Solutions to Mid-Term Exam for GP I, MATH,
Fall 2012.
1.
25%
A ball rolls horizontally off the top of a stairway with a speed of 1.52 m/s. The
steps are 20.3 cm high and 20.3 cm wide. Which step does the ball hit first?
ANS.
With the origin placed at the top of the stairway, and the x- & y- axes pointing to
the right & up, respectively, we have (see figure)
x  v0 t
1
y   g t2
2
The conditions for the ball to land on the nth step are
xnw
y  n h
Eliminating t, we have
1  x
n h  g 
2  v0 

2 v02 n h
 n 2 w2
x 
g
2
2 v02 h
n
g w2

Using

2
v0  1.52 m / s , g  9.81 m / s 2 , and h  w  20.3 cm , we have
n
2  1.52 
2
9.81   20.3  102 
 2.31
Therefore, the ball lands on the 3rd step.
2. 25%
A slab of mass m1  40 kg rests on a
frictionless floor, and a block of mass
m 2  12 kg rests on top of the slab (see figure).
Between block and slab, the
coefficients of static and kinetic friction are 0.61 and 0.40, respectively. A
horizontal force F of magnitude 120 N begins to pull directly on the block. In
unit-vector notation, what are the resulting accelerations of (a) the block and (b)
the slab?
ANS.
The free-body diagrams for the slab and block are shown below.
Let the +x direction to be to the right and the +y direction to be up.
Applying Newton’s second law, we have:
 f  ms as
f  F  mb ab
Slab:
Block:
FNs  FNs  ms g  0
FNb  mb g  0
The maximum possible static friction magnitude is therefore
f s, max  s FNb  s mb g  (0.60)(12 kg)(9.8 m/s2 )  71 N
If the block does not slide on the slab, then as = ab . The x-components of the force
equations give
ms F
(40 kg)(120 N)
f 

 92 N
ms  mb
40 kg  12 kg
which is greater than fs,max .
(a)
Using
ab 
Hence, the block is sliding across the slab (as  ab).
f  k FNb  k mb g , we have
k mb g  F
mb

(0.40)(12 kg)(9.8 m/s2 )  120 N
 6.1 m/s2
12 kg
That is, ab  ( 6.1 m/s2 ) ˆi . (to the left)
(b) We also obtain
as  
k mb g
ms

(0.40)(12 kg)(9.8 m/s 2 )
  1.2 m/s2
40 kg
That is, as  ( 0.98 m/s2 ) ˆi . (to the left)
3. 25%
A block slides along a path that is
frictionless until it reaches the section
of length L  0.65 m , which begins at
height h  2.0 m on a ramp of angle
  30 . In that section, the
coefficient of kinetic friction is 0.40. The block passes through point A with a
speed of 8.0 m/s. If the block can reach point B (where the friction ends), what
is its speed there? If it cannot, what is its greatest height above A.
ANS.
Let the point where the block first encounters the “rough region” as point C (this is
the point at a height h above the reference level). Energy conservation gives
1 2 1 2
mv A  mvC  mgh
2
2

vC  v A2  2 gh 
8.0
2
 2  9.8 2.0  4.98  m / s 
Its kinetic energy right at the beginning of its “rough slide” is
KC 
1
m vC2  12.4 m  J 
2
where m is the mass of the block.
Let d be the displacement along the slope beyond point C. For d  L , the work
done by the block against gravity & friction is
U  d   m g d sin   k mgd cos  m g d sin   k cos 
In particular,
1
3
U  L   m  9.8  0.65   0.40 
  5.39 m  J   KC
2 
2
Thus,
KC  KB  U  L
vB 

vC2 
2U  L 
m

 4.98
2
 2  5.39  3.74  m / s 
4. 25%
Due to air drag from Earth’s atmosphere, the radius of a satellite decreases from r
to r  r, where 0  r
r.
(a) Show that the increase of the orbital speed is v  
r
2
G mE
, where mE
r3
is the mass of the Earth.
(b) Show that the work done by the force of the air drag is W  
G mE m
r ,
2r 2
where m is the mass of the satellite.
ANS.
(a) For a circular orbit around Earth,
v2
m
 G 2E
r
r


v  v 

v 
r
2
v
G
mE

G
r  r
G
mE
r
m
G E
r
 r 
1  
r 

1/2

G
mE
r
 r 
1 

2r 

mE
r3
(b)
K

1
m v2
2

K  K 
K  mv v  m G
U  G
mmE
r
U  G

m mE
r
r2
mE r
r 2
1
1
2
m  v  v   mv 2  mvv
2
2
G
U  U  G
mmE
mE
G
r
3
2r 2
r
m mE
m mE  r 
 G
1  
r  r
r 
r 
1
Energy work theorem:
W  K  U  
G mE m
r
2r 2