chapter 3 matriculation stpm

PHYSICS CHAPTER 3

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MOMENTUM AND IMPULSE

CHAPTER 3

PHYSICS CHAPTER 3

3.0 MOMENTUM AND IMPULSE

3.1 Momentum and impulse

3.2 Conservation of linear momentum

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PHYSICS CHAPTER 3

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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:

DefineDefine momentum.

DefineDefine impulse and use F-t graph to determine impulse

Use Use

Learning Outcome:3.1 Momentum and impulse

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3.1.1 Linear momentum,

is defined as the product between mass and velocitythe product between mass and velocity. is a vector quantity. Equation :

The S.I. unit of linear momentum is kg m skg m s-1-1. The direction of the momentumdirection of the momentum is the samesame as the direction direction

of the velocityof the velocity. It can be resolve into vertical (y) component and horizontal (x)

component.

p

vmp =

xp

pyp

θ

θmvθppx coscos ==θmvθppy sinsin ==

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3.1.2 Impulse, Let a single constant force, constant force, FF acts on an object in a short time

interval (collision), thus the Newton’s 2nd law can be written as

is defined as the product of a force, the product of a force, FF and the time, and the time, tt OR the change of momentumthe change of momentum.

is a vector quantityvector quantity whose directiondirection is the samesame as the constant forceconstant force on the object.

J

constant ===∑ dtpdFF

12 pppddtFJ −===

momentum final: 2pwheremomentum initial: 1p

force impulsive :F

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The S.I. unit of impulse is N sN s or kg m skg m s−−11. If the forceforce acts on the object is not constantnot constant then

Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :

dtFdtFJ av

t

t

== ∫ 2

1

where force impulsive average :avF

( ) ( )xxx1x2xavx uvmppdtFJ −=−==

( ) ( )yyy1y2yavy uvmppdtFJ −=−==

( ) ( )zzz1z2zavz uvmppdtFJ −=−==

consider 2-D consider 2-D collision onlycollision only

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When two objects in collision, the impulsive force, F against time, t graph is given by the Figure 3.20.

1t 2tFigure 3.20Figure 3.20 t0

F

Shaded area under the F−t graph = impulse

Picture 3.1

Picture 3.2

Picture 3.3

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A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s−1 and it bounces off with a speed of 70 m s−1 in the opposite direction.a. Calculate the magnitude of impulse delivered to the ball by the wall,b. If the ball is in contact with the wall for 10 ms, determine the magnitude of average force exerted by the wall on the ball.Solution :Solution :

Example 3.1 :

Wall (2)11

1s m 100 −=1u

111s m 70 −=1v

022 == uv

kg 0.201 =m

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Solution :Solution :a. From the equation of impulse that the force is constant,

Therefore the magnitude of the impulse is 34 N s34 N s.

b. Given the contact time,

12 ppdpJ −==( )111 uvmJ −=

dtFJ av=

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An estimated force-time curve for a tennis ball of mass 60.0 g struck by a racket is shown in Figure 3.21. Determinea. the impulse delivered to the ball,b. the speed of the ball after being struck, assuming the ball is being served so it is nearly at rest initially.

Example 3.2 :

0.2 1.8 ( )mst0

( )kNF

1.0

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Figure 3.21Figure 3.21

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Solution :Solution :a. From the force-time graph,

b. Given the ball’s initial speed,

graph under the area tFJ −=

0=u( )uvmdpJ −==

kg 1060.0 3−×=m

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1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m.a. Calculate the impulse delivered to the ball during impact.b. If the ball is in contact with the slab for 2.00 ms, determine the average force on the ball during impact.

ANS. : 0.47 N s; 237. 1 NANS. : 0.47 N s; 237. 1 N2. A golf ball (m = 46.0 g) is struck with a force that makes an

angle of 45° with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.)

ANS. : 293 NANS. : 293 N

Exercise 3.1 :

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3.

A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s−1 strikes a wall at a 45° angle and rebounds with the same speed at 45° as shown in Figure 3.22. Calculate the impulse given by the wall.

