HW4 Solutions Impulse and Momentum

Introduction to Dynamics (N. Zabaras)

HW4 Solutions Impulse and Momentum

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

April 15, 2016

1

mailto:[email protected]

http://www.zabaras.com/


Problem 1

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.

SOLUTION:

• Apply the principle of impulse and

momentum to the package-cart

system to determine the final

velocity.

• Apply the same principle to the

package alone to determine the

impulse exerted on it from the

change in its momentum.

2


Problem 1

SOLUTION:

• Apply the principle of impulse and momentum to the package-cart

system to determine the final velocity.

2211 vmmvm cpp

Imp

x

y

x components:

1 2

2

cos30 0

10 kg 3 m/s cos30 10 kg 24 kg

p p cm v m m v

v

2 0.764 m/sv

3


Problem 1

• Apply the same principle to the package alone to determine the

impulse exerted on it from the change in its momentum.

x

y

2211 vmvm pp

Imp

x components:

2

21

kg 1030cosm/s 3kg 10

30cos

vtF

vmtFvm

x

pxp

sN56.18 tFx

y components:

030sinm/s 3kg 10

030sin1

tF

tFvm

y

yp

sN15 tFy

sN 9.23sN 51sN 56.1821 tFjitF

Imp

4


Problem 1

To determine the fraction of energy lost,

J 63.9sm742.0kg 25kg 10

J 45sm3kg 10

2

212

221

1

2

212

121

1

vmmT

vmT

cp

p

786.0J 45

J 9.63J 45

1

21

T

TT

5


Problem 2

m=100 kg

At rest smooth

t=10 s

V2=?

N=?

2

1

1 2( ) ( ) ( )

t

x x x

t

m F dt m

2

2

0 200 cos 45 (10 ) (100 )

14.1 /

oN s kg

m s

2

1

1 2( ) ( )

t

y y y

t

m F dt m

0 (10 ) 981 (10 ) 200 (10 )sin 45 0

840

o

c

c

N s N s N s

N N

0yF

6

The 100-kg crate shown is originally

at rest on the smooth horizontal

surface. If a towing force of 200 N.

acting at an angle of 45°, is applied for

10 s, determine the final velocity and

the nominal force which the surface

exerts on the crate during this time

interval.


Problem 3

W=50 Ib

P=(20t) Ib

V2=?

T=2 sec.

V1=3 ft/s

mk=0.3

2

1

1 2( ) ( )

t

x x x

t

m F dt m

2

2

0

2

50 50(3) 20 0.3 (2) (50sin 30 )(2)

32.2 32.2

4.66 40 0.6 50 1.55

o

c

c

tdt N

N

0yF 50cos30 0o

cN Ib 2

43.3

44.2 /

cN Ib

ft s

7

The 50-lb crate shown is acted

upon by a force having a variable

magnitude P= (20t) lb, where t is in

secs. Determine the crate's velocity

2 s after P has been applied. The

initial velocity is v1=3 ft/s down the

plane, and the coefficient of kinetic

friction is mk=0.3.


Problem 4WC =1200-Ib

Wp = 8-Ib

vp=1500ft/s

t = 0.03 s.

vc2= ?

Favg= ?

1 1 2 2( ) ( ) ( ) ( ) ( )c c p p c c p pm m m m

2

1200 80 0 ( ) (1500)

32.2 32.2c

2( ) 10 /c ft s

1 2( ) ( ) ( ) ( )p avg pm F t m

80 (0.03) (1500)

32.2avgF

Favg

12400avgF Ib

8

The 1200-lb cannon shown

fires an 8-lb projectile with a

muzzle velocity of 1500 ft/s

measured relative to the

cannon. If firing takes place in

0.03 sec, determine the recoil

velocity of the cannon just

after firing. The cannon support

is fixed to the ground, and the

horizontal recoil of the cannon

is absorbed by two springs.

Cannon + Projectile

Projectile


Problem 5

mp = 800 kg

mH = 300 kg

From rest

Impulse = ?

