QUIZES GIVEN IN PHYSICS 103, WITH SOLUTIONS: SPRING 1998
INDEX OF QUIZES
Quiz 1: 28 January 1998
An engineer standing on a bridge drops a penny toward the water and sees
the penny splashing into the water 3.0 seconds later. [Neglect the time
associated with the travel of light.]

How high above the water was the man?

What was the speed of the penny when it hit the water?
Solution:
The height h through which the penny falls is given by the formula
h = ½gt^{2}, where t is the time in flight. Thus,
h = ½´9.8
´(3.0)^{2} = 44.1 meters.
The splash speed v is given as v = v_{0} + gt, where
v_{0} = 0. Thus v = 9.8´3
= 28.4 meters/sec.
Quiz 2: 04 February 1998
A particle launched with initial speed v_{0} at an angle
q to the horizontal travels to a
height of 4 meters at its apex, and travels a horizontal distance of 12
meters before coming back to the [horizontal] ground.

With what speed v_{0} was the particle launched? [in meters/second]

What was the launch angle q? [in degrees]
Solution:
The x and y components of position at time t are given in terms of the
initial velocity v_{0} and the launch angle
q by the formulas.
x = v_{0}cosq t ;
y = v_{0}sinq t 
½gt^{2} .
We solve the second of these for t; t =
x/v_{0}cosq; insert that
value into the first equation; and get
y = xtanq 
gx^{2}/(2v_{0}^{2}
cos^{2}q)
Setting y = 0 in that formula and solving for x, we obtain the range
R = v_{0}^{2} sin2q/g .
To get the maximum height, we set dy/dt = 0, and obtain
t = v_{0}sinq/g. Inserting that
time into the equation for y, we obtain
H = v_{0}^{2}sin^{2}
q/(2g).
Thus,
H/R = sin^{2}q/(2
sin2q) =
tanq/4. Inserting the numbers, we get
tanq = 16/12, or
q= 53°. Going to the Range equation,
we get v_{0}^{2} = Rg/sin2q
= 12´9.8/0.96 = 122.5. Thus,
v_{0}^{2} = 11.1 meters/sec.
Quiz 3: 11 February 1998
Two heavy boxes of masses 20 kg and 35 kg sit on a smooth, frictionless
horizontal surface. The boxes are in contact, and a force of 60 Nt
pushes horizontally against the lighter [less massive] box.

What is the acceleration of the two boxes? [in meters/second^{2}]

What is the force that the smaller box exerts on the heavier box
[magnitude and direction]?

What is the force that the larger box exerts on the lighter box
[magnitude and direction]?
Solution:
Let us suppose that the lighter box is on the left of the heavier box.
A force of 60 Nt [to the right] acts on the lighter one, whereas
actionreaction forces of magnitude P act between the boxes, to the
left on the lighter box and to the right on the heavier box. Let us
take the positive direction for force and acceleration to the right, and
apply Newton's second law to each box:
60  P = 20 a [left box] ;
P = 35 a [right box] ;
where a is the rightward acceleration of each box. Adding the equations,
we get 60 = 55 a; or a = 1.1 meters/sec^{2}. The internal
force P is found from the second equation; P = 35
´1.1 = 38.5 Nt.
Quiz 4: 18 February 1998
A roadway with a radius of curvature of R = 200 meters is banked at an
angle q = 10° to the horizontal.
A vehicle of mass m = 1000 kg moves with constant speed v around this
roadway on a circular path in the horizontal plane without slipping off .
Assume that the coefficient of static friction on this roadway is
m_{S} = 0.3.

Draw a diagram showing all the forces involved, and determine the
horizontal and vertical components of each of them.

Determine the greatest speed v_{critical} with which the vehicle
will go around the curve without slipping off. [You may express your
answer in terms of R, q,
m_{S}, m, and g, if you wish.]
Solution:
The three forces that act on the vehicle are the weight W [downward];
the normal force N [perpendicular to track at angle
q to the vertical]; and the static
friction force f [down the track at angle
q below the horizontal]. The resultant
force R must be horizontal and toward the center of the track,
with magnitude mv^{2}/R.
Here are the horizontal [x] and vertical [y] components of the forces:
Force
 xcomponent
 ycomponent

W
 0
 mg

N
 N sinq
 N cosq

f
 f cosq
 f sinq

R
 mv^{2}/R
 0

Applying Newton's second law to horizontal and vertical components,
we obtain
N cosq 
f sinq mg = 0
 N sinq
 f sinq = mv^{2}/R
For the critical case, f = m_{S}N,
we get
N [cosq 
m_{S}
sinq] = mg
N [sinq +
m_{S}
cosq] = mv^{2}/R
Taking the ratio of these equations, we eliminate N and m to get
v^{2}/(Rg) = [m_{S}
+ tanq]/[ 1 
m_{S}
tanq]
Putting in the numbers, we have
v^{2} = 200´9.8
´[0.30 + 0.18]/
[10.30´0.18] = 990.
Thus, v_{critical} = 31 meters/sec, or 110 kilometers/hour.
Quiz 5: 04 March 1998
A box of mass 50 kilograms is initially at rest on a level floor. The
coefficient of kinetic friction between the box and the floor is
m_{k} = 0.5. A
woman pushes horizontally against the box [to the right] with a force of
300 Newtons until the box attains a speed of 2.0 meters/sec.