ANS. : 2.4 N s to the left or ANS. : 2.4 N s to the left or −−2.4 N s2.4 N s

Exercise 3.1 :

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3.2 Conservation of linear momentum

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3.2.1 Principle of conservation of linear momentum states “In an isolated (closed) system, the total momentum In an isolated (closed) system, the total momentum

of that system is constantof that system is constant.” OR“When the net external force on a system is zero, the total When the net external force on a system is zero, the total momentum of that system is constantmomentum of that system is constant.”

In a Closed system,

From the Newton’s second law, thus

0 ==∑ dtpdF

0=∑ F

0=pd

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According to the principle of conservation of linear momentum, we obtain

OR

The total of initial momentum = the total of final momentumThe total of initial momentum = the total of final momentum

∑∑ = fi pp

constant=p

constant=∑ xpconstant=∑ yp

Therefore then

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Linear momentum in one dimension collisionLinear momentum in one dimension collisionExample 3.3 :

Figure 3.14 shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed of 2 m s-1 to the left. Determine the velocity of A after Collision.SolutionSolution : :

1s m 6 −=Au

AB

1s m 3 −=Bu


∑∑ = fi pp

1s m 6 ;kg 0.100 ;kg 0.200 −−=== ABA umm11 s m 2 ;s m 3 −− −== BB vu

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Linear momentum in two dimension collisionLinear momentum in two dimension collisionExample 3.4 :

A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the tennis ball is deflected 50° from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s−1 and v1 = 4 m s−1. Calculate the magnitude and direction of soccer ball after the collision.


1u

Before collision After collision

m1 m2

m1 1v

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Simulation 3.3

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Solution :Solution :

From the principle of conservation of linear momentum,

The x-component of linear momentum,

∑∑ = fi pp

x22x11x22x11 vmvmumum +=+

;s m 20 ;kg 0.900 ;kg 0.250 1−=== 121 umm0 ;s m 4 ;0 1 5θvu 112 === −

∑∑ = fxix pp

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Solution :Solution :The y-component of linear momentum,

Magnitude of the soccer ball,

Direction of the soccer ball,

yy vmvm 22110 +=∑∑ = fyiy pp

( ) ( )2y2

2x22 vvv +=

=

= −−

4.840.851tantan 1

x2

y212 v

vθ

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1. An object P of mass 4 kg moving with a velocity 4 m s−1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s−1 towards it.a. Determine the total momentum before collision.b. If P immediately stop after the collision, calculate the final velocity of Q.c. If the two objects stick together after the collision, calculate

the final velocity of both objects.ANS. : 10 kg m sANS. : 10 kg m s−−11; 5 m s; 5 m s−−11 to the right; 1.7 m s to the right; 1.7 m s−−11 to the right to the right2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his

hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1. Determinea. the recoil velocity of the rifle,b. the final momentum of the system.

ANS. : ANS. : −−0.5 m s0.5 m s−−11; U think.; U think.

Exercise 3.2 :

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3.

In Figure 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s−1 and 1.40 m s-1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determinea. the speed of the bullet immediately after it emerges from the first block andb. the initial speed of the bullet.ANS. : 721 m sANS. : 721 m s−−11; 937.4 m s; 937.4 m s−−11


1.20 kg 1.80 kg

0.630 m s-1 1.40 m s-1

Before

After

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4. A ball moving with a speed of 17 m s−1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45° from its original direction, and the struck ball moves off at 30° from the original direction as shown in Figure 3.17. Calculate the speed of each ball after the collision.

ANS. : 8.80 m sANS. : 8.80 m s−− 1 1; 12.4 m s; 12.4 m s−−11

Exercise 3.2 :

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3.2.2 Collision

is defined as an isolated event in which two or more bodies an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each (the colliding bodies) exert relatively strong forces on each other for a relatively short timeother for a relatively short time.

