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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.

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Presentation on theme: "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."— Presentation transcript:

1 Physics

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

3 Session Objectives

4 Session Objective 1.Impulse – Momentum Theorem 2.Elastic Collisions 3.Inelastic Collisions

5 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

6 Linear Momentum of a System of Particles vivi mimi

7 P i =m i v i If total mass is M, then Hence,

8 Questions

9 u vv Illustrative Example

10 Newtons Second Law and Linear Momentum Rate of change of momentum is equal to the net force acting on the particle. F ext a Since, and Also, Hence,

11 Kinetic Energy and Linear Momentum Then,

12 Conservation of Momentum If no net external force acts on a system,the linear momentum of the system remains constant. F ext a If F ext =0 then,

13 Question

14 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?

15 Solution Let the masses be m and 2m after explosion. In an explosion the momentum remains constant. So. 3m x V=m x 0+2mV 2

16 Newtons Third Law and Conservation of Linear momentum F ext

17 Impulse-Momentum Theorem for a System of Particles From Newtons second law: We can write,

18 Impulse-Momentum Theorem

19 Class Test

20 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

21 Solution We know that if P is the linear momentum of the particle, then Hence answer is (a)

22 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

23 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)

24 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

25 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)

26 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

27 Solution Since there are no external forces acting on the system, hence P f = P i A(u) = (A – 4)v´ + 4v A(0) = (A – 4)v´ + 4v Before emission: u = 0 A After emission V v A - 4 4 a

28 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?

29 Solution At the highest point velocity has only horizontal component. Since there are no external forces acting in horizontal direction, momentum is conserved. P f = P i v´ = 3v cos x y

30 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?

31 Solution Change in momentum P = m(v f – v i ) = 15 N-s Impulse = P = F· t

32 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?

33 Solution m = 2 kgv i = 25 m/s v f = 10 m/s Change in momentum P = m(v f – v i ) = 2(10 + 25) = 70 (a) Impulse = +70 N-s (b) Now, impulse = F· t 70 N-S = F × (0.020) Force on the floor from the ball = 3500 N

34 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 with approximately the same speed perpendicular to the wall as was in the case of impact. What is F max, the maximum magnitude of the force on the ball from the wall during the collision?

35 Solution Impulse = Area under F-t graph = 4 × F max Now, impulse = P = m(v f – v i ) = 10 N-S

36 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?

37 Solution (a)Change in KE = 10500 × 110 = 1155 × 10 3 J = 1155 kJ

38 Solution (b)P i = mv i

39 Thank you


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