momentumlinearinchange

G

G

impulselinear

t

t

GGGGddtF 12

2

1

2

1

GddtFFdt

Gd

The product of force and time is defined as the linear impulse of the force, and this equation

states that the total linear impulse on m equals the corresponding change in linear

momentum of m.

We may write the basic equation of motion for the particle, as

vmamF

or

GFvmGvmdt

dvmF

Alternatively, we may write

21 GdtFG

I

The initial linear momentum of the body plus the linear impulse

applied to it equals its final linear momentum.

m v1+ =

11 vmG

dtF

22 vmG

xxxxx

t

t

x GGGmvmvdtF 12

2

1

12

yyyyy

t

t

y GGGmvmvdtF 12

2

1

12

zzzzz

t

t

z GGGmvmvdtF 12

2

1

12

Conservation of Linear Momentum

If the resultant force on a particle is zero during an interval of time, its linear

momentum G remains constant. In this case, the linear momentum of the particle is

said to be conserved. Linear momentum may be conserved in one direction, such as x,

but not necessarily in the y- or z- direction.

210 GGG

21 vmvm

This equation expresses the principle of conservation of linear momentum.

PROBLEMS

1. The 9-kg block is moving to the right with a velocity of 0.6 m/s on a horizontal

surface when a force P is applied to it at time t=0. Calculate the velocity v of the block

when t=0.4 s. The kinetic coefficient of friction is mk=0.3.

SOLUTION

)3.88(3.0

3.88)81.9(900

NF

N N mgN F

kf

y

m

motionx

y

P

W=mg

N Ff=mkN

smvv

vdtdtdt

mvmvFdt

direction x in

t

/823.14.59)4.0(49.26)2.0(36)2.0(72

)6.0(9)3.88(3.03672

22

2

4.0

0

4.0

2.0

2.0

0

10

2

Impulse-

Momentum2. Car B is initially stationary and is struck by car Amoving with initial speed v1 = 30 km/h. The carsmove together with speed v′ after the collision. If thetime duration of the collision is 0.1 s, determine thecommon final speed v′, (b) the magnitude R of theaverage force exerted by each car on the other carduring the impact. brakes are released during thecollision.

x

y

R

W=mBg

NB

h/kms/m.v

v.

vmmvm BAAA

205555

900180063

301800

B

NR

..RvmFdtvmt

BBB

50000

55559001000

(a)

(b)

3.

smvv

vmmvGGdtFx

/91.1901.0314.060014.0 22

2112

0

121212 eegg VVVVTTE

kJE

JTTE

18.17

74.1718160014.02

191.19044.0

2

1 22

12

Impulse-

Momentum

PROBLEMS

4. A tennis player strikes the tennis ball with her racket while the ball is still rising.

The ball speed before impact with the racket is v1=15 m/s and after impact its speed

is v2=22 m/s, with directions as shown in the figure. If the 60-g ball is in contact with

the racket for 0.05 s, determine the magnitude of the average force R exerted by the

racket on the ball. Find the angle b made by R with the horizontal.

N R R

tR

mvmvdtF

xx

x

x

t

xx

53.42127.205.0

10cos1506.020cos2206.005.0

0

10

2

SOLUTION

Rx

Ry

R

W=mg

Rx

Ry R b

in x direction

in y direction

x

1v

2v

10°

20°

xv1

yv1

xv2yv2

y

N R

N R R

ttR

mvmvdtF

yy

y

y

t

yy

02.43

49.6325.005.0

10sin1506.020sin2206.0)81.9(06.005.0

0

05.0

0

10

2

68.8tan bb R

R

x

y

v1=15 m/s

v2=22 m/s

PROBLEMS

5. The 40-kg boy has taken a running jump from the upper surface and lands on his 5-

kg skateboard with a velocity of 5 m/s in the plane of the figure as shown. If his impact

with the skateboard has a time duration of 0.05 s, determine the final speed v along the

horizontal surface and the total normal force N exerted by the surface on the

skateboard wheels during the impact.

(mB+mS)g

N

y

x

s/m.vvcos

vmmvmvm SBSxSBxB

853540030540

Linear momentum is conserved in x-direction;

kNNorNN

N

dtgmmNvmvm SBSySByB

44.22440

005.081.94505.0030sin540

0

05.0

0

in y direction

y

GrvmrHo

Moment of the resultant force about the origin O is the vector cross productF

vmrFrM o

We now differentiate with time, using the rule for the differentiation

of a cross product and obtain

vmrHo

oM

amrmr

o vmrvmrvmrdt

dH

0

oo HM

21

2

1

o

t

too HdtMH

o

momentumangularinchange

impulseangulartotal

t

t

o HvmrvmrdtM

1122

2

1

F

Conservation of Angular Momentum

210 OOo HHH

This equation expresses the principle of conservation of angular momentum.

PROBLEMS

1. The assembly starts from rest and reaches an angular speed of

150 rev/min under the action of a 20 N force T applied to the

string for t seconds. Determine t. Neglect friction and all masses

except those of the four 3-kg spheres, which may be treated as

particles.

st

t

HHdtM

sphere

linkspherepulley

v

rmrT

t

tzzz

08.15

4.060

21504.0341.020

2

112

SOLUTION

z

v

v

v

v

PROBLEMS

2. A pendulum consists of two 3.2 kg concentrated masses positioned as shown on a

light but rigid bar. The pendulum is swinging through the vertical position with a

clockwise angular velocity w=6 rad/s when a 50-g bullet traveling with velocity v=300

m/s in the direction shown strikes the lower mass and becomes embedded in it.

Calculate the angular velocity w which the pendulum has immediately after impact and

find the maximum deflection q of the pendulum.

SOLUTION

Angular momentum is conserved during impact;

21

0

0

,02112

vmrvmrM

HH HHdtM

O

OOOO

t

O

)(/77.2

2.02.02.34.04.02.3050.0

4.04.062.32.02.062.34.020cos300050.0

22

11

ccwsrad

BA

AB

vv

vv

w

ww

(1)

(3)

w

w’

O

1Av

A

B(3)

(1)

B

A

(1) (2)

(2)(1)

2Av

2Bv1Bv

(2)

(2)

Energy considerations after impact;

3322 gg VTVT (Datum at O)

Reference line

o.

cos....cos...

..............

152

819400502381920230

8194005023819202377220232

17724023050

2

1 22

q

qq

q

q

w’

O

2Av(2)

(3)

(3)(2)

2Bv