Couple together

9

An 800-kg rigid pile shown above is driven into the

ground using a 300-kg hammer. The hammer falls

from rest at a height y0=0.5 m and strikes the top of

the pile. Determine the impulse which the pile

exerts on the hammer if the pile is surrounded

entirely by loose sand so that after striking, the

hammer does not rebound off the pile.


1 1 2 2T V T V

1( ) 3.13 /H m s

1 1 2( ) ( ) ( ) ( )H H p p H pm m m m 2(300)(3.13) 0 (300 800) 2 0.854 /m s

2

1

t

y avg

t

F dt R dt F t Impulse

(300)(3.13) (300)(0.854)R dt 683 .R dt N sImpulse =

2

1

10 (300)(9.81)(0.5) (300)( ) 0

2H

1 2( ) ( )H Hm Rdt m

Problem 5mp = 800 kg

mH = 300 kg

From rest

Impulse = ?

Couple together

10

The velocity at which H strikes the pile

can be determined using conservation of

energy applied to H.

Conservation of momentum for H+P

Principle of impulse for H

2 2

0 0 1 1

1 1( ) ( )

2 2H H H H H Hm W y m W y


Problem 6

The magnitude and direction of

the velocities of two identical

frictionless balls before they

strike each other are as shown.

Assuming e = 0.9, determine the

magnitude and direction of the

velocity of each ball after the

impact.

SOLUTION:

• Resolve the ball velocities into

components normal and tangential to

the contact plane.

• Tangential component of momentum

for each ball is conserved.

• Total normal component of the

momentum of the two ball system is

conserved.

• The normal relative velocities of

the balls are related by the

coefficient of restitution.

• Solve the last two equations

simultaneously for the normal velocities

of the balls after the impact.

11


Problem 6

SOLUTION:

• Resolve the ball velocities into components normal and

tangential to the contact plane.

sft0.2630cos AnA vv sft0.1530sin AtA vv

sft0.2060cos BnB vv sft6.3460sin BtB vv

• Tangential component of momentum for each ball

is conserved.

sft0.15tAtA vv sft6.34

tBtB vv

• Total normal component of the momentum of the

two ball system is conserved.

0.6

0.200.26

nBnA

nBnA

nBBnAAnBBnAA

vv

vmvmmm

vmvmvmvm

12


Problem 6

6.557.23

6.34tansft9.41

6.347.23

3.407.17

0.15tansft2.23

0.157.17

1

1

B

ntB

A

ntA

v

v

v

v

t

n

• The normal relative velocities of the balls are related

by the coefficient of restitution.

4.410.200.2690.0

nBnAnBnA vvevv

• Solve the last two equations simultaneously for the

normal velocities of the balls after the impact.

sft7.17nAv sft7.23

nBv

13


Problem 7

Ball B is hanging from an

inextensible cord. An identical ball

A is released from rest when it is

just touching the cord and acquires

a velocity v0 before striking ball B.

Assuming perfectly elastic impact (e

= 1) and no friction, determine the

velocity of each ball immediately

after impact.

SOLUTION:

• Determine orientation of impact line of

action.

• The momentum component of ball A

tangential to the contact plane is

conserved.

• The total horizontal momentum of the

two ball system is conserved.

• The relative velocities along the line

of action before and after the impact

are related by the coefficient of

restitution.

• Solve the last two expressions for the

velocity of ball A along the line of

action and the velocity of ball B which

is horizontal.

14


Problem 7SOLUTION:

• Determine orientation of impact line of

action.

30

5.02

sin

r

r

• The momentum component of ball A

tangential to the contact plane is

conserved.

0

0

5.0

030sin

vv

vmmv

vmtFvm

tA

tA

AA

• The total horizontal (x component)

momentum of the two ball system is

conserved.

0

0

433.05.0

30sin30cos5.00

30sin30cos0

vvv

vvv

vmvmvm

vmvmtTvm

BnA

BnA

BnAtA

BAA

15

Since B is constrained to move in a

circle of center C, its velocity after

impact needs to be horizontal.Bv


Problem 7

• The relative velocities along the line of action

before and after the impact are related by the

coefficient of restitution.

