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.
PROBLEMS1. The 200 kg lunar lander is descending onto the moon’s surface with a velocity of 6 m/s when its
retro-engine is fired. If the engine produces a thrust T for 4 s which varies with the time as shown
and then cuts off, calculate the velocity of the lander when t=5 s, assuming that it has not yet landed.
Gravitational acceleration at the moon’s surface is 1.62 m/s2.
SOLUTION
PROBLEMS
2. 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
PROBLEMS
3. 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
4. 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;
2211 gg VTVT (Datum at O)
Reference line
o.
cos....cos...
..............
152
819400502381920230
8194005023819202377220232
17724023050
2
1 22
q
q
q
w’
O
2Av(2)
(3)
(3)(2)
2Bv
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