QUIZES GIVEN IN PHYSICS 103, WITH SOLUTIONS: SPRING 1998


INDEX OF QUIZES

QUIZ 1
QUIZ 2
QUIZ 3
QUIZ 4
QUIZ 5
QUIZ 6
QUIZ 7
QUIZ 8
QUIZ 9
QUIZ 10
QUIZ 11
QUIZ 12
TEST 1
TEST 2
EQUATION SHEET
FINAL EXAM

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.]
  1. How high above the water was the man?
  2. 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 = ½gt2, 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 = v0 + gt, where v0 = 0. Thus v = 9.8´3 = 28.4 meters/sec.
Quiz 2: 04 February 1998

A particle launched with initial speed v0 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.
  1. With what speed v0 was the particle launched? [in meters/second]
  2. 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 v0 and the launch angle q by the formulas.
x = v0cosq t ;
y = v0sinq t - ½gt2 .
We solve the second of these for t; t = x/v0cosq; insert that value into the first equation; and get
y = xtanq - gx2/(2v02 cos2q)
Setting y = 0 in that formula and solving for x, we obtain the range
R = v02 sin2q/g .
To get the maximum height, we set dy/dt = 0, and obtain t = v0sinq/g. Inserting that time into the equation for y, we obtain H = v02sin2 q/(2g).
Thus, H/R = sin2q/(2 sin2q) = tanq/4. Inserting the numbers, we get tanq = 16/12, or q= 53°. Going to the Range equation, we get v02 = Rg/sin2q = 12´9.8/0.96 = 122.5. Thus, v02 = 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.
  1. What is the acceleration of the two boxes? [in meters/second2]
  2. What is the force that the smaller box exerts on the heavier box [magnitude and direction]?
  3. 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 action-reaction 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/sec2. 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 mS = 0.3.
  1. Draw a diagram showing all the forces involved, and determine the horizontal and vertical components of each of them.
  2. Determine the greatest speed vcritical with which the vehicle will go around the curve without slipping off. [You may express your answer in terms of R, q, mS, 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 mv2/R.
Here are the horizontal [x] and vertical [y] components of the forces:
Force x-component y-component
W 0 -mg
N -N sinq N cosq
f -f cosq f sinq
R -mv2/R 0
Applying Newton's second law to horizontal and vertical components, we obtain
N cosq - f sinq -mg = 0
- N sinq - f sinq = -mv2/R
For the critical case, f = mSN, we get
N [cosq - mS sinq] = mg
N [sinq + mS cosq] = mv2/R
Taking the ratio of these equations, we eliminate N and m to get
v2/(Rg) = [mS + tanq]/[ 1 - mS tanq]
Putting in the numbers, we have
v2 = 200´9.8 ´[0.30 + 0.18]/ [1-0.30´0.18] = 990.
Thus, vcritical = 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 mk = 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.
  1. How far does the box move in this process?
  2. What is the change in kinetic energy of the box [in Joules]?
  3. Determine the work done by the friction force on the box [in Joules].
  4. 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 = mkN [to the left]. Applying Newton's second law to horizontal and vertical components, we get
N = mg ; F - mkN = ma
or 300 - 0.5´50 ´9.8 = 55 = 50 a. The acceleration is thus 1.1 meters/second2. The distance travelled is calculated from the formula
v2 = v02 + 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 ½mv2 = ½´50 ´22 = 100 Joules.
The work done by the friction force [245 Newtons acting in opposition to a displacement of 1.82 meters] is Wfric = - 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 mk = 0.5. The block is moved up the plane by a constant force F parallel to the plane.
  1. Draw a diagram showing all the forces acting on the box.
  2. Determine the magnitude of the applied force F [in Newtons].
  3. Determine the power expended by the applied force F [in Watts].
  4. How much power would be required to move the box up the plane if there were no friction [mk= 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 fk [down the plane]. The net force acting on the box must be zero. Let us separate them into components up the plane [x-direction] and out of the plane [y-direction]:
