13.10 Principle of Impulse and Momentum p p

Preview of 13.10- 13.15

13.10 Principle of Impulse and Momentum p p

• From Newton’s second law, d ( )

r

• What do we want to know?

( ) =

= m v m v

dt

F r d r r

linear momentum

( ) m v d

dt

F r = r

• Definition of Impulse of a force?

( )

1 2

2

1

v m v

m dt

F

t t

r r = r −

• Relationship of the impulse with the

∫

motion?

F dt

F

t

r r

force the

of impulse

2 1

2

=

=

∫ ^Imp

t1

t₁

• The final momentum of the particle ÆThe final momentum of the particle Æ

Initial momentum + Impulse of the force during the time interval.

• Units/Dimension for the impulse of a force are

( ^{kg m s} ) ^{s kg m s}

s

N ⋅ = ⋅

⋅ = ⋅ m v r

= m v r

+ Imp

13.11 Impulsive Motion p

Impulsive force:

• Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum.

2

1

F t m v

v

m v r

+ ∑ F r Δ t = m v r

m + ∑ Δ

• Nonimpulsive forces are forces for which

is small and therefore, may be neglected.

t F r Δ

• Examples?

Direct Newton’s Method: Work and Energy Method: Impulse and Momentum Method:

Sample Problem 13.11 p

SOLUTION:

• Apply the principle of impulse and momentum in terms of horizontal and vertical component equations. p q

A 0.5 kg baseball is pitched with a

velocity of 80 m/s. After the ball is hit y

by the bat, it has a velocity of 120 m/s

in the direction shown. If the bat and

ball are in contact for 0.015 s, deter-

ball are in contact for 0.015 s, deter

mine the average impulsive force

exerted on the ball during the impact.

Sample Problem 13.11 p

SOLUTION:

• Apply the principle of impulse and momentum in

t f h i t l d ti l t ti

terms of horizontal and vertical component equations.

2 2

1

1 mv

v

mr +Imp _→ = r

t ti

x component equation:

( ) ( ) 0 . 5 ( 120 cos 40 )

15 0 5 80

. 0

40

2

cos

1

° +

°

= Δ +

−

F

mv t

F mv

( ) ( ) ( )

N 3 . 57

40 cos 10 120

15 . 0 10 80

=

°

= +

−

x

x

F

F

y

y component equation:

( ) ^{0 5} ( )

40 sin 0 + F

Δ t = mv

°

x

( ) ( )

N 64 . 30

40 cos 10 120

5 . 15 0

. 0

=

°

=

y y

F F

( ^{57 . 3 N} ) ( ^{+ 30 . 64 N} ) ^{, = 64 . 9 N}

= i j F

F r r r

Sample Problem 13.12 p

SOLUTION:

• Apply the principle of impulse and

• Apply the principle of impulse and momentum to the package-cart system to determine the final velocity.

A 10 k k d f h t i t

• Apply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum.

A 10 kg package drops from a chute into a 24 kg cart with a velocity of 3 m/s.

Knowing that the cart is initially at rest

and can roll freely, determine (a) the

final velocity of the cart, (b) the impulse

exerted by the cart on the package, and

(c) the fraction of the initial energy lost

in the impact.

Sample Problem 13.12 p

SOLUTION:

• Apply the principle of impulse and momentum to the package-cart t t d t i th fi l l it

system to determine the final velocity.

y

x

( )

2 1

1 m m v

v

m_pr +

∑

x components:

( )

(

)( ) (

)

0 30

cos m m v

v

m_p °+ = _p + _c

(

)(

)

(

)

m/s 742 .

2 = 0 v

13.12 Impact p

• Impact: Collision between two bodies which occurs during a small time interval and during which the bodies exert large forces on each

other

bodies exert large forces on each

other.

• Line of Impact: Common normal to the surfaces in contact during impact.g p

• Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact

Direct Central Impact an eccentric impact.

Direct Central Impact

Di t I t I t f hi h th l iti f th

• Direct Impact: Impact for which the velocities of the two bodies are directed along the line of impact.

• Oblique Impact: Impact for which one or both of the q p p bodies move along a line other than the line of impact.

Oblique Central Impact

13. 13 Direct Central Impact p

Bodies moving in the same straight line, v_A> v_B.

• Wish to determine the final velocities ofWish to determine the final velocities of the two bodies. The total momentum of the two body system is preserved

,

′

′

B B

B A

A

v m v m v m v

m + = ′ + ′

• Two unknowns!: A second relation between the final velocities is required

final velocities is required.

• Upon impact the bodies undergo a

period of deformation, at the end of which, they are in contact and moving at a

• A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed

they are in contact and moving at a common velocity.

or remain permanently deformed.

u m Pdt

v

m

− ∫ =

^m

^{u −} ∫ ^{Rdt = m}

^{v ′}

13.13 Direct Central Impact p

∫

• Coefficient of restitution

− ′

=

=

=

∫

∫ ^{Rdt u v}

n restitutio of

t coefficien e

A

• Period of deformation:

∫

1 0 ≤ ≤

∫ − e

u v

Pdt

• Period of restitution:

A A

Au m v

m

Rdt = − ′

∫

• A similar analysis of particle B yields

(Change of sign in the impulsive force) _B B

v u

u e v

−

′ −

=

• Combine the relations Æ Second relation between

the final velocities.

