0
F1  F2
Impulse-Momentum Theorem for a
System of Particles
From Newton’s second law:
dPsystem
Fext 
dt
We can write,
dPsystem  F ext dt
Or,
Pf
t2
Pi
t1
 dP   F  t dt
 P f  Pi 
t2
 F  t dt
t1
t2
J  Pf  Pi   F  t  dt
t1
Impulse-Momentum Theorem


J  P f  Pi 
tf

 F dt
ti
 Area under F  t curve
area
tf
  Fdt
ti
Class Test
Class Exercise - 1
Two masses of 1 g and 4 g are moving
with KE in the ratio of 4 : 1. The ratio
of their linear momentum is
(a) 1 : 1
(b) 1 : 2
(c) 4 : 1
(d) 16 : 1
Solution
We know that if P is the linear
momentum of the particle, then
P2
KE 
2m

KE2
KE1
2
 P2 
m
   1
 P1  m2
2
4  P2 
4
   
1  P1 
1
2
 P2 
   1
 P1 
P2

1
Hence answer is (a)
P1
Class Exercise - 2
A bullet hits a block kept at rest on a
smooth horizontal surface and gets
embedded into it. Which of the
following does not change?
(a) Linear momentum of the block
(b) PE of the block
(c) KE of the block
(d) Temperature of the block
Solution
Since in the absence of external forces on
the system (bullet + block) linear
momentum of system does not change.
But since external force acted on the block
during collision, the block changes its
momentum. KE is also not conserved as
some amount of heat energy is lost when
bullet penetrated into the block.
Temperature also increases during the
process.
Hence answer is (b)
Class Exercise - 3
Consider the following two statements.
I. The linear momentum is independent
of frame of reference.
II. The kinetic energy is independent of
frame of reference.
(a) Both I and II are true (b) Both I and II are false
(c) II is true but I is false (d) I is true but II is false
Solution
Velocity of any body is dependent upon
the choice of frame of reference. For
example, a man sitting in train finds his
co-passenger at rest and hence, having
neither momentum nor kinetic energy.
But the same man has both momentum
and kinetic energy with respect to a man
standing on a bus-stand.
Hence answer is (b)
Class Exercise - 4
A nucleus of mass number A originally
at rest, emits an  - particle with
speed v. The recoil speed of daughter
nucleus is equal to
4v
(A – 4)
v
(c)
(A – 4)
(a)
4v
(A  4)
v
(d)
(A  4)
(b)
Solution
u=0
Before emission:
A
V’
After emission
v
a
A-4
4
Since there are no external forces
acting on the system, hence
Pf = Pi
 A(u) = (A – 4)v´ + 4v
 A(0) = (A – 4)v´ + 4v
 v'  –
4v
(A – 4)
Class Exercise - 5
A shell is fired from a cannon with a
velocity v making an angle  with the
horizontal. At the highest point in its
path it explodes into two pieces of equal
masses. One of the particle retraces its
path to the cannon. What is the speed of
the other piece?
Solution
At the highest point velocity has only
horizontal component.
Since there are no external forces
acting in horizontal direction,
momentum is conserved.
\ Pf = Pi
 Mv cos  

M
M
(v cos )  (v ')
2
2
3
M
Mv cos   v´
2
2
 v´ = 3v cos
y
x
Class Exercise - 6
A 150 g cricket ball, bowled at a
speed of 40 m/s is hit straight back to
the bowler at a speed of 60 m/s.
What is the magnitude of the average
force on the ball from the bat if the
bat is in contact with the ball for 5.0
ms?
Solution
Change in momentum DP = m(vf – vi)

150
(60  40)
1000
= 15 N-s
Impulse = DP = F·Dt
F
DP
15

 3000 N
–3
Dt 5  10
Class Exercise - 7
A 2 kg ball drops vertically onto a
floor, hitting with a speed of 25
m/s. It rebounds with an initial
speed of 10 m/s.
(a) What impulse acts on the ball during the contact?
(b) If the ball is in contact with the floor for 0.020 s,
what is the magnitude of the average force on the floor
from the ball?
Solution
m = 2 kg
vi = 25 m/s
vf = 10 m/s
Change in momentum DP = m(vf – vi)
= 2(10 + 25)
= 70
(a) \ Impulse = +70 N-s
(b) Now, impulse = F· t
70 N-S = F × (0.020)
F
70
 3500 N
0.020
\ Force on the floor from the ball = 3500 N
Class Exercise - 8
Figure above shows an approximate
plot of force magnitude versus time
during the collision of 100 g ball with a
wall. The initial velocity of the ball is 50
m/s and it rebounds directly back with
approximately the same speed perpendicular to the wall
as was in the case of impact. What is Fmax, the maximum
magnitude of the force on the ball from the wall during the
collision?
Force
Force (N)
0
2
4
Time
6
Solution
Impulse = Area under F-t graph
1
1
  2  Fmax  Fmax  2   2  Fmax
2
2
= 4 × Fmax
Now, impulse = DP = m(vf – vi)

100
(50  50)
1000
 Fmax 
10
4  103
= 10 N-S
 2500 N
Class Exercise - 9
A 2100 kg truck traveling north at
50 m/s turns east and accelerates
to 60 m/s.
(i) What is the change in kinetic energy
of the truck?
(ii) What is the change in linear
momentum of the truck?
Solution
(a)
Change in KE
1
1
2
 mv f – mvi2
2
2
1
 m(602 – 502 )
2

1
 2100  10  110
2
V
6
0
m
/s
f=
Y
= 10500 × 110
= 1155 × 103 J
= 1155 kJ
V
5
0
m
/s
i=
X
Solution
(b)
Pi = mvi

 2100  50 j

P f  2100  60 i

 
\ DP  Pf – Pi  21 6 i – 5 j  KN-S




Thank you