Force x-comp [up plane] y-comp [out of plane]
W -mg sinq -mg cosq
N 0 N
F F 0
fk -fk 0
Setting the total force [x-component and y-component] to zero, we get
N = mg cosq
F = mg sinq + fk
We have fk = mkN = mkmg cosq, so that
F = mg(sinq + mk 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.
  1. Determine the speed v with which the particles move in the orbit, expressed in terms of G [Newtonian gravitational constant], m, and R.
  2. 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 m2/(4 R2) .
Each mass moves with speed v in a circle of radius R, with centripetal acceleration ac = v2/R.
Thus F = G m2/(4 R2) = m ac = mv2/R.
We may solve for the velocity to obtain v2 = Gm/(4R), or v = Ö[Gm/(4R)].
The angular velocity of motion of each mass is given by w = v/R = Ö[Gm/(4R3)].
The period of revolution is T = 2p/w = 2p Ö[4R3/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.
  1. [a] Determine the magnitude and direction of the velocity of the composite object immediately after the collision [in meters/second].
  2. How much mechanical energy is dissipated during the initial collision [in Joules]?
Solution:
Momentum [a vector] is conserved during the collision. Define the x-direction to be East and the y-direction to be North. Thus
Pinit = 2000 ´20x + 1000 ´30y = 40000x + 30000y ; Pfin = 3000v .
As a consequence of Pinit = Pfin ,
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 Ki = ½1000´(30)2 + ½2000´(20)2 = 8.5 ´105 Joules.
The final kinetic energy is Kf = ½3000´(16.67)2 = 4.2 ´105 Joules.
Thus the energy dissipated is 4.3 ´105 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 x-direction, and let it recoil at an angle q above the x-axis, so that the target will recoil at an angle q below the x-axis. The momentum [a vector] and the kinetic energy of motion [a scalar] are conserved in this collision. The initial momentum is Pinit = mvx, whereas the final momentum [expressed in terms of the final speeds v1 and v2] is
Pfin = mv1 [ cosq x + sinq y] + mv2 [cosq x - sinq y ] = m [v1 + v2] cosq x + m [v1 - v2] sinq y
From momentum conservation, we get v1 = v2 and v = 2 v1 cosq.
Using energy conservation, we obtain ½mv2 = ½mv12 + ½mv22; or v2 = v12 + v22.
Consequently, v12 = v22 = ½v2.
Finally, v1 = v2 = 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/second2] and the number of revolutions made in this process. Also, compute the tangential and radial components of acceleration [in meters/second2] 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 = w0 + at
q = v0t + ½at2
q = ½(w + w0)t
w2 = w02 + 2aq
The initial angular velocity is w0 = 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/sec2.
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 ar = w2 R = [1.5p]2 ´ 0.10 = .225p2 = 2.22 meters/second2
The [uniform] tangential acceleration is at = a R = 0.15p ´ 0.1 = 0.015p = .0471 meters/second2.
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.]
  1. What is the angular velocity w of the beam immediately after the bullet went through it?
  2. How much mechanical energy is converted into heat during the collision?
  3. 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 + I0w0, where I0 = Ml2/3 is the moment of inertia of the rod about the pivot point. Thus w0 = 3m(V - v)/[Ml].
With initial angular velocity w0 at the bottom, the rod system has kinetic energy K = ½I0 w02.
The rod swings up to an angle q, where Ml2w02/6 = 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 Ki = ½mV2, whereas its final kinetic energy is Kf = ½mv2 + ½I0 w02.
Thus the kinetic energy lost is DE = ½m[V2 -v2] -½I0 w02 = ½m [V - v] [(1+3m/M)V+(1-3m/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 mS. The ladder is in equilibrium when it is tilted at an angle q to the vertical direction.
  1. Draw a diagram showing and labelling all forces acting on the ladder.
  2. Obtain the conditions for static equilibrium, and determine all the forces acting on the ladder.
  3. If the maximum angle q for equilibrium is 30° , what is the coefficient of static friction mS?

