Answer key for second exam version B
1
1. This problem is solved in three steps. When the hockey puck leaves the top of
the cliff, it is traveling horizontally with a speed
v
1
. It will then be a projectile
falling to the ground from an initial height
y
0
= 8
.
5 meters and with vertical
acceleration
g
. This projectile will also travel at constant horizontal speed
v
1
.
We can find
v
1
by computing the fall time
t
, and then divided the horizontal
distance traveled (6.2 meters) by that time
t
. Then we can use conservation of
energy to find the energy at the top of the cliff, which must be all kinetic energy
when the puck started with speed
v
0
.
Falling body with no initial speed
y
(
t
) = 0 =
y
0

1
2
gt
2
Time to fall 8.5 meters =
⇒
t
=
v
u
u
t
2
y
0
g
=
v
u
u
t
17
9
.
8
= 1
.
32 seconds
Horizontal speed at the top of the cliff
v
1
=
6
.
50
1
.
32
= 4
.
84 m/s
Total Energy at the top of the cliff =
mgh
+
1
2
mv
2
1
Total Energy at the start =
1
2
mv
2
0
Conservation of energy =
1
2
mv
2
0
=
mgh
+
1
2
mv
2
1
=
⇒
v
2
0
= 2
gh
+
v
2
1
= 2
*
9
.
8
*
8
.
5 + (4
.
84)
2
(m/s)
2
=
⇒
v
0
= 13
.
8 m/s
Answer is D
2.
First calculate how much total energy is added to the 2000 kg of sand in
60 seconds.
The power will be that total energy divided by 60 seconds.
The
total energy is the potential energy increase in rising 12 meters plus the kinetic
energy in acquiring a speed of 5 m/s.
E
total
=
Mgh
+
1
2
Mv
2
= 2000(9
.
8
×
12 + 0
.
5
×
(5)
2
) = 2
.
602
×
10
5
Joules
P
=
E
t
=
2
.
602
×
10
5
Joules
60 seconds
= 4340 Watts
Answer is C
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Answer key for second exam version B
2
3.
This is a static equilibrium situation where the net forces on each of the
two objects
A
and
B
are both zero.
For object
A
that means the tension in
the connecting rope is equal to the weight of
A
. That tension
T
=
w
A
acts to
the right on object
B
. The force
Q
acts to the left on
B
, attempting to move
B
to the left. That means there is a static friction force
f
max
s
which is at its
maximum acting to the right on
B
, preventing
B
from moving to the left. In
turn, the static friction force is given by the normal force of the table, which
is the sum of the weight of
B
and the downward push force
P
which is acting
vertically on
B
.
f
max
s
=
μ
s
N
=
μ
s
(
w
B
+
P
)
Q
=
T
+
f
max
s
=
w
A
+
μ
s
N
=
w
A
+
μ
s
(
w
B
+
P
)
Q
= 19
*
9
.
8 + 0
.
40
*
(15
*
9
.
8 + 60) = 269 N
Answer = B
4. The average force is found from the impulse equation using the net momentum
change
F
Δ
t
= Δ
p
=
m
Δ
v
.
In this case, taking the positive direction as the
initial direction of the ball, Δ
v
= 11

(

11) = 22 m/s.
So
F
= (0
.
8)
*
(22)
/
0
.
05 = 352 N.
Answer = A
5. Doubling the compression of a spring means that it contains four times the
amount of potential energy. When all of that potential energy is transferred to
the mass, the mass will have four times the kinetic energy.
Quadrupling the
kinetic energy means that the speed has doubled.
Answer = A
6. Simple plugin solution from the banked curve speed equation
v
max
=
√
Rg
tan
θ
=
⇒
θ
= tan

1
(
v
2
max
/Rg
) = tan

1
((78)
2
/
(190
*
9
.
8))
θ
= 73
.
0
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 Spring '08
 Maguire
 Physics, Force, Friction, Kinetic Energy, wA, 1 m/s, 0.3 kg

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