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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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.
Introduction to Dynamics (N. Zabaras)
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.
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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.
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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
Introduction to Dynamics (N. Zabaras)
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.
Introduction to Dynamics (N. Zabaras)
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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