How far does the box move in this process?

What is the change in kinetic energy of the box [in Joules]?

Determine the work done by the friction force on the box [in Joules].

Determine the work done by the woman on the box [in Joules].
Solution:
The four forces acting on the box are the weight W
[magnitude mg; downward], the normal force N [upward],
the applied force F [to the right], and the friction
force f = m_{k}N [to the left].
Applying Newton's second law to horizontal and vertical components, we
get
N = mg ; F  m_{k}N = ma
or 300  0.5´50
´9.8 = 55 = 50 a.
The acceleration is thus 1.1 meters/second^{2}. The distance
travelled is calculated from the formula
v^{2} = v_{0}^{2} + 2ad = 4 = 0 +
2´1.1´d,
so that d = 1.82 meters.
The initial kinetic energy of the box is 0, so its final kinetic energy
is ½mv^{2} = ½´50
´2^{2} = 100 Joules.
The work done by the friction force [245 Newtons acting in opposition to a
displacement of 1.82 meters] is W_{fric} =
 245 ´1.82 = 445 Joules.
The work done by the applied force [300 Newtons over a displacement of
1.82 meters] is 300 ´1.82 = 545 Joules.
Quiz 6: 11 March 1998
A box of mass 100 kilograms is dragged up a
rough plane inclined at an angle
q = 20° to the
horizontal at a steady speed of 5.0 meters/sec.
The coefficient of kinetic friction between the
box and the plane is m_{k}
= 0.5. The block is moved up the plane by a constant force F
parallel to the plane.

Draw a diagram showing all the forces acting on the box.

Determine the magnitude of the applied force F [in Newtons].

Determine the power expended by the applied force F [in Watts].

How much power would be required to move the box up the plane if there
were no friction [m_{k}= 0]?
Solution:
The four forces on the box are the weight W [magnitude mg; downward],
the normal force N [perpendicular to the plane], the applied
force F [up the plane], and the kinetic friction force
f_{k} [down the plane]. The net force acting on the box
must be zero. Let us separate them into components up the plane
[xdirection] and out of the plane [ydirection]:
Force
 xcomp [up plane]
 ycomp [out of plane]

W
 mg sinq
 mg cosq

N
 0
 N

F
 F
 0

f_{k}
 f_{k}
 0

Setting the total force [xcomponent and ycomponent] to zero,
we get
N = mg cosq
F = mg sinq + f_{k}
We have f_{k} = m_{k}N
= m_{k}mg
cosq, so that
F = mg(sinq +
m_{k}
cosq) =
100´9.8´
(sin 20° + 0.5´cos 20°) = 796 Nt.
The power expended by this force is P = F · v =
796´5 = 3978 Watts [over 5 HP].
Without friction, the applied force would be
F = mg sinq = 100
´ 9.8 ´
sin 20° = 335 Nt, and the power P = F v =
335 ´ 5 = 1676 Watts [2.2 HP].
Quiz 7: 25 March 1998
Two point particles, each of mass m, subject
to mutual gravitational attraction, move at
diametrically opposite points on circular orbits
of radius R. The only
force between the particles is their mutual
gravitational attraction. The particles each
execute uniform circular motion.

Determine the speed v with which the particles move in the orbit, expressed
in terms of G [Newtonian gravitational constant], m, and R.

Determine the period of revolution in terms of G, m, and R.
Solution:
The masses are separated by a distance 2R, so that their gravitational
attraction is
F = G m^{2}/(4 R^{2}) .
Each mass moves with speed v in a circle of radius R, with centripetal
acceleration a_{c} = v^{2}/R.
Thus
F = G m^{2}/(4 R^{2}) = m a_{c} = mv^{2}/R.
We may solve for the velocity to obtain
v^{2} = Gm/(4R), or v = Ö[Gm/(4R)].
The angular velocity of motion of each mass is given by
w = v/R =
Ö[Gm/(4R^{3})].
The period of revolution is T =
2p/w =
2p
Ö[4R^{3}/Gm] .
Quiz 8: 08 April 1998
An automobile of mass 1000 kg going North at 30 meters/second collides at
an intersection with a minivan of mass 2000 kg going East at 20 meters/sec.
The two vehicles stick together after the collision.

[a] Determine the magnitude and direction of the velocity of the
composite object immediately after the collision [in meters/second].