0

0

866.05.0

030cos30sin

vvv

vvv

vvevv

nAB

nAB

nBnAnAnB

• Solve the last two expressions for the velocity of

ball A along the line of action and the velocity of

ball B which is horizontal.

00 693.0520.0 vvvv BnA

0

10

00

693.0

1.16301.46

1.465.0

52.0tan721.0

520.05.0

vv

vv

vvv

B

A

ntA

16


Problem 8

A 30 kg block is dropped from a

height of 2 m onto the 10 kg pan of

a spring scale. Assuming the

impact to be perfectly plastic,

determine the maximum deflection

of the pan. The constant of the

spring is k = 20 kN/m.

SOLUTION:

• Apply the principle of conservation of

energy to determine the velocity of

the block at the instant of impact.

• Since the impact is perfectly plastic,

the block and pan move together at

the same velocity after impact.

Determine that velocity from the

requirement that the total momentum

of the block and pan is conserved.

• Apply the principle of conservation

of energy to determine the

maximum deflection of the spring.

17


Problem 8SOLUTION:

• Apply principle of conservation of energy to

determine velocity of the block at instant of

impact.

sm26.6030 J 5880

030

J 588281.9300

2222

1

2211

2222

1222

12

11

AA

AAA

A

vv

VTVT

VvvmT

yWVT

• Determine velocity after impact from requirement

that total momentum of the block and pan is

conserved.

sm70.41030026.630 33

322

vv

vmmvmvm BABBAA

18


Problem 8

Initial spring deflection due to

pan weight:

m1091.4

1020

81.910 3

33

k

Wx B

• Apply the principle of conservation of energy to

determine the maximum deflection of the spring.

2

43

213

4

24

3

21

34

242

14

4

233

212

321

3

2

212

321

3

10201091.4392

1020392

0

J 241.01091.410200

J 4427.41030

xx

xxx

kxhWWVVV

T

kx

VVV

vmmT

BAeg

eg

BA

m 230.0

10201091.43920241.0442

4

24

3

213

4

4433

x

xx

VTVT

m 1091.4m 230.0 334

xxh m 225.0h

19


2

0 1 0 2

1 1 2 2

12 1

2

Angular momentum about O is conserved,

since, none of the forces produce an angular

impulse about this axis.

angular momentum v

( ) ( )

'

0.5' 1.5 1.07 ( / )

0.7

r m v r m v

rv v m s

r

H H

1 1 2 2

2 2 2

2

2

2

2

2 2 2 2

2 2 2

Conservation of Energy

1 1 12 (1.5) 0 2 ( ) 20 (0.2)

2 2 2

2.25 0.4 ( )

1.36 /

" ' 1.36 1.07 0.839 /

T V T V

v

v

v m s

v v v m s

Problem 920.7 ?

rate at which the cord is being stretched

r v

20

A 2-kg disk rests on a smooth horizontal

surface and is attached to an elastic cord

with k=20 N/m and is initially unstretched.

If the disk is given a velocity (VD)1 = 1.5

m/s, perpendicular to the cord, determine

the rate at which the cord is being stretched

and the speed of the disk when the cord is

stretched 0.2 m.


1 2( ) ( )Z Z ZH M dt H

4

2

0

50 (0.5 0.8) 4[( )(0.6 )]

32.2t dt v

2

2

7.2 0.3727

19.3 /

v

v ft s

Problem 10

m = 5-Ib

v1 = 0 ft/s

M = ( 0.5 t + 0.8 ) Ib.ft

v2= ?

t = 4 s

21

The four 5-lb spheres are rigidly attached

to the crossbar frame having a negligible

weight. If a couple moment M=(0.51 +

0.8t) lb · ft, where t is in secs, is

applied, determine the speed of each of the

spheres in 4 secs starting from rest.

Neglect the size of the spheres.

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HW4 Solutions Impulse and Momentum

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