PHY 1351
General Physics I
First Hour Exam
September 21, 2001
Statistics:
High: 80
Mean: 46
Low: 18
You simply cannot survive this course without doing the homework. This is a homeworkintensive, laborintensive, timeintensive course. it is not difficult, per se. But it does require time!
Possibly useful information:
v = v_{i} + a t
x = x_{i} + v_{i} t + (^{1}/_{2}) a t^{2}
v^{2} = v_{i}^{2} + 2 a x
a_{c} = v^{2}/r
F = m a
F_{12} = – F_{21}
sin = ^{opp}/_{hyp}
cos = ^{adj}/_{hyp}
sin = ^{opp}/_{adj}
Conceptual Questions:
Besides a “yes” or “no”, be sure to explain your
answers.
C1: If the velocity of a particle is nonzero, can its acceleration be zero?
YES. Acceleration is the change in the velocity divided by the
change in time. So if the velocity is nonzero, it could readily be changing
(or not changing!). A particle with nonzero velocity could have any kind
of an acceleration.
C2: The magnitudes of two vectors A and B are A = 5 units and B = 2 units.
What are the largest and smallest values possible for the resultant vector R
= A + B ?
When vectors A and B point in the same direction, their resultant R will have a magnitude of (5 + 2 = ) 7 units; this is maximum.
When vectors A and B point in the opposite direction, their resultant R will have a magnitude of (5  2 = ) 3 units; this is minimum.
C3: Can an object accelerate if its velocity is constant?
NO. Acceleration is the change in the velocity divided by the
change in time. So if the velocity is constant, it is not changing
which means the acceleration is zero.
C4: What forces cause a propellerdriven airplane to move?
As the propeller pushes on the air, the air pushes back on the propeller. This
force of the air on the propeller cause the propeller  and the airplane 
to move. This is an example of the actionreaction force pairs of Newton's Third
Law.
P1: A car brakes from 100 ^{km}/_{h} to rest in a distance
of 75 m.
a) What was the initial speed in units of ^{m}/_{s}?
v_{i} = 100 ^{km}/_{h} [^{1 000 m}/_{km}]
[^{h}/_{3 600 s}] = 27.8 ^{m}/_{s}
b) What is the acceleration of the car?
v^{2} = v_{i}^{2} + 2 a x
0 = (27.8 m/s)^{2} + 2 a (75 m)
2 a (75 m) =  (27.8 m/s)^{2}
a =  (27.8 m/s)^{2}/150 m
a =  5.14 m/s^{2}
c) How much time does the car require to stop?
v = v_{i} + a t
0 = 27.8 ^{m}/_{s} + (  5.14 m/s^{2}) t
t = [27.8 m/s]/[5.14 m/s^{2}]
t = 5.41 s
Almost by intuition we can easily find that
T_{3} = mg = 980 N
Now we turn our attention to the "knot" where the three ropes meet. That is not moving so we know F_{net} = 0
F_{net} = T_{1} + T_{2} + T_{3} = 0
Remember(!) this is a vector equation; this really means
F_{net,x} = T_{1x} + T_{2x}
+ T_{3x} = 0

F_{net,y} = T_{1y} + T_{2y}
+ T_{3y } = 0

 T_{1} cos 60^{o} + T_{2} cos 20^{o} + 0 = 0  T_{1} (0.500) + T_{2} (0.940) + 0 = 0 0.50 T_{1}= 0.94 T_{2} T_{1}= 0.94 T_{2}/0.50 T_{1}= 1.88 T2 T_{1}= 1.88 (497 N) T_{1} = 935 N 
T_{1} sin 60^{o} + T_{2} sin 20^{o}  T3 = 0 T_{1} (0.866) + T_{2} (0.342)  980 N = 0 0.866 T_{1} + 0.342 T_{2} = 980 N 0.866 (1.88 T_{2}) + 0.342 T_{2} = 980 N 1.97 T_{2} = 980 N T_{2} = 497 N