How much mechanical energy is dissipated during the initial collision
[in Joules]?
Solution:
Momentum [a vector] is conserved during the collision. Define the
xdirection to be East and the ydirection to be North. Thus
P_{init} = 2000 ´20x +
1000 ´30y =
40000x + 30000y ;
P_{fin} = 3000v .
As a consequence of P_{init} = P_{fin} ,
v = 13.33x + 10.00y ;
or 16.67 meters/sec at an angle of 37° to the horizontal; i e, North of
East.
The initial kinetic energy is K_{i} =
½1000´(30)^{2} +
½2000´(20)^{2} =
8.5 ´10^{5} Joules.
The final kinetic energy is K_{f} =
½3000´(16.67)^{2} =
4.2 ´10^{5} Joules.
Thus the energy dissipated is
4.3 ´10^{5} Joules.
Quiz 9: 15 April 1998
Two objects of equal mass m undergo an elastic collision. Initially, the
first mass is moving to the right with a speed v, and the second mass is
at rest. After the collision, the two masses both leave at equal angles
q to the direction of motion of the first
mass. Determine the final speeds of each mass, as well as the angle q.
Solution:
Let the projectile be moving initially in the xdirection, and let it recoil
at an angle q above the xaxis, so that
the target will recoil at an angle q below
the xaxis. The momentum [a vector] and the kinetic energy
of motion [a scalar] are conserved in this collision. The initial momentum
is P_{init} = mvx, whereas the final momentum [expressed in
terms of the final speeds v_{1} and v_{2}] is
P_{fin} = mv_{1} [
cosq x +
sinq y] + mv_{2}
[cosq x 
sinq y ] =
m [v_{1} + v_{2}]
cosq x +
m [v_{1}  v_{2}]
sinq y
From momentum conservation, we get v_{1} = v_{2} and
v = 2 v_{1} cosq.
Using energy conservation, we obtain
½mv^{2} = ½mv_{1}^{2} +
½mv_{2}^{2}; or
v^{2} = v_{1}^{2} + v_{2}^{2}.
Consequently, v_{1}^{2} = v_{2}^{2} =
½v^{2}.
Finally, v_{1} = v_{2} =
v/Ö2 and
q = 45°.
Quiz 10: 22 April 1998
A small phonograph record of radius 10 centimeters undergoes uniform
angular acceleration from rest to its final rotational rate of 45
revolutions per minute in 10 seconds. Determine its angular acceleration
[in radians/second^{2}] and the number of revolutions made
in this process. Also, compute the tangential and radial components of
acceleration [in meters/second^{2}] of a point on the edge of
the record when it is rotating at its final rate.
Solution:
The relevant formulas for uniform angular acceleration are
w =
w_{0} +
at
q =
v_{0}t +
½at^{2}
q =
½(w +
w_{0})t
w^{2} =
w_{0}^{2} +
2aq
The initial angular velocity is w_{0}
= 0, whereas the final angular velocity is w
= 0.75 revolutions/second = 1.50p rad/sec.
Since this occurs over 10 seconds, we use the first equation to obtain the
angular acceleration:
a = 0.15p
= 0.471 rad/sec^{2}.
The number of revolutions is then determined from the fourth formula to be
q =
(1.50p)^{2}/
2(0.15p) =
7.5p radians = 3.75 revolutions.
The final radial acceleration is
a_{r} = w^{2} R
= [1.5p]^{2}
´ 0.10 =
.225p^{2} =
2.22 meters/second^{2}
The [uniform] tangential acceleration is
a_{t} = a R =
0.15p ´
0.1 = 0.015p = .0471 meters/second^{2}.
Quiz 11: 29 April 1998
A wooden beam of mass M and length l is hung vertically at one end, and is
free to rotate about the pivot. A bullet of mass m and velocity V is aimed at
the lower end. The bullet goes through the beam and emerges with reduced
velocity v. [Make the simplifying assumption that the beam remains still until
after the bullet passes through it.]

What is the angular velocity w of the
beam immediately after the bullet went through it?

How much mechanical energy is converted into heat during the collision?

What is the maximum angle q with
respect to the vertical to which the stick will swing after the collision.
You should express the answers in terms of M, m, V, v, g , and l.
Solution:
The angular momentum of the system [bullet and rod] about the pivot point
of the rod is conserved during the collision, since the force at that point
produces no torque, and the collision forces are internal. Thus
mVl = mvl + I_{0}w_{0},
where I_{0} = Ml^{2}/3 is the moment of inertia of the
rod about the pivot point. Thus
w_{0} = 3m(V  v)/[Ml].
With initial angular velocity w_{0}
at the bottom, the rod system has kinetic energy
K = ½I_{0}
w_{0}^{2}.
The rod swings up to an angle q, where
Ml^{2}w_{0}^{2}/6
= ½Mgl[1  cos q]. Note that the
center of mass of the rod rises only by half the height of the end of
the rod. The angle q is thus given
by the relation
1 cos q =
3 [m/M]^{2} (V v)^{2}/[gl]
The initial kinetic energy of the system is
K_{i} = ½mV^{2}, whereas its final kinetic energy is
K_{f} = ½mv^{2} + ½I_{0}
w_{0}^{2}.
Thus the kinetic energy lost is
DE =
½m[V^{2} v^{2}] ½I_{0}
w_{0}^{2} =
½m [V  v] [(1+3m/M)V+(13m/M)v] .
Quiz 12: 06 May 1998
A uniform ladder of length l and mass m and
length l is placed against a smooth [frictionless]
vertical wall, and sits on a rough horizontal
surface with coefficient of static friction
m_{S}.
The ladder is in equilibrium when it is tilted at
an angle q to the vertical direction.

Draw a diagram showing and labelling all forces acting on the ladder.

Obtain the conditions for static equilibrium, and determine all
the forces acting on the ladder.