You should express the answers to [b] in terms of m, l, g, and mS.
Solution:
The four forces on the ladder are its weight W [downward; at its center], the normal force of the wall N1 [horizontal; away from wall], the normal force of the floor N2 [vertical], and the static friction force f2of the floor [horizontal; toward the wall]. The net force [a vector] must be zero, so that N2 = mg and N1 = f2.
We require the net torque about the contact point of the ladder with the floor to be zero, so that N1l cos q - ½ mgl sin q = 0, so that N1 = ½mg tan q.
In summary, f2 = N1 = ½mg tan q, and N2 = mg, so that mS = f2/N2 = ½tan q.
Inserting q = 30°, we obtain mS = 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 mk = 0.3.
  1. Draw a diagram showing all forces acting on the car.
  2. Determine the deceleration of the car [in meters/second2].
  3. 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 fk = mkN = 2940 Nt.
Therefore, the acceleration is a = - fk/m = - 2.94 meters/sec2.
The initial velocity is v0 = (100/3.6) meters/sec = 27.8 meters/sec.
The distance travelled is d = v02/[-2a] = [27.8]2/ [2 ´2.94] = 131 meters. Tragedy is thus averted.
The time required to stop is t = v0/|a| = 27.8/2.94 = 9.45 sec.

#2. Two masses [m1 = 14 kg; m2 = 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.

  1. Draw a diagram showing all the forces involved.
  2. Determine the tension in the cord [in Newtons].
  3. Determine the acceleration of each mass [in meters/second2].

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 = m1g [downward] and the tension in the string [upward]. Thus, from Newton's second law,
m2g - T = m2a
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 = m2g [downward]. Applying Newton's second law we obtain
N = m2g cos45° ;
T - m2g sin45° = m2a .
We obtain two equations for N and T:
m2g - T = m2a and T - m2g sin45° = m2a.
Solving these equations, we get
a = [m2 - m1 sin45°]/[m2+ m1] = [12 - 14/Ö2] ´9.8/[14 + 12] = 0.79 meters/sec2 ;
T = m1m2g[1 + sin45°]/ [m1 + m2] = 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.

  1. How far does the ball travel horizontally before returning to the ground [in meters]?
  2. What is the maximum altitude of the ball [in meters]?
  3. How long does the ball stay in the air[in seconds]?

Solution:
The relevant trajectory equation is
y = xtanq - gx2/(2v02 cos2q)
Setting y = 0 and solving for x, we obtain the range R = v02sin2q/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 = v02sin2q/[2g] = R/2. Then, insert this in the trajectory equation to get ymax = v02 sin2q/[2g] = (40)2´sin230°/ [2´9.8] = 20.4 meters.
the flight time is given by tflight = R/[v0cosq] = 141/[40 cos30°] = 4.1 seconds.

#4. A person of mass 60 kilograms sits on a merry-go-round of radius 10 meters. The merry-go-round [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.

  1. Draw a diagram labeling all the forces acting on the person.
  2. What is the coefficient of static friction between the person and the merry-go-round?

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 fS [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 fS = mac = mv2/R. for the critical case fS = mSN, or mS = fS/N = v2/[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 x-axis to the East and the y-axis to the North. The displacement R1 is 40 meters along the positive x-axis, and the displacement R2 is 50 meters at 37° below the negative y-axis. Here are the components of R1, R2, and R = R1 + R2:

Displacement x-component y-component
R1 50 0
R2 -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.]
  1. From how high above the window did the cat fall [in meters]?
  2. Determine the speed of the cat at the window base, [in meters/second].

Solution:
The cat starts falling at time t = 0. At time t1 it reaches the top of the window, with speed v1 = gt1, having fallen a distance z1 = ½gt12. It reaches the bottom of the window at time t2, with speed v2 = gt2, having fallen a total distance z2 = ½gt22. We are given
Dt = t2 - t1 = 0.1 seconds .
Dz = 2 meters = z2 - z1 = ½g(t22 - t12) = ½g[(t1+ 0.1)2 - t12] = ½g( 0.2 t1 + 0.01) . Thus t1 = (2.0 - ½´9.8 ´0.1)/ [½´9.8] = 1.99 seconds, and t2 = 2.09 seconds.
The speeds are v1 = gt1 = 9.8 ´ 1.99 = 19.5 meters/sec and v2 = gt2 = 9.8 ´ 2.09 = 20.5 meters/second.
The distances are z1 = ½gt12 = ½ ´ 9.8 ´ (1.99)2 = 19.4 meters and z2 = ½gt22 = 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.

  1. What is the stopping force [in Newtons]?
  2. How much time elapsed between the instant the bullet entered the tree and the instant it stopped [in seconds]?

Solution:
The bullet has velocity v0 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 v2 = v02 + 2ad. Thus a = -v02/[2d] = - (1000)2/[2 ´ 0.1] = - 5 ´106 meters/second2.
By Newton's second law, the stopping force on the bullet [opposite to direction of motion of bullet] is F = ma = 0.01 ´ 5´106 = 5´104 meters/second.
The stopping time is t = v0/a = 1000/ [5 ´106] = 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.]

