Physics 150

Fall 2003

Homework 3 Solution


Question 1

Newton’s theory of gravitation postulated the existence of an attractive force, between all massive objects, that depended on the product of the two masses, the inverse square of the distance between the objects and a universal constant G.  His theory accounts for both the motion of heavenly bodies (including Kepler’s laws) and the motion of everyday objects under the influence of Earth’s gravitational field.  In principal, it can account for the motion of any set of bodies interacting by gravitational forces.  Newton’s theory is indeed a scientific theory because it is testable and falsifiable.  Kepler gave a mathematical and geometric relationship that fit the observations of planetary motion.  Newton ventured beyond the observation and postulated a cause to the mathematical relationships of Kepler: the gravitational force.  Moreover, he made the conceptual leap that linked the force that gave Kepler’s laws with the force that acted on the falling apple.


Chapter 2

Problem 10

  1. momentum (p) = mass (m) * velocity (v) || p=m*v

m=3 kg, v=16 m/s || p=48 kg*m/s or N*s

  1. Velocity is zero for 5 kg ball therefore p=0.
  2. Momentum is conserved for this collision because no external forces act on the balls, therefore p=48 N*s.
  3. Since the balls stuck to one another after the collision, there combined mass is m=8 kg.  The total momentum of 48 N*s is conserved thus the velocity of the combined balls is v=48 N*s/8 kg=6 m/s.
  4. The 3 kg ball is now traveling at 6 m/s, as calculated in part (d).  This means that its new momentum is p=m*v=3 kg * 6 m/s = 18 N*s.
  5. Since the momentum of the 3 kg ball went from 48 N*s to 18 N*s, there was a net transfer of (48-18) N*s = 30 N*s from the 3 kg ball to the 5 kg ball.


Chapter 3

Question C

Newton’s 3rd law is the most important feature in describing how a rocket works.  The rocket burns fuel which exhausts at the bottom.  The rocket pushes off the exhaust with an equal and opposite force that the exhaust pushes off it.  This causes the rocket to accelerate in the forward direction while the exhaust accelerates in the reverse direction.  The acceleration of the rocket at any given moment is generally given by Newton’s second law (F=ma), but the problem is more complicated because as the rocket burns fuel its mass changes.

Another way of looking at the rocket/exhaust exchange is to understand that a rocket “throwing away” exhaust is similar to a reverse collision, where the two objects fly apart rather than crash together.  So long as no external forces act on the rocket and exhaust, the momentum of both remain constant.  On Earth, gravity acts on both, therefore the total momentum is not absolutely conserved.  However, if you look at the rocket in deep space, away from heavy heavenly bodies like planets, stars and galaxies, then you would see indeed that the momentum gained by the rocket going forward by pushing off the exhaust is balanced by the momentum gained by the exhaust by pushing off the rocket.


Problem 2

The net force here is 10 N.  Therefore the acceleration is given by a=F/m=10 N/5 kg = 2 m/s2.


Problem 6

When objects move in a circle at constant speed, they are actually accelerating.  That is because it is changing direction, which means that the velocity (speed and/or direction) is changing.  In uniform circular motion, the speed stays constant but the direction always changes.  This means the car in this problem accelerates, called centripetal acceleration, a=v2/r, where v is the speed and r is the radius of turning.  In this problem, v=15 m/s and r=75 m: a = 152/75 m/s2= 3 m/s2.


Problem 8

We know by Newton’s 2nd law that an accelerating object must have a net force that causes them to accelerate.  This rule holds applies to objects executing circular motion because they are accelerating.  The net force required to accelerate an object of mass m is given by F=m*a.  In uniform circular motion, like in problem 6 and in this one, a = v2/r.  Therefore, F=m*v2/r = 6 kg * (20 m/s) 2/ 10 m = 240 kg*m/s2= 240 N.


Chapter 4

Question A

There are three items that were verified by the observation of Kepler’s laws applied to the satellites of Jupiter:

1)      there is a force

2)      that is attractive

3)      and depends of 1/r2

One could argue that since the force exists between Jupiter and its satellites that the force must exist between all objects.  It, at least, verifies that the force exists outside of the Sun, Moon, and Earth.  Or, at most, it verifies that the force exists between objects beyond a certain size, but this is not a verification that the force exists between a glass and a book say.


Problem 5

There are two ways of getting this answer.

Method 1 (Kepler’s 3rd law as a ratio):

R3T2= r3t2where r=6500 km and t=80 minutes for the known satellite and T=24*60=1440 minutes for the geosynchronous orbit (has to revolve at same period as the Earth rotating about its axis).

Solving the equation for the radius R: R = r (T2/t2)1/3  = 6500*18 2/3 km = 44600 km.

Method 2 (Using Newton’s gravitational force equation):

Force on the satellite is given by: GMm/R2 here G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the satellite and R is the radius of orbiting for the satellite.  Since gravity is the only force acting on the satellite we can set this force equal to the mass times acceleration: GMm/R2  m*a, where a is the acceleration of the satellite as it circles the Earth.  Notice that the mass of the satellite appears on both sides of the equation, canceling out.  This means that the radius of the geosynchronous orbit does not depend on the mass of the satellite (provided that the mass is not so great as to effect the Earth’s orbit).  The acceleration is given by the formula for uniform circular motion: a=v2/R.  How fast does the satellite orbit?  Well, it must move one circumference of its circular orbit in one day: v=2*pi*R/T.  So, GM/R2=v2/R=(2*Pi*R/T) 2/R or after a little algebraic manipulation R3T2= GM/(2*Pi) 2 (can you see where Kepler’s 3rd law comes from?).  Using the numbers for each variable, G=6.67E-11 m3(kg*s2), T=24*60*60=86400 s, M=5.98E24 kg and Pi=3.14159…, we find R=42300 km.


Chapter 5

Question B

As soon as the ball leaves the bat, it has its maximum KE.  As it flies through the air, the ball loses KE to air resistance which dissipates in the form of heat.  Also, the ball converts some (but not all) of its KE into gravitational PE as it rises.  When it reaches the top of its flight, its PE is at a maximum, but the ball still has KE because it is moving forward as well (unless the ball was hit straight up).  On the way down, it reconverts PE into KE and continues to lose KE due to air resistance.  Finally, the ball lands in the mitt and all of its KE is lost into heat, sound, and elastic vibrations in the glove.


Problem 2

Work is equal to the force times the distance that is parallel to the force (remember force has both magnitude and direction).  In this simple case, the force and distance are exactly parallel to one another so there is no need to involve angles into the problem.  The answer is simply W=F*d=m*g*d=5 kg*10 m/s2*30 m = 1500 kg*m/s2 r J (joules).  Continuing the problem, since there is no friction or air resistance in the problem all of the work is reconverted back into KE when the ball falls back to the original height.  The answer is 1500 J.