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Physics
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A boxer wisely moves his head backward just before receiving a punch. How does this tactics help reduce the force of impact?
Session Opener
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Session Objectives
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Session Objective
1. Impulse – Momentum Theorem
2. Elastic Collisions
3. Inelastic Collisions
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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 = mvP = mv
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Linear Momentum of a System of Particles
vimi
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Linear Momentum of a System of Particles
Pi=mivi
n n
total i i ii 1 i 1
P P mv
n
i i n ni 1
CM i i i CMni 1 i 1
ii 1
mvV mv m V
m
n
total i CMi 1
P m V
If total mass is M, then total CMP MV
system CMP MV,,,,,,,,,,,,, ,
system CMP MV,,,,,,,,,,,,, ,
Hence,
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Questions
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uv v
Illustrative Example
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Newton’s Second Law and Linear Momentum
“Rate of change of momentum is equal to the net force acting on the particle.”
Fext
a
ni, j
i 1 i jF 0
Since,
and
Also,
systemCM CM
dp dM V Ma
dt dt
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Hence, systemext
dPF
dt
systemext
dPF
dt
system CMP MV,,,,,,,,,,,,, ,
ext CMF Ma
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Kinetic Energy and Linear Momentum
P mv [Linear Momentum],,,,,,,,,,,,, ,
Then, P.P mv . mv,,,,,,,,,,,,,,,,,,,,,,,,,, ,,
2 2 2
2 2 2
22
P m v
P m v2 2P 1
mv KE2m 2
2PKE [Kinetic Energy]
2m
2PKE [Kinetic Energy]
2m
P 2m KE [Linear Momentum] P 2m KE [Linear Momentum]
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Conservation of Momentum
If no net external force acts on a system,the linear momentum of the system remains constant.
Fext
a
ext CMF Ma
If Fext=0 then,
ext
Since,
dP dPF 0
dt dt
P cons tant [Conservation of Linear Momentum]P cons tant [Conservation of Linear Momentum]
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Question
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Illustrative problem
3v2
)d(3v4
)c(
v23
)b(v4)a(
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?
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Solution
V23
V2
Let the masses be m and 2m afterexplosion.
In an explosion the momentum
remains constant.
So. 3m x V=m x 0+2mV2
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Newton’s Third Law and Conservation of Linear momentum
Fext
1 2P P cons tant
1 2dP dP0
dt dt
1 2
1 2
1 21 2
dv dvm m 0
dt dt
m a m a 0
,,,,,,,,,,,,,,,,,,,,,,,,,,,,
1 2F F 0
1 2F F
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Impulse-Momentum Theorem for a System of Particles
From Newton’s second law:
systemext
dPF
dt
systemext
dPF
dt
We can write,
system extdP F dt
tPf 2
tPi 1
t2
fi
t1
Or,
dP F t dt
P P F t dt
t2
fit1
J P P F t dt
t2
fit1
J P P F t dt
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Impulse-Momentum Theorem
tf
if
ti
J P P F dt
Area under F t curve
,,,,,,,,,,,,,, tf
if
ti
J P P F dt
Area under F t curve
,,,,,,,,,,,,,,
tf
ti
area
Fdt
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Class Test
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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
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Solution
We know that if P is the linear momentum of the particle, then
2P
KE2m
Hence answer is (a)
22 2 1
1 1 2
KE P m
KE P m
22
1
P4 4
1 P 1
22
1
P1
P
2
1
P1
P
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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
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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)
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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
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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)
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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 4v(a) (b)
(A – 4) (A 4)
v v(c) (d)
(A – 4) (A 4)
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Solution
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
4vv ' –
(A – 4)
Before emission:
u = 0
A
After emissionV’ v
A - 4 4
a
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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?
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Solution
At the highest point velocity has only horizontal component.
Since there are no external forces acting in horizontal direction, momentum is conserved.
M M
Mv cos ( v cos ) (v ')2 2
Pf = Pi
3 M
Mv cos v´2 2
v´ = 3v cos
x
y
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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?
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Solution
Change in momentum P = m(vf – vi)
150
(60 40)1000
= 15 N-s
Impulse = P = F·t
–3P 15
F 3000 Nt 5 10
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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?
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Solution
m = 2 kg vi = 25 m/s vf = 10 m/s
Change in momentum P = m(vf – vi)
= 2(10 + 25)
= 70
(a) Impulse = +70 N-s
(b) Now, impulse = F· t
70 N-S = F × (0.020) 70
F 3500 N0.020
Force on the floor from the ball = 3500 N
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Class Exercise - 8
0 2 4 6
Force
Force (N)
Time
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 withapproximately 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?
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Solution
Impulse = Area under F-t graph
max max max1 1
2 F F 2 2 F2 2
= 4 × Fmax
Now, impulse = P = m(vf – vi)
100
(50 50)1000
= 10 N-S
max 310
F 2500 N4 10
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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?
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Solution
V = 60 m /sf
V = 50 m /si
Y
X
(a) Change in KE
2 2f i
1 1mv – mv
2 2
2 21m(60 – 50 )
2
1
2100 10 1102
= 10500 × 110
= 1155 × 103 J
= 1155 kJ
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Solution
(b) Pi = mvi
2100 50 j
fP 2100 60 i
f iP P – P 21 6 i – 5 j KN-S
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Thank you