  1. Determine the least speed of the particle, at the "top" of the circle [in meters/second] for the string to remain taut.
  2. For that case, determine the maximum speed of the particle at the "bottom" of the circle [in meters/second].
  3. 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, Ttop [downward]. If it is moving with speed v, it experiences a downward centripetal acceleration ac = v2/l at that point. Thus mg + Ttop = mac = mv2/l. For the minimum speed at the top, Ttop = 0 and vtop = Ö[gl].
The velocities at the top and bottom of the swing, vtop and vbot, are related by conservation of energy: ½mvtop2 + mgl = ½mvbot2 - mgl, so that ½mvbot2 = 5mgl/2. Thus for the minimum case, vbot = Ö[5gl].
The tension in the string at the bottom Tbot acts upward, whereas the weight acts downward. Thus, Tbot - mg = mvbot2/l = 5mg, so that Tbot = 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.

  1. Draw a diagram labeling all the forces acting on the mass just after release.
  2. How far does the mass m travel before stopping? Express the answer in terms of d, m, k, g, and mk.

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 fk [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 ½kd2, and after release of the block this is the kinetic energy of the block. The energy is all dissipated by the friction force fk = mk N = mk mg. If the block travels a distance l, the work done by friction is - mk mgl. Thus, ½kd2 = mk mgl, or l = kd2/[2mk 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:


Solution:
The potential energy of a mass m on earth is Ue = - GMem/Re, whereas its potential energy on the moon is Um = - GMmm/Rm.
The difference in potentials, DE = Um - Ue, is the energy necessary to take the mass to the moon.
The energy required per unit mass is DE /m = G[Me/Re - Mm/Rm] = 6.67´10-11[ 5.98´1024/ (6.38´106) - 7.95´1022/ (1.74´106)] = 5.95´107 Joules/kilogram .
Equation Sheet Phys 103-004 Spring 1998
Trigonometry:
a2 + b2 = c2    sinq = b/c
cosq = a/c       tanq = b/a
Kinematics in one dimension:
v = v0 + at
x = v0t + ½at2
x = ½(v + v0)t
v2  =  v02 + 2ax 
Vectors:
A = axx + ayy + azz
B = bxx + byy + bzz
A + B = (ax+ bx)x + (ay + by)y + 
                 (az + bz)z
A · B = axbx + ayby  +  azbz 
                   = |a||b|cosq
|A ´ B|  = |A||B|sinq [right hand rule]
Projectile Motion: [g = 9.8 m/sec2] 
x = v0cosqt
y = v0sinqt - ½gt2
vy = v0sinq- gt
y = xtanq - gx2/(2v02cos2q)
Centripetal Acceleration: 
|ac| = |v|2/R     [into center]
Newton's Laws:
    1.  Principle of Inertia
    2.  Fnet = ma = dp/dt
    3.  Action - Reaction
Momentum:  p  = mv
Kinetic Energy:  K = ½mv2
Static Friction:  |fs|£  ms|N|
Kinetic Friction:   |fk| =  mk|N|
Work-Energy Theorem:
  W = DK =  ò F·dr 
Power:    P = dW/dt  = F·v
Potential Energy:
  gravity[uniform]: U = mgh
  spring:   U = ½kx2
Gravitation
  gravitational force [attractive]
    F = Gm1m2/d2
    G = 6.67 ´ 10-11  Ntm2/kg2
  Potential Energy
    U = - Gm1m2/d
Kepler's Laws
1.  Orbit ellipse with sun at focus
2.  Equal areas in equal times
3.  Period2 µ [semi-major axis]3
Center of Mass:
Rcm = 1/Måimiri
Vcm = 1/Måimivi
Fnet = Macm
Rotational Motion:
s = rq ; vT = rq ; aT = ra
Uniform Angular Acceleration:
w =  w0  + at
q =  w0t  + ½at2
q = ½(w + w0)t
w2 = w02 + 2aq
Angular Momentum:
L = m r ´ v = I  w
Torque:  t = r ´ F = dL/dt = Ia
Kinetic Energy:  K = ½Iw2
Power:  P = | t´w|
Moment of Inertia:
I = åi miri2 = ò r2dm
Parallel Axis Theorem:
I' = Icm + Md2
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 ´ 1024 kilograms; radius of earth = 6.378 ´ 106 meters]. At closest approach [perigee] the satellite lies 1000 kilometers above the earth's surface and is moving at 10.0 kilometers/second.

