Physics. A boxer wisely moves his head backward just before receiving a punch. How does this tactics...

Physics

A boxer wisely moves his head backward just before receiving a punch. How does this tactics help reduce the force of impact?

Session Opener

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 = mvP = mv

Linear Momentum of a System of Particles

vimi

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,

Questions

uv v

Illustrative Example

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

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]

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]

Question

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?

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

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

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

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

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

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

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 4v(a) (b)

(A – 4) (A 4)

v v(c) (d)

(A – 4) (A 4)

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

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.

M M

Mv cos ( v cos ) (v ')2 2

Pf = Pi

3 M

Mv cos v´2 2

v´ = 3v cos

x

y

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 P = m(vf – vi)

150

(60 40)1000

= 15 N-s

Impulse = P = F·t

–3P 15

F 3000 Nt 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 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

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?

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

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

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

Solution

(b) Pi = mvi

2100 50 j

fP 2100 60 i

f iP P – P 21 6 i – 5 j KN-S

Thank you