If the maximum angle q for equilibrium is 30° , what is the coefficient
of static friction m_{S}?
You should express the answers to [b] in terms of m, l, g, and m_{S}.
Solution:
The four forces on the ladder are its weight W [downward; at its
center], the normal force of the wall N_{1} [horizontal; away
from wall],
the normal force of the floor N_{2} [vertical], and the
static friction force f_{2}of the floor [horizontal;
toward the wall]. The net force [a vector] must be zero, so that
N_{2} = mg and N_{1} = f_{2}.
We require the net torque about the contact point of the ladder with the
floor to be zero, so that
N_{1}l cos q  ½ mgl
sin q = 0, so that N_{1} =
½mg tan q.
In summary, f_{2} = N_{1} =
½mg tan q, and N_{2} = mg,
so that m_{S} =
f_{2}/N_{2} =
½tan q.
Inserting q = 30°, we obtain
m_{S} = 0.289 .
Test 1: 25 February 1998
#1. An automobile of mass 1000 kg going at 100 kilometers/hour on level
ground is heading toward a brick wall 200 meters away, when the driver
locks the brakes. Assume that the car slides to rest on the road with
coefficient of kinetic friction
m_{k} = 0.3.

Draw a diagram showing all forces acting on the car.

Determine the deceleration of the car [in meters/second^{2}].

Determine the distance [in meters] and the time [in seconds]
required to stop the car.
Solution:
The net vertical force is zero, so that N = mg =
1000´9.8 = 9800 Nt.
The net horizontal force is f_{k} =
m_{k}N = 2940 Nt.
Therefore, the acceleration is a =  f_{k}/m =
 2.94 meters/sec^{2}.
The initial velocity is v_{0} = (100/3.6) meters/sec =
27.8 meters/sec.
The distance travelled is d = v_{0}^{2}/[2a] =
[27.8]^{2}/
[2 ´2.94] = 131 meters. Tragedy is thus
averted.
The time required to stop is t = v_{0}/a = 27.8/2.94 = 9.45 sec.
#2. Two masses [m_{1} = 14 kg;
m_{2} = 12 kg] are connected across
a massless, frictionless pulley by
a massless cord. Mass #1 sits on
a frictionless surface, tilted 45°
to the horizontal, whereas mass #2 hangs
freely.

Draw a diagram showing all the forces involved.

Determine the tension in the cord [in Newtons].

Determine the acceleration of each mass [in meters/second^{2}].
Solution:
Take the positive direction of motion for the hanging mass #2 to be down,
and for the sliding mass #1 take it up the plane. The two forces on mass
#2 are its weight W = m_{1}g [downward] and the tension
in the string [upward]. Thus, from Newton's second law,
m_{2}g  T = m_{2}a
The forces on #1 are the tension T in the string [up the plane],
the normal force N [perpendicular to and away from the plane], and
its weight W = m_{2}g [downward]. Applying Newton's second
law we obtain
N = m_{2}g cos45° ;
T  m_{2}g sin45° = m_{2}a .
We obtain two equations for N and T:
m_{2}g  T = m_{2}a and T  m_{2}g sin45° =
m_{2}a.
Solving these equations, we get
a = [m_{2}  m_{1} sin45°]/[m_{2}+ m_{1}]
= [12  14/Ö2]
´9.8/[14 + 12] =
0.79 meters/sec^{2} ;
T = m_{1}m_{2}g[1 + sin45°]/
[m_{1} + m_{2}] = 12´14
[1 + 1/Ö2]/[12+14] = 108 Nt .
#3. On level ground, a ball of mass 200 grams is thrown at an angle of 30° to
the horizontal with an initial speed of 40 meters per second.

How far does the ball travel horizontally before returning
to the ground [in meters]?

What is the maximum altitude of the ball [in meters]?

How long does the ball stay in the air[in seconds]?
Solution:
The relevant trajectory equation is
y = xtanq 
gx^{2}/(2v_{0}^{2}
cos^{2}q)
Setting y = 0 and solving for x, we obtain the range
R = v_{0}^{2}sin2q/g
= (40)^{2}´sin60°/9.8
= 141 meters.
To get the maximum altitude, we set dy/dx = 0, and solve for x to get
x = v_{0}^{2}sin2q/[2g]
= R/2. Then, insert this in the trajectory equation to get
y_{max} = v_{0}^{2}
sin^{2}q/[2g] =
(40)^{2}´sin^{2}30°/
[2´9.8] = 20.4 meters.
the flight time is given by
t_{flight} = R/[v_{0}cosq]
= 141/[40 cos30°] = 4.1 seconds.
#4. A person of mass 60 kilograms sits on a merrygoround of radius 10
meters. The merrygoround [it rotates in a horizontal plane!] is slowly and
steadily increased, until the person is moving at a speed of 10 meters/second
at which point s/he slides off.

Draw a diagram labeling all the forces acting on the person.