Two types of collisions :

Elastic collisionElastic collision

Inelastic (non-elastic) collisionInelastic (non-elastic) collision

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Elastic collisionElastic collision is defined as one in which the total kinetic energy (as well as one in which the total kinetic energy (as well as

total momentum) of the system is the same before and after total momentum) of the system is the same before and after the collisionthe collision.

Figure 3.18 shows the head-on collision of two billiard balls.

11 22

Before collision

At collision

After collision

11 2222um11um

11 2222vm11vm


Simulation 3.4

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The properties of elastic collisionproperties of elastic collision are

a. The total momentum is conservedtotal momentum is conserved.

b. The total kinetic energy is conservedtotal kinetic energy is conserved.

OR

∑∑ = fi pp

∑∑ = fi KK

222

211

222

211 vmvmumum

21

21

21

21 +=+

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Inelastic (non-elastic) collisionInelastic (non-elastic) collision is defined as one in which the total kinetic energy of the one in which the total kinetic energy of the

system is not the same before and after the collision (even system is not the same before and after the collision (even though the total momentum of the system is conserved)though the total momentum of the system is conserved).

Figure 3.19 shows the model of a completely inelastic completely inelastic collisioncollision of two billiard balls.

11 22At collision

After collision (stick together)

11 22v


Before collision 11 2211um 0=2u

2m

Simulation 3.5

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Caution: Not allNot all the inelastic collision is stick togetherstick together. In fact, inelastic collisions include many situationsmany situations in which

the bodies do not stickbodies do not stick

The properties of inelastic collisionproperties of inelastic collision area. The total momentum is conservedtotal momentum is conserved.

b. The total kinetic energy is not conservedtotal kinetic energy is not conserved because some of the energy is converted to internal energyinternal energy and some of it is transferred away by means of sound or heatsound or heat. But the total total energy is conservedenergy is conserved.

OR

∑∑ = fi pp

∑∑ = fi EE energy losses += ∑∑ fi KK

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Ball A of mass 400 g and velocity 4 m s-1 collides with ball B of mass 600 g and velocity 10 m s-1. After collision, A and B will move together. Determine the final velocity of both balls if A and B moves in the opposite direction initiallySolution : Solution : mmAA = 0.4 kg, u = 0.4 kg, uAA = 4 m s = 4 m s-1 -1 , m, mB B = 0.6 kg, u= 0.6 kg, uB B = -10 m s= -10 m s-1-1, , inelastic collisioninelastic collision

By using the principle of conservation of linear momentum, thus

Example 3.5 :

Before collision AA BBBuAu

After collision AA BB?=v

∑∑ = fi pp

vmmumum BABBAA )( +=+

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Final velocity of both balls is - 4.4 m s -1

BA

BBAA

mmumumv

++=

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A ball A of mass 1 kg moving at a velocity of 4 m s-1 collides with ball B of mass 2 kg which at rest. Calculate the velocity of both balls after collision if the collision is an elastic collision. Solution :Solution :

Given mA = 1 kg, uA = 4 m s-1, mB = 2 kg, uB = 0 m s-1, elastic collision

Example 3.6 :

Before collision AA BB

10 −= msuBAu

After collision AA BB?=Bv?=Av

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Solution :Solution :Apply principle of conservation of momentum,

Apply principle of conservation of kinetic energy,

BBAABBAA vmvmumum +=+)(2)(1)0(2)4(1 BA vv +=+

222 )(21)(

21)(

21

BBAAAA vmvmum +=

222 )()()( BBAAAA vmvmum +=

1B s m v2 -4 −=Av ……..(1)

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222 )(2)(1)4(1 BA vv +=Solution :Solution :

2216 BA vv += ………..(2)

Substitute equation (1) into equation (2)22 2)24(16 BB vv +−=

22 2)2(416 BB vv +−=22)2(28 BB vv +−=

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Substitute into equation (1),

238 BB vv =

167.2 −= msvB

(2.67) 2 -4=Av


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THE END…Next Chapter…

CHAPTER 4 :Forces

chapter 3 matriculation stpm

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