  1. Determine the speed and altitude at apogee [greatest distance from earth].
  2. 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 r1 = 7.378´106 meters and its speed is v1= 10000 meters/sec, whereas at the apogee these quantities are r2 and v2, respectively.
Energy conservation yields ½mv12 - GMem/r1 = ½mv22 - GMem/r2; or
v12 - v22 = 2GMe/r1 [1 - r1/r2] .
For angular momentum conservation, we have m v1 r1 = m v2 r2, or v2 = v1 r1/r2. Inserting this into the energy relation, we obtain
v12 [1 - r12/r22] = 2GMe/r1 [1 - r1/r2] ; or
1 + r1/r2 = 2GMe/[r1v12] = 2 ´ (6.67 ´ 10-11)(5.98´1024) /[7.378´106(10000)2] = 1.081.
Thus, r2 = r1/0.081 = 9.11´107 meters.
The semi-major axis is R = [r1 + r2]/2 = 4.93´107 meters.
The period of revolution is T = 2p Ö[R3/Gm] = 2pÖ [(4.93´107)3/ {(6.67´10-11) ´ 5.98´1024)}] = 1.09´105 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.

  1. Calculate the range of the ball, in meters.
  2. Calculate the time of flight, in seconds.
  3. 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 = v0cosqt and y = v0sinqt - ½gt2.
We set y = 0 in the second formula and obtain the time the ball returns to earth:
t = 2v0sinq/g = 2´30´ sin30°/9.8 = 3.06 seconds.
The value of x at this time [range] is R = v0cosqt = 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 mk = 0.20. The box starts from rest and slides down the plane a distance of 8 meters.

  1. Determine kinetic energy of the box [in Joules].
  2. Determine the time required for the box to travel [in seconds].
  3. 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 fk [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° - fk = mg sin30° - mkN = mg[sin30° - mk cos30°] .
Thus, a = g[sin30° - mk cos30°] = 9.8[0.5 - 0.3´0.866] = 3.02 meters/second2
Starting from rest and travelling a distance d = 8 meters, the box has speed vf = Ö[2ad] = Ö[2´ 3.20´8] = 7.16 meters/second.
The box travels for a time t = vf/a = 7.16/3.20 = 2.24 seconds.
Its final kinetic energy is K = ½mvf2 = 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 Impulse-Momentum Theorem, the Change in momentum of the ball Dp is given by the [average] force F multiplied by the time interval Dt:
Dp = pf - pi = FDt = 9000 ´.003 = 27 kilogram meters/second.
The initial momentum is pi = mvi = 0.125´ (-40) = -5 kilogram meters/second.
Thus pf = 22 kilogram meters/second, and vf = pf/m = 22/0.125 = 176 meters/second [more than half the speed of sound!].

#5. A point mass m moves with speed v0 directly toward a point mass M, which is initially at rest. After their elastic collision the mass m moves with speed v0/2 at an angle of 90° to its initial direction of motion.

  1. Determine the magnitude and direction of the recoil velocity of the mass M.
  2. Determine the mass M, in terms of m and v0.

Solution:
The momentum [a vector quantity] and energy are conserved during the collision, so that mv0 = mvf + MV. Taking the initial direction to be the x-axis, and the recoil direction to be the y-axis, and the direction of the mass M an angle q below the x-axis, here are the momentum components:
Momentum x-component y-component
p0 mv0 0
p 0 mv0/2
P MVcosq MVsinq

From momentum conservation, we have
MVcosq = mv0
MVsinq = mv0/2
Thus tanq = MVsinq/ [MVcosq] = mv0/2/[mv0] = 1/2, or q = 26.56°
Also from momentum conservation, we have [MV]2 = 5[mv0]2/4
From energy conservation, ½mv02 = ½m(v0/2)2 + ½MV2 ; or MV2 = 3mv02/4 .
From these relations, we obtain M = 5m/3 and V = 3v0/[2 Ö5] = 0.671v0 .

#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.

  1. How many revolutions did it make over the full interval of 20 seconds?
  2. 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/second2.
  3. How much torque [in Newton-meters] is required to drive the turntable during acceleration?

Solution:
The initial speed is w0 = 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/second2.
For the first 10 seconds, the turntable rotates through an angle q = ½ at2 = 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 ar = w2R = (4.71)2 ´0.15 = 3.33 meters/second2.
The tangential acceleration is at = aR = 0.47 ´0.15 = .071 meters/second2.
The moment of inertia of the turntable [a uniform disk] is I = ½MR2 = ½ ´ 0.3 ´(.15)2 = 0.00337 kilograms meter2. 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/second2], 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 = macent = mv2/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:

  1. Determine the tension T in the guy wire;
  2. Determine the Force F on the hinge [magnitude and direction].
  3. 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 x-axis as horizontal [away from the wall] and the y-axis as vertical [up], the forces have these components:
Force x-component y-component
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