What is the coefficient of static friction between the person
and the merrygoround?
Solution:
The three forces acting on the person are the weight W [mg; downward],
the normal force of the platform N [upward], and the friction force
of the platform f_{S} [inward]. According to Newton's second
law, the weight and normal force balance [N = mg], and the [static] friction
force equals the mass times the centripetal acceleration f_{S} =
ma_{c} = mv^{2}/R. for the critical case f_{S} =
m_{S}N, or
m_{S} = f_{S}/N =
v^{2}/[Rg] = (10)^{2}/
[9.8´10] = 1.02 .
#5. On an old map there is a set of instructions telling where to find a buried
treasure. Beginning at a certain point, one is directed to walk 50 meters due
East, and then walk 40 meters at 37° South of West. One could reach the
same position by a single vector displacement. What is that displacement?
Solution:
Take the xaxis to the East and the yaxis to the North. The displacement
R_{1} is 40 meters along the positive xaxis, and the
displacement R_{2} is 50 meters at 37° below the negative
yaxis. Here are the components of R_{1},
R_{2}, and R = R_{1} +
R_{2}:
Displacement
 xcomponent
 ycomponent

R_{1}
 50
 0

R_{2}
 40cos37°=32
 40sin37°=24

R
 18
 24

Thus, R is 30 meters at 53° South of East.
Test 2: 01 April 1998
#1. A black cat breaks a mirror, walks under a ladder, destroys a rabbit foot,
and accidentally falls [from rest] over the edge of a building. The cat is seen
passing a window 2.0 meters in height, with its base near the ground. The
cat took 0.1 seconds to pass the window. [Remark: The cat is likely to
survive the fall.]

From how high above the window did the cat fall [in meters]?

Determine the speed of the cat at the window base,
[in meters/second].
Solution:
The cat starts falling at time t = 0. At time t_{1} it reaches
the top of the window, with speed v_{1} = gt_{1}, having
fallen a distance z_{1} = ½gt_{1}^{2}. It
reaches the bottom of the window at
time t_{2}, with speed v_{2} = gt_{2}, having
fallen a total distance z_{2} = ½gt_{2}^{2}.
We are given
Dt = t_{2}  t_{1} = 0.1 seconds .
Dz = 2 meters = z_{2}  z_{1}
= ½g(t_{2}^{2}  t_{1}^{2}) =
½g[(t_{1}+ 0.1)^{2}  t_{1}^{2}] =
½g( 0.2 t_{1} + 0.01) .
Thus t_{1} = (2.0  ½´9.8
´0.1)/
[½´9.8] = 1.99 seconds, and
t_{2} = 2.09 seconds.
The speeds are v_{1} = gt_{1} =
9.8 ´ 1.99 = 19.5 meters/sec and v_{2} = gt_{2}
= 9.8 ´ 2.09 = 20.5 meters/second.
The distances are z_{1} = ½gt_{1}^{2} =
½ ´ 9.8
´ (1.99)^{2} = 19.4 meters and
z_{2} = ½gt_{2}^{2} = ½
´ 9.8 ´
(2.09)^{2} = 21.4 meters.
#2. A bullet of mass 10 grams and speed of 1000 meters/second strikes a
tree and penetrates a tree to a depth of 0.1 meters. Assume that the force
stopping the bullet is a constant friction force.

What is the stopping force [in Newtons]?

How much time elapsed between the instant the bullet entered
the tree and the instant it stopped [in seconds]?
Solution:
The bullet has velocity v_{0} when it enters the tree, and
is uniformly decelerated to rest, so that when it stops [speed 0] it
has travelled a distance d, where v^{2} = v_{0}^{2}
+ 2ad. Thus a = v_{0}^{2}/[2d] =  (1000)^{2}/[2
´ 0.1] =
 5 ´10^{6} meters/second^{2}.
By Newton's second law, the stopping force on the bullet [opposite to
direction of motion of bullet] is F = ma =
0.01 ´
5´10^{6} =
5´10^{4} meters/second.
The stopping time is t = v_{0}/a = 1000/
[5 ´10^{6}] =
2 ´10^{4} seconds.
#3. A point particle of mass m is
attached to a string of length l, which
is pivoted around a point O. The
mass is swung in a vertical circle.
The string remains taut at
every point on the circle. [You may
express answers in terms of m, l, and g.]

Determine the least speed of the particle, at the "top"
of the circle [in meters/second] for the string to remain taut.

For that case, determine the maximum speed of the particle at
the "bottom" of the circle [in meters/second].

Determine the tension in the string at the bottom of the circle,
[in Newtons].
Solution:
At the top of its swing, the forces acting on the mass are gravity
[mg; downward] and the tension in the strong, T_{top} [downward].
If it is moving with speed v, it experiences a downward centripetal
acceleration a_{c} = v^{2}/l at that point. Thus
mg + T_{top} = ma_{c} = mv^{2}/l. For the
minimum speed at the top, T_{top} = 0 and
v_{top} = Ö[gl].
The velocities at the top and bottom of the swing, v_{top} and
v_{bot}, are related by conservation of energy:
½mv_{top}^{2} + mgl =
½mv_{bot}^{2}  mgl, so that
½mv_{bot}^{2} = 5mgl/2. Thus for the minimum case,
v_{bot} = Ö[5gl].
The tension in the string at the bottom T_{bot} acts upward,
whereas the weight acts downward. Thus, T_{bot}  mg =
mv_{bot}^{2}/l = 5mg, so that T_{bot} = 6mg.
#4. A mass m slides on a rough horizontal plane, the coefficient
of kinetic friction being m_{}.
An ideal spring of spring constant k is compressed by a distance d,
and the mass is placed in front of the spring, which is then released.
Note that the spring initially pushes the mass, but is not attached to it.

