miet2097- exam example 1
TRANSCRIPT
Question 1 (Projectile motion)
You throw a ball such that, at the top of its path, it
just clears the 3 m high net at B. The angle at which
the ball is thrown θA = 50º.
a. Determine the initial speed of the ball vA.
b. Determine the distance from the net d that
you need to stand.
c. Determine the speed at which the ball hits
the ground at C.
d. State all assumptions you make.
h m
θA
vA
1 m C
A
d m
B
Question 2 (Rectilinear motion)
Stone A is thrown vertically with velocity 20 m/s, and stone B – with velocity 25 m/s 1
s later.
a. Determine distance between stones in 2 s after second stone is thrown.
b. Determine maximum height and time traveled before hitting the ground for each
stone.
c. State all assumptions you make
Question 3 (Relative motion)
If the end of the cable at A is pulled down with
acceleration a = 2t m/s2
1. Determine the acceleration at which block B
rises (as a function of time)
2. Find velocity of B in 2 seconds
Question 4 (Conservation of energy, force-acceleration)
A man has a mass of 100[kg] and from rest swings
from the cliff by rigidly holding on the tree vine,
which is 10[m] measured from supporting limb A
to his centre of mass. Determine his velocity just
after the vine strikes to the lower limb at B. Also,
with what force must he hold on the vine just
before and just after the vine contacts the limb at
B?
Question 5 (work-energy, impulse-momentum or force acceleration)
The masses A 40 kg and B 60 kg. The collar A
slides on smooth horizontal bar. The system is
released from rest.
1. Determine the velocity of collar A when it
has moved 0.5 m to the right.
2. Determine velocity of A when it has moved
for 1 s.
3. State all assumptions you made.
Question 6 (Cons of energy, impact, cons of energy)
The 3 kg block A is released from rest at 60
degrees position shown and subsequently strikes
the 1 kg cart B. If the coefficient of restitution for
collision is e=0.7, determine the maximum
displacement s of cart B beyond point C. Neglect
friction.
Question 1 (Projectile motion)
You throw a ball such that, at the top of its path, it
just clears the 3 m high net at B. The angle at which
the ball is thrown θA = 50º.
• Determine the initial speed of the ball vA.
• Determine the distance from the net d that
you need to stand.
• Determine the speed at which the ball hits
the ground at C.
• State all assumptions you make.
h m
θA
vA
1 m C
A
d m
B
Solution:
Initial speed of the ball
Analyse in y direction from A to B
2
002
1tatvss c++=
At max height there is zero component of velocity in y direction
( ) ( ) 281.92
150sin13 ttvA −++= (1)
tavv c+= 0
( )tvA 81.950sin0 −+= (2)
Solving simultaneously
smvA /18.8= and st 639.0=
Distance to the net
Time to net st 639.0=
Analyse in x direction from A to B
tvss 00 +=
( )( )639.050cos18.80 +=d md 36.3=
Velocity at point C
Analyse from A to C
( )0
2
0
2 2 yyavv c −+=
( ) ( )( )1081.9250sin18.822
−−+=v ( ) smvyC /67.7=
Total velocity
( ) ( ) ( ) ( ) smvvvyCxCc /29.967.750cos18.8
2222=+=+=
Assumptions:
Motion of particle
No air-resistance
Motion with constant acceleration
Question 2 (Rectilinear motion)
Stone A is thrown vertically with velocity 20 m/s, and stone B – with velocity 25 m/s 1
s later.
• Determine distance between stones in 2 s after second stone is thrown.
• Determine maximum height and time traveled before hitting the ground for
each stone.
• State all assumptions you make
Solution:
Distance
Stone A
?
/81.9
/20
3
2
0
=
−=
=
=
s
sma
smv
st A
( )( ) ( )( )
ms
s
tatvss
A
A
c
81.15
381.92
13200
2
1
2
2
00
=
−++=
++=
Stone B
?
/81.9
/25
2
2
0
=
−=
=
=
s
sma
smv
stB
( )( ) ( )( )
ms
s
tatvss
B
B
c
4.30
281.92
12250
2
1
2
2
00
=
−++=
++=
Difference m6.1481.154.30 =−
Max Height:
Stone A
?
/81.9
0
/20
2
0
=
−=
=
=
s
sma
v
smv
( )
( ) ( )( )
ms
s
yyavv
A
A
c
4.20
081.92200
2
2
0
2
0
2
=
−−+=
−+=
Time:
( )
st
t
atvv
08.4
81.92020
0
=
−+=−
+=
Stone B
?