v_{ix} = (20 ^{m}/_{s}) cos 60^{o} = (0.500)(20^{ m}/_{s}) = 10.0 ^{m}/_{s}
v_{iy} = (20 ^{m}/_{s}) sin 60^{o} = (0.866)(20 ^{m}/_{s}) = 17.32 ^{m}/_{s}
The trajectory of the football is given by
x = x_{i} + v_{ix} t + (^{1}/_{2}) a_{x} t^{2}
where
a_{x} = 0
a_{y} =  g
and
y_{i} = 0
x_{i} = 0
which then means
x = 0 + v_{ix} t + 0
x = (10.0 ^{m}/_{s}) t
y = 0 + v_{iy} t  (^{1}/_{2}) g t^{2}
y = (17.32 ^{m}/_{s}) t  (^{1}/_{2}) (9.8 m/s^{2}) t^{2}
These are "parametric equations" in the parameter t. For an equation in only x and t, we can solve for t from the xequation (because it is a simpler, linear equation),
t = x/(10 ^{m}/_{s})
and then substitute that into the yequation,
y = (17.32 ^{m}/_{s}) [x/(10 ^{m}/_{s})]  (^{1}/_{2}) (9.8 m/s^{2}) [x/(10 ^{m}/_{s})]^{2}
y = 1.732 x  0.049 x^{2}
Remember, this is the equation of the trajectory, the path the football follows.
(I have not been entirely careful with the units. x and y are measured in meters(m) so x^{2} has units of m^{2}. I have chosen not to carry that through; so this equation now requires that x and y be measured in m).
The inclined plane can be described by the equation
y = x tan 20^{o} = 0.364 x
y = 0.364 x
Where do these two "curves" or lines intersect? We can set the two equations for y equal to each other,
y = 0.364 x = 1.732 x  0.049 x^{2} = y
0.364 x = 1.732 x  0.049 x^{2}
0.049 x^{2}  1.368 x = 0
(x)(0.049 x  1.368) = 0
x = 0
True, but not very interesting, . . . or
0.049 x  1.368 = 0
0.049 x = 1.368
x = 1.368/0.049
x = 27.92 m
y = 0.364 x = 0.364 (27.92 m)
y = 10.16 m
R = SQRT[x^{2} + y^{2}] = SQRT[(27.92)^{2} + (10.16)^{2}] m
R = 29.71 m
P4: A 150 g mass is suspended as shown in the diagram.
It is attached to a string which runs over a pulley and is then attached to
a 250g block which slides along an inclined plane as shown in the diagram.
The inclined plane is inclined at 20° with the horizontal. The coefficient
of friction between the block and the plane is 0.15. (this was corrected on
the board in class).
Find the acceleration of the masses and the tension in
the string.
Always(!) begin with good, clear "free body diagrams" showing all the forces on each of the masses.
I am expecting this block to come down  so I have written the acceleratio pointing down and have taken down as positive: F = m a F_{net} = m a F_{ = w  T = m a} 1.47 N  T = (0.150 kg) a 

I am expecting this block to accelerate in the direction shown  so I have taken the positive xdirection as shown, up and to the left. As we discussed in class, there are other ways to do this. There is seldom (if ever!) only a single "right way" to solve a Physics problem.

F = m a


F_{net,x} = m a T  F_{fr}  0.838 N = (0.250 kg) a F_{fr} = (0.15)(2.302 N) = 0.345 N T  0.345 N  0.838 N = (0.250 kg) a T  1.183 N = (0.250 kg) a 
F_{net,y} = m a_{y} n  2.302 N = m (0) n = 2.302 N 
We now have two equations with two unknowns, the tension T and the acceleration a,
T  1.183 N = (0.250 kg) a
1.47 N  T = (0.150 kg) a
Solve for T from the first one,
T = 1.183 N + (0.250 kg) a
and subsitute that into the second one for T,
1.47 N  [1.183 N + (0.250 kg) a] = (0.150 kg) a
1.47 N  1.183 N = (0.150 kg) a + (0.250 kg) a
0.29 N = (0.400 kg) a
(0.400 kg) a = 0.29 N
a = 0.725 m/s^{2}
Now we can use this acceleration to find the tension T,
T = 1.183 N + (0.250 kg) a
T = 1.183 N + (0.250 kg) (0.725 m/s^{2})
T = (1.183 + 0.181) N
T = 1.364 N
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