Draw a diagram labeling all the forces acting on the mass just after release.

How far does the mass m travel before stopping? Express the answer in terms
of d, m, k, g, and m_{k}.
Solution:
The forces acting on the block while it slides on the plane are its
weight W [mg; downward], the normal force N [upward],
and the friction force f_{k} [opposite the direction of
motion of the block]. Because there is no vertical motion, N = mg.
The potential energy initially stored in the compressed spring is
½kd^{2}, and after release of the block
this is the kinetic energy of the block. The energy is all
dissipated by the friction force f_{k} =
m_{k} N =
m_{k} mg. If the block
travels a distance l, the work done by friction is
 m_{k} mgl. Thus,
½kd^{2} = m_{k} mgl,
or l = kd^{2}/[2m_{k} mg].
#5. How much energy [in Joules] is required to take a 1.0 kilogram mass
from the surface of the earth to the surface of the moon? You should neglect
air resistance and the rotational or revolutional motion of the earth or moon,
and ignore energy wasted in launching the propulsion system. Instead, you
should merely count the energy necessary to take the payload from the earth
to the moon.
Conceivably useful information:
 Mass of Earth: 5.98 ´ 10^{24} kilograms.
 Mass of Moon: 7.95 ´ 10^{22} kilograms.
 Radius of Earth: 6.38 ´ 10^{6} meters.
 Radius of Moon: 1.74 ´ 10^{6} meters.
 Distance from Earth to Moon: 3.88 ´ 10^{8} meters.
Solution:
The potential energy of a mass m on earth is
U_{e} =  GM_{e}m/R_{e}, whereas its potential
energy on the moon is U_{m} =  GM_{m}m/R_{m}.
The difference in potentials,
DE = U_{m}  U_{e}, is
the energy necessary to take the mass to the moon.
The energy required per unit mass is DE /m
= G[M_{e}/R_{e}  M_{m}/R_{m}] =
6.67´10^{11}[
5.98´10^{24}/
(6.38´10^{6}) 
7.95´10^{22}/
(1.74´10^{6})]
= 5.95´10^{7} Joules/kilogram .
Equation Sheet Phys 103004 Spring 1998
Trigonometry:
a^{2} + b^{2} = c^{2} sinq = b/c
cosq = a/c tanq = b/a
Kinematics in one dimension:
v = v_{0} + at
x = v_{0}t + ½at^{2}
x = ½(v + v_{0})t
v^{2} = v_{0}^{2} + 2ax
Vectors:
A = a_{x}x + a_{y}y + a_{z}z
B = b_{x}x + b_{y}y + b_{z}z
A + B = (a_{x}+ b_{x})x + (a_{y} + b_{y})y +
(a_{z} + b_{z})z
A · B = a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z}
= abcosq
A ´ B = ABsinq [right hand rule]
Projectile Motion: [g = 9.8 m/sec^{2}]
x = v_{0}cosqt
y = v_{0}sinqt  ½gt^{2}
v_{y} = v_{0}sinq gt
y = xtanq  gx^{2}/(2v_{0}^{2}cos^{2}q)
Centripetal Acceleration:
a_{c} = v^{2}/R [into center]
Newton's Laws:
1. Principle of Inertia
2. F_{net} = ma = dp/dt
3. Action  Reaction
Momentum: p = mv
Kinetic Energy: K = ½mv^{2}
Static Friction: f_{s}£ m_{s}N
Kinetic Friction: f_{k} = m_{k}N
WorkEnergy Theorem:
W = DK = ò F·dr
Power: P = dW/dt = F·v
Potential Energy:
gravity[uniform]: U = mgh
spring: U = ½kx^{2}
Gravitation
gravitational force [attractive]
F = Gm_{1}m_{2}/d^{2}
G = 6.67 ´ 10^{11} Ntm^{2}/kg^{2}
Potential Energy
U =  Gm_{1}m_{2}/d
Kepler's Laws
1. Orbit ellipse with sun at focus
2. Equal areas in equal times
3. Period^{2} µ [semimajor axis]^{3}
Center of Mass:
R_{cm} = 1/Må_{i}m_{i}r_{i}
V_{cm} = 1/Må_{i}m_{i}v_{i}
F_{net} = Ma_{cm}
Rotational Motion:
s = rq ; v_{T} = rq ; a_{T} = ra
Uniform Angular Acceleration:
w = w_{0} + at
q = w_{0}t + ½at^{2}
q = ½(w + w_{0})t
w^{2} = w_{0}^{2} + 2aq
Angular Momentum:
L = m r ´ v = I w
Torque: t = r ´ F = dL/dt = Ia
Kinetic Energy: K = ½Iw^{2}
Power: P =  t´w
Moment of Inertia:
I = å_{i} m_{i}r_{i}^{2} = ò r^{2}dm
Parallel Axis Theorem:
I' = I_{cm} + Md^{2}
Static Equilibrium:
1. Net Force must be zero.
2. Net Torque must be zero.
Final Exam: 13 May 1998
#1. A small satellite is orbiting the earth [mass of earth = 5.98 ´ 10^{24}
kilograms; radius of earth = 6.378 ´ 10^{6} meters]. At closest approach
[perigee] the satellite lies 1000 kilometers above the earth's surface and is
moving at 10.0 kilometers/second.

