MAE 208: Homework 4 Solution
Problem 1:


a) Suppose that we have x  t  , y  t  , and f x  t  , y  t  , t . What is
Solution:
df
?
dt
df f f dx f dy



dt t x dt y dt
b) Suppose that we have x  t   2t  3 , y  t   t 2 , and
f  x  t  , y  t  , t   x t   t  y t   t exp  t 3  .
2
What is
Solution:
f
?
y
f
2
 3ty  t   3t 5
y
What is
Solution:
f
?
x
f
 2 x  t   4t  6
x
What is
Solution:
3
f
?
t
f
3
  y  t   exp  t 3   3t 3 exp(t 3 )  t 6  exp  t 3   3t 3 exp(t 3 )
t
What is
df
?
dt
Solution:
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Homework 4 Solution
df f f dx f dy



dt t x dt y dt
 t 6  exp  t 3   3t 3 exp  t 3    4t  6  x  t   3t 5 y  t 
 t 6  exp  t 3   3t 3 exp  t 3    4t  6  2  3t 5  2t 
 t 6  exp  t 3   3t 3 exp  t 3   8t  12  6t 6
 7t 6  exp  t 3   3t 3 exp  t 3   8t  12
c) Suppose that we have x  , t   x*  t     t  , y   , t   y*  t     t  , and some function
f  x  , t  , y   , t  , t  .
What is
df
?
d
Solution:
df
f f dx f dy



d  x d y d
df
f
f
 0   t    t 
d
x
x
What is
df
?
d
Solution:
df
f f dx f dy



d   x d  y d 
df
f
f
 0   t    t 
d
y
y
What is
df
?
dt
Solution:
df f f dx f dy



dt t x dt y dt
df f f *
f *

  x  t     t   
y  t     t 
dt t x
y

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
Homework 4 Solution
What is
Solution:
f
?

f
0

Problem 15-7:
The total impulse is found by integrating the forces with respect to time (area under the curves).
a)
b)
I  4  3  0   4  6  3  0.5  8  4    6  3  0.5  8  0   10  6 
I  46 lb  s
I  6 8  0   0.5   6  0 10  8
I  54 lb  s
Problem15-24:
The net forces on the block are:
F
x
  F2  4F1
Applying the principle of impulse and momentum:
mv1    Fx dt  mv2
40 1.5   10  2  0   20  4  2   40  6  4  
 4  30  4  0   10  6  4    40v2
v2  12.0 m / s
Problem 15-53:
We can apply conservation of energy to the crate, but note that this speed is the speed of the crate
relative to the ramp.
T1  U1  T2  U 2
0  mC g 3.5sin  30o  
1
1
mC vC2  mR vR2 ,2
2
2
1
1
1
0  mC g 3.5sin  30o   mC vC2 , x  mC vC2 , y  mR vR2 ,2
2
2
2
(1)
We can relate the velocity of the crate and the ramp using the relative velocity:
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Homework 4 Solution
vC  v R  v C / R  vR i  vC / R cos  30o  i  vC / R sin  30o  j
 vC , x  vC / R cos  30o   vR
(2)
 vC , y  vC / R sin  30o 
Applying conservation of momentum in the x direction, we get:
mC vC ,1  mR vR ,1  mC vC ,2  mR vR ,2
0  mC vC ,2  mR vR ,2
Substituting for the relative velocity we have:


0  mC vC / R cos  30o   vR,2  mR vR,2
(3)
Solving (1), (2), and (3) for vR ,2 and vC / R :
vC / R  6.36 m / s
vR  1.1 m / s
 vC 
v
R
 
 vC / R cos  30o   vC / R sin  30o 
2

2
 5.43 m / s
Problem 15-67:
We need to first find the velocity of the block right after impact. Applying conservation of momentum
and using the coefficient of restitution, we have:
mAvA,1  mB vB ,1  mAvA,2  mB vB ,2
2  4   2vA,2  20vB ,2
e
(4)
vB ,2  v A,2
v A,1  vB ,1
0.8 
vB ,2  v A,2
(5)
4
Solving (4) and (5) yields:
vA,2  2.545 m / s
vB ,2  0.6545 m / s
Applying conservation of energy on the block after impact, we get:
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Homework 4 Solution
T1  U1  T2  U 2
1 2
mvB ,2  0  0  mgh
2
vB2 ,2 0.65452
h 

 0.0218 m
2g
2  9.81
Problem 15-83:
Note that in this problem we do not have a coefficient of restitution. We know the final directions of the
velocities, we just don’t know the magnitudes. Conservation of momentum is a vector equation, which
gives us two different equations in this case. Two equations, two unknowns.
Applying conservation of momentum along both axes:
mAvAx ,1  mB vBx ,1  mAvAx ,2  mB vBx ,2
4
0.5    0.8sin  30o   vA,2 cos  45o   vB ,2 cos  30o 
5
(6)
mAvAy ,1  mB vBy ,1  mAvAy ,2  mB vBy ,2
3
0.5    0.8cos  30o   vA,2 sin  45o   vB ,2 sin  30o 
5
(7)
Solving (6) and (7) yields:
vA,2  0.766 ft / s
vB ,2  0.298 ft / s
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Homework 4 Solution