^{v ′}

^{− v ′}

^{= e} ( ^v

^{− v}

)

• Perfectly plastic impact, e = 0:

v

′ = v ′

= v ′ m

v

+ m

v

= ( m

+ m

) v ′

• Perfectly elastic impact e = 1:

′ ′

• Perfectly elastic impact, e = 1:

Total energy and total momentum conserved.

v ′

− v ′

= v

− v

13.14 Oblique Central Impact q p

• Final velocities are unknown in magnitude g and direction. Æ Four equations are required.

• No tangential impulse component;

( ) ( ) ^v

= ^v ′

( ) ( ) ^v

= ^v

^′

each particle is conserved.

( ) ( )

( ) ( )

N l t f t t l t

( ) ⁺ ( ) ( ) ^{′ +} ( ) ^′

• Normal component of total momentum of the two particles is conserved.

( )

( )

( )

( )

A

v m v m v m v

m + = ′ + ′

• Normal components of relative

velocities before and after impact are related by the coefficient of restitution

( ) ( ) ^v

^′

^{− v ′}

^{= e} [ ( ) ( ) ^v

^{− v}

]

related by the coefficient of restitution.

13.14 Oblique Central Impact q p

Block constrained to move along horizontal surface.

• Impulses?

Internal forces along the n axis

External force exerted by horizontal surface d di t d l th ti l t th f

F

Fr r

− and Frext

and directed along the vertical to the surface.

• Final velocity of ball unknown in direction and magnitude and unknown final block velocity g y magnitude. Æ Three equations required.

13.14 Oblique Central Impact q p

• Tangential momentum of ball is conserved

( ) ( ) ^v

= ^v

^′

conserved.

• Total horizontal momentum of block and ball is conserved.

( )

( )

( )

( )

A

v m v m v m v

m + = ′ + ′

• Normal component of relative

velocities of block and ball are related b ffi i t f tit ti

( ) ( ) ^{v ′}

^{− v ′}

^{= e} [ ( ) ( ) ^v

^{− v}

]

by coefficient of restitution.

13.15 Problems Involving Energy and Momentum g gy

• Three methods for the analysis of kinetics problems:

- Direct application of Newton’s second law Direct application of Newton s second law - Method of work and energy

- Method of impulse and momentum

• Select the method best suited for the problem or part of a problem

d id i

under consideration.

Sample Problem 13.14 p

A ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has a

magnitude v and forms angle of 30^o with the horizontal. Knowing that

e = 0.90, determine the magnitude and direction of the velocity of the ball as it rebounds from the wall.

Sample Problem 13.16 p

SOLUTION:

• Determine orientation of impact line of p action.

• The momentum component of ball A tangential to the contact plane is conserved.

Th t t l h i t l t f th

Ball B is hanging from an inextensible cord.

• The total horizontal momentum of the two ball system is conserved.

• The relative velocities along the line of

when it is just touching the cord and

acquires a velocity v₀ before striking ball B.

A i f tl l ti i t ( 1)

The relative velocities along the line of action before and after the impact are related by the coefficient of restitution.

Assuming perfectly elastic impact (e = 1) and no friction, determine the velocity of each ball immediately after impact.

• Solve the last two expressions for the velocity of ball A along the line of action and the velocity of ball B which is

13 - 18

and the velocity of ball B which is

horizontal.

Sample Problem 13.16 p

SOLUTION:

• Determine orientation of impact line of action.

5 0 i θ ^r

°

=

=

= 30

5 . 2 0

sin θ

θ r

• The momentum component of ball A tangential to the contact plane is conserved

conserved.

0sin30 0 m

( )

v m t F v

m

A t A A

= ′ +

°

= ′ Δ

+ r r

r

( )

• The total horizontal ( x component)

• The total horizontal ( x component) momentum of the two ball system is conserved.

r

( ) ( )

(

)

( )

0

30 sin 30

cos 0

v v

v

v m v

m v

m

v m v

m t T v

m

B A

n B t A

A

B A

A

− ′

′ °

−

°

=

− ′

′ °

−

′ °

=

+ ′

= ′ Δ

+ r r

r

( ) ( )

( )

5 . 0

30 sin 30

cos 5

. 0 0

v v

v

v v

v

n B A

n B A

′ =

′ +

Sample Problem 13.17 p

A 30 kg block is dropped from a height of 2 A 30 kg block is dropped from a height of 2 m onto the the 10 kg pan of a spring scale.

Assuming the impact

to be perfectly plastic, determine the p y p , maximum deflection of the pan. The constant of the spring is k = 20 kN/m.

Sample Problem 13.17 p

SOLUTION:

• Apply principle of conservation of energy to

d t i l it f th bl k t i t t f i t determine velocity of the block at instant of impact.

( )( )( )

( ) ( )( )

J 588 2

81 . 9 30 0

1 2 1 2

1

1 = = _A = =

V T

y W V

T

( ) ( )( )

( )( )

( )

J 588 0

0 30

1 2 2 2

1 1

2 2 2

1 2 2

2 1

= +

= +

+

= +

=

=

=

A A

A A

A

v v

V T

V T

V v

v m T

( )( )

( )

J 588

0 _{2 2}

2 + =

=

+ v_A v_A

• Determine velocity after impact from requirement that y p q total momentum of the block and pan is conserved.

( )

( ) (

)

m_{A A B B A B}

( )(

)

(

)