Determine the speed and altitude at apogee [greatest distance
from earth].

Determine the period of revolution of the satellite.
Solution:
The energy and angular momentum are the same at the perigee and apogee,
as at all points on the orbit. At the perigee the satellite distance [from
the center of the earth] is r_{1} =
7.378´10^{6} meters and its speed
is v_{1}= 10000 meters/sec,
whereas at the apogee these quantities are r_{2} and v_{2},
respectively.
Energy conservation yields
½mv_{1}^{2}  GM_{e}m/r_{1} =
½mv_{2}^{2}  GM_{e}m/r_{2};
or
v_{1}^{2}  v_{2}^{2} =
2GM_{e}/r_{1} [1  r_{1}/r_{2}] .
For angular momentum conservation, we have m v_{1} r_{1} =
m v_{2} r_{2}, or v_{2} =
v_{1} r_{1}/r_{2}. Inserting this into the
energy relation, we obtain
v_{1}^{2}
[1  r_{1}^{2}/r_{2}^{2}] =
2GM_{e}/r_{1} [1  r_{1}/r_{2}] ; or
1 + r_{1}/r_{2} =
2GM_{e}/[r_{1}v_{1}^{2}] =
2 ´ (6.67 ´
10^{11})(5.98´10^{24})
/[7.378´10^{6}(10000)^{2}]
= 1.081.
Thus, r_{2} = r_{1}/0.081 =
9.11´10^{7} meters.
The semimajor axis is R = [r_{1} + r_{2}]/2 =
4.93´10^{7} meters.
The period of revolution is T = 2p
Ö[R^{3}/Gm] =
2pÖ
[(4.93´10^{7})^{3}/
{(6.67´10^{11})
´
5.98´10^{24})}] =
1.09´10^{5} sec = 30.3 hours.
#2. A football is passed to a receiver with an initial velocity of 30
meters/second at an angle of 30° above the horizontal. The receiver is
initially 30 meters from the point of impact of the ball.

Calculate the range of the ball, in meters.

Calculate the time of flight, in seconds.

How fast must the receiver run [in meters/second] to catch the ball?
Solution:
The x [horizontal] and y [vertical] components of position of the ball
[from launch] are
x = v_{0}cosqt
and
y = v_{0}sinqt 
½gt^{2}.
We set y = 0 in the second formula and obtain
the time the ball returns to earth:
t = 2v_{0}sinq/g =
2´30´
sin30°/9.8 = 3.06 seconds.
The value of x at this time [range] is
R = v_{0}cosqt = 30
´cos30°
´3.06 = 79.5 meters.
The receiver must thus travel 30 meters in 3.06 seconds, corresponding to
V = 30/3.06 = 9.8 meters/sec
[Note: this corresponds to 100 meters in 10.2 seconds; a record pace.]
#3. A box of mass 10 kilograms lies on
a plane inclined at 30° to the horizontal.
The coefficient of kinetic friction between
the box and the plane is m_{k}
= 0.20. The box starts from rest and slides down the
plane a distance of 8 meters.

Determine kinetic energy of the box [in Joules].

Determine the time required for the box to travel [in seconds].

Determine the final speed of the box, in meters/second.
Solution:
The forces acting on the box are its weight W [mg, downward],
the normal force of the plane N [out of the plane] and the friction
force of the plane f_{k} [up the plane]. Since there
is no motion perpendicular to plane, the net component of force in that
direction is zero; N = mg cos30°. From Newton's second law, the
acceleration down the plane a is given through the relation
ma = mg sin30°  f_{k} = mg sin30° 
m_{k}N =
mg[sin30°  m_{k}
cos30°] .
Thus, a = g[sin30°  m_{k}
cos30°] = 9.8[0.5  0.3´0.866] =
3.02 meters/second^{2}
Starting from rest and travelling a distance d = 8 meters, the box has
speed v_{f} = Ö[2ad] =
Ö[2´
3.20´8] = 7.16 meters/second.
The box travels for a time t = v_{f}/a = 7.16/3.20 = 2.24 seconds.
Its final kinetic energy is K = ½mv_{f}^{2} =
½ 10´(7.26)^{2} =
256 Joules.
#4. A baseball of mass 0.125 kilograms travels horizontally toward a batter
at 40 meters/second. The batter hits the ball, imparting an average force
[horizontal] of 9000 Newtons to the ball for 3.0 milliseconds. If the ball
reverses direction, determine its final speed [in meters/second].
Solution:
By the ImpulseMomentum Theorem, the Change in momentum of the ball
Dp is given by the [average] force F
multiplied by the time interval Dt:
Dp = p_{f}  p_{i} =
FDt = 9000
´.003 = 27 kilogram meters/second.
The initial momentum is p_{i} = mv_{i} =
0.125´ (40) = 5 kilogram meters/second.
Thus p_{f} = 22 kilogram meters/second, and v_{f}
= p_{f}/m = 22/0.125 = 176 meters/second [more than half the
speed of sound!].
#5. A point mass m moves with speed v_{0} directly toward a point
mass M, which is initially at rest. After their elastic collision the
mass m moves with speed v_{0}/2 at an angle of 90° to its
initial direction of motion.