/81.9
0
/25
2
0
=
−=
=
=
s
sma
v
smv
( )
( ) ( )( )
ms
s
yyavv
A
B
c
9.31
081.92250
2
2
0
2
0
2
=
−−+=
−+=
Time:
( )
st
t
atvv
1.5
81.92525
0
=
−+=−
+=
Assumptions:
Motion of particle
No air-resistance
Motion with constant acceleration
Question 3 (Relative motion)
If the end of the cable at A is pulled down with
acceleration a = 2t m/s2
• Determine the acceleration at which block B
rises (as a function of time)
• Find velocity of B in 2 seconds
Solution:
Position equation
( )CBB
CA
sssL
ssL
−+=
+=
2
1 2
Differentiate once to get velocities:
( )CBB
CA
vvv
vv
−+=
+=
0
20
Differentiate twice to get accelerations:
( )CBB
CA
aaa
aa
−+=
+=
0
20
BA aa 4−=
taA 2= m/s2 taB
2
1−= m/s
2
Now have to integrate to get velocity of B in 2 seconds:
taB2
1−= m/s
2
tdt
dvaB
2
1−==
dtt
tdv ∫∫
−=
−=
4
2tvB smvB /1−=
Question 4 (Conservation of energy, force-acceleration)
A man has a mass of 100[kg] and from rest swings
from the cliff by rigidly holding on the tree vine,
which is 10[m] measured from supporting limb A
to his centre of mass. Determine his velocity just
after the vine strikes to the lower limb at B. Also,
with what force must he hold on the vine just
before and just after the vine contacts the limb at
B?
Solution:
Find velocity in vertical position – just before hitting B
Conservation of energy-
( )( )( ) ( )
smv
v
mghmvmghmv
VTVT
/67.7
01002
145cos101081.91000
2
1
2
1
2
2
2
2
2
21
2
1
2211
=
+=−+
+=+
+=+
Force just before hitting limb B
smv /67.72 =
( )( ) ( )( )
NT
T
r
vmmgT
maF nn
1569
10
67.710081.9100
2
2
=
=−
=−
=∑
Force just after hitting limb B
smv /67.72 =
( )( ) ( )( )
NT
T
r
vmmgT
maF nn
2942
3
67.710081.9100
2
2
=
=−
=−
=∑
Question 5 (work-energy, impulse-momentum or force acceleration)
The masses A 40 kg and B 60 kg. The collar A
slides on smooth horizontal bar. The system is
released from rest.
• Determine the velocity of collar A when
it has moved 0.5 m to the right.
• Determine velocity of A when it has
moved for 1 s.
• State all assumptions you made.
Solution:
Determine the velocity of collar A when it has moved 0.5 m to the right
Work-energy
( ) ( )
( )( )( ) ( )
smv
v
ghmvmmghmvmm
VTVT
BBABBA
/43.2
060402
15.081.9600
2
1
2
1
2
2
2
2
2
21
2
1
2211
=
++=+
++=++
+=+
Determine velocity of A when it has moved for 1 s.
Impulse-momentum
( ) ( ) ( )
( )( ) ( )
smv
vdt
vmmdtgmvmm BABBA
/89.5
604081.9600
2
2
1
0
2
1
0
1
=
+=+
+=++
∫
∫
Assumptions
No friction
Motion with constant acceleration
Constant tension
Question 6 (Cons of energy, impact, cons of energy)
The 3 kg block A is released from rest at 60
degrees position shown and subsequently strikes
the 1 kg cart B. If the coefficient of restitution for
collision is e=0.7, determine the maximum
displacement s of cart B beyond point C. Neglect
friction.
Solution:
Velocity of A just before impact
Conservation of energy
( ) ( )
( )( )( ) ( )
smv
v
ghmvmghmvm
VTVT
AAAA
/2.4
032
160cos8.18.181.930
2
1
2
1
2
2
2
2
2
21
2
1
2211
=
+=−+
+=+
+=+
Impact
Conservation of linear momentum
2211 BBAABBAA vmvmvmvm +=+
( )( ) ( ) ( ) 22 1302.43 BA vv +=+ (1)
Coefficient of restitution
21
22
BA
AB
vv
vve
−
−=
02.47.0 22
−
−= AB vv
(2)
Solving (1) and (2) simultaneously smvA /42.22 = smvB /36.52 =
Velocity of block B at point C
Conservation of energy
( ) ( )
( )( ) ( )( )( )
smv
ghmvmghmvm
VTVT
BBBB
/74.4
30cos4.24.281.910036.512
1
2
1
2
1
2
2
2
2
21
2
1
2211
=
−+=+
+=+
+=+
Max displacement of block B from point C
Conservation of energy
( ) ( )
( )( ) ( )( )( )
mh
s
h
h
ghmvmghmvm
VTVT
BBBB
3.230sin
15.1
30sin
15.1
81.910074.412
1
2
1
2
1
2
2
2
21
2
1
2211
===
=
+=+
+=+
+=+