Determine the magnitude and direction of the recoil velocity of
the mass M.

Determine the mass M, in terms of m and v_{0}.
Solution:
The momentum [a vector quantity] and energy are conserved during the
collision, so that
mv_{0} = mv_{f} + MV. Taking
the initial direction to be the xaxis, and the recoil direction to be
the yaxis, and the direction of the mass M an angle
q below the xaxis, here are the
momentum components:
Momentum
 xcomponent
 ycomponent

p_{0}
 mv_{0}
 0

p
 0
 mv_{0}/2

P
 MVcosq
 MVsinq

From momentum conservation, we have
MVcosq = mv_{0}
MVsinq = mv_{0}/2
Thus tanq =
MVsinq/
[MVcosq] = mv_{0}/2/[mv_{0}]
= 1/2, or q = 26.56°
Also from momentum conservation, we have
[MV]^{2} = 5[mv_{0}]^{2}/4
From energy conservation,
½mv_{0}^{2} = ½m(v_{0}/2)^{2}
+ ½MV^{2} ; or
MV^{2} = 3mv_{0}^{2}/4 .
From these relations, we obtain M = 5m/3 and V = 3v_{0}/[2
Ö5] = 0.671v_{0} .
#6. A phonograph turntable is a uniform disk of mass 300 grams and radius
15 cm. Initially at rest, it undergoes uniform angular acceleration to a final
rotational rate of 45 revolutions per minute over a period of 10 seconds. It
then rotates uniformly at that rate for an additional 10 seconds.

How many revolutions did it make over the full interval of 20
seconds?

Determine the radial and tangential components of acceleration
of a point at the edge of the turntable, just before the end of 10
seconds, in meters/second^{2}.

How much torque [in Newtonmeters] is required to drive the
turntable during acceleration?
Solution:
The initial speed is w_{0} = 0,
whereas the final speed is w =45 rev/min
´2p
rad/rev ´ 1min/[60 sec] =
4.71 radians/second.
This angular velocity is acquired under uniform angular acceleration for
10 seconds, so that a = 4.71/10 =
0.47 radians/second^{2}.
For the first 10 seconds, the turntable rotates through an angle
q = ½
at^{2} = ½
´0.47´
(10)^{2} = 23.56 radians. For the next 10 seconds, it
rotates through an angle of 4.71´10 =
47.12 radians. The total rotational angle is 70.69 radians, or 11.25
revolutions.
The radial acceleration is
a_{r} = w^{2}R
= (4.71)^{2} ´0.15 =
3.33 meters/second^{2}.
The tangential acceleration is
a_{t} = aR
= 0.47 ´0.15 = .071 meters/second^{2}.
The moment of inertia of the turntable [a uniform disk] is
I = ½MR^{2} = ½ ´
0.3 ´(.15)^{2} =
0.00337 kilograms meter^{2}. From the torque relation, we get
t = Ia =
0.00337´0.471 = .00159 Newton meters.
#7. A 1.0 kilogram mass is connected to a
string that makes an angle q
with the vertical rotates uniformly in a horizontal
circle of radius 0.1 meters. If its speed is
1.1 meters/second, determine the tension
T in the string [in Newtons], the acceleration
of the mass [in meters/second^{2}], and the
angle q, in degrees.
Solution:
The two forces acting on the mass are its weight W [mg downward]
and the tension T in the string [in the direction of the string
toward the pivot point]. The net force in the vertical direction is
zero, since there is no motion in that direction. Thus,
Tcosq = mg =
1.0´9.8 = 9.8 Newtons.
The horizontal component of force is
Tsinq = ma_{cent} = mv^{2}/R
= 1.0 ´(1.1)^{2}/(0.1) = 12.1 Newtons.
Thus T = 15.6 Newtons and tanq = 12.1/9.8 =
1.23, so that q = 51°.
#8. A uniform bar is hung horizontally
from a vertical wall, to which it is
attached at a hinged joint at the wall,
as well as a by massless "guy wire"
at the other end. The wire
is inclined at an angle q to the vertical
direction. In terms of the weight W
and length l of the bar:

Determine the tension T in the guy wire;

Determine the Force F on the hinge [magnitude and direction].

If the maximum tension in the wire is 5 times the weight of
the bar, determine the maximum value of q.
Solution:
The three forces on the bar are its weight W [mg downward, at its
center], the tension T in the guy wire [along the wire toward the
wall], and the contact force F [horizontal component H and
vertical component V]. Taking the xaxis as horizontal [away from the wall]
and the yaxis as vertical [up], the forces have these components:
Force
 xcomponent
 ycomponent

W
 0
 mg

T
 Tsinq
 Tcosq

F
 H
 V

Resultant
 0
 0

Thus, we have Tsinq = H and
Tcosq + V = mg.
Balancing torques about the end of the rod, we have
mgl/2 = Tcosq l, or
Tcosq = mg/2.
Consequently, T =mg/[2cosq]; V = mg/2; and
H = T/sinq =
mg tanq/2.
The magnitude of the force at the hinge is F =
mg/[2cosq].
For the critical case we have T = 5mg =
mg/[2cosq]; or
cosq = 0.1; or
q = 84.26°.
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Last update 11 January 1999 by
Porter Johnson