Kepler's Third Law

Question: Determine the period of an object that orbit the sun at an average distance of 5 AU.

The equation to the find the planet orbit is given by T^2 / R^3 = K

• T = the time it taken for the planet to complete it orbit in years
• R = average distance of the planet from the sun in Astronomical Unit (AU)
• K= Kepler's constant = 1, Earth has a period of 1 year and an average distance of 1 AU from the sun

T^2 / (5^3) = 1, T = 11.2 years

Newton's Law of Universal Gravitation

Question: A 7500 kg object and a 9500 kg object are separated by a distance of 5 m.

a) What amount of gravitational force exists between them?

   The equation for gravitational force between any two objects is given by F = (G m1 m2) / r^2

• F = the force of gravity (N)
• G = the universal gravitational constant = 6.67 x 10^-11 (Nm^2) / kg^2
• m1 = mass of first object
• m2 = mass of second object
• r = the distance between the centers of the two objects

   F = [(6.67 x 10^-11) (7500) (9500) ] / 5^2 = 1.9 x 10^-4 N

b) The distance between two object is increased to 10 m. What amount of force? Divide the answer into the first answer. What do you get?

   F = [(6.67 x 10^-11) (7500) (9500) ] / 10^2 = 4.7524 x 10^-5 N

    1.9 x 10^-4 N / 4.7524 x 10^-5 N = 4

c) What would the force be if you increase the distance to 15 m?
he
   F = [(6.67 x 10^-11) (7500) (9500) ] / 15^2 = 2.1122 x 10^-5 N

   1.9 x 10^-4 N / 2.1122 x 10^-5 N = 9

This is an example of an inverse square, the force decreases by a factor of 4 when the distance decreases by a factor or 2....
    r = 1, 1^2 = 2
    r = 2, 2^2 = 4
    r = 3, 3^2 = 9
    r = 4, 4^2 = 16
    r = 5, 5^2 = 25
    ......


Question: Two object of equal mass M, and M2 are separated by a distance of 8 m. The gravitational force between these object is 6 N. What is the mass of  each object?

     F = (G m1 m2) / r^2 , m1 = m2 = m
     F = (G m^2) / r^2
     6 N = [(6.67 x 10^-11) m^2 ] / 8^2
     2.399 x 10^6 kg = m

Momentum and Collisions

Question: The 6000kg 787 airplane speed on the runway up to 85 m/s before takeoff. Determine the momentum of the airplane right before takeoff.

The equation of momentum is given by p = m*v

• p, momentum in kg * m/s
• m, mass in kg
• v, velocity of the object in m/s

p = 6000 * 85 = 5.1 x 10^5 kg-m/s

Question: A space craft forces it engines in deep space, the engine exert a force of 9500 N for a period of 20 s. The mass of the spaceship is 3500kg. What is the change in velocity of the spaceship? What is the final speed?

Impulse is a force applied over  a given amount of time, F * t

• F, force applied in Newton (N)
• t, time in seconds

Impulse is the change in momentum induced in an object, 

F * t = m * delta V

• t = 20 s
• F = 9500 N
• m = 3500 kg
• Si, initial speed = 0

(9500) * (20) = (3500) * delta V, delta V = 54.28 m/s

The equation for final speed is give by Sf = Si + a*t

Find a by using Newton's Second Law, F = m*a

(9500 N) = (3500 kg) a, a = 2.714 m/s^2

Sf = 0 + (2.714 m/s^2)* 20 s, Sf = 54.28 m/s

Question: A 10,000 kg freight train car travelling at 10 m/s collide with a second freight train car of mass 9500 kg travelling 2 m/s. After collision the first car is travelling at 4 m/s. What is the velocity of the second car? (This is an example of an elastic collision, which the object do not stick together after collision.)

The equation of conservation of momentum is given by

m1*v1 + m2*v2 = m1*v1' + m2*v2' 

• the left side of the equation is the system before collision
• the right side of the  equation is the system after collision
• m1, mass of first object 
• m2, mass of second object
• v1, velocity of the first object before collision
• v2, velocity of the second object before collision
• v1', velocity of the first object after collision
• v2', velocity of the second object after collision

(10,000) (10) + (9500) (2) = (10000) (4) + (9500) (v2')

solve for v2', v2' = 8.32 m/s

Question: A car of mass 1500 kg travelling at 20 m/s collides with a second stationary car of mass 2000 kg. The car bumper lock after collision. What is the velocity of the wreckage immediately after impact? (This is an example of an inelastic collision, which the object stick together after collision).

                       m1*v1 + m2*v2 = m1*v1' + m2*v2' 

After the collision, the mass is combined, and only one velocity

                        m1*v1 + m2*v2 = (m1 + m2) * V 

                   1500 (20) + 2000 (0) = (1500 + 2000) V 

solve for V, V = 8.57 m/s

Question: A bullet of mass 50 g and travelling at 450 m/s impact and embeds in a 50 kg wooden block suspended by a two sting as shown. a) What is the speed of the block and bullet immediately after collision? b) How much momentum is lost by the bullet? c) How high does the block rise above its initial rest position?

a) 50 g = .050 kg

    m1*v1 + m2*v2     =  (m1 + m2) * V 
(.050) (450) + (5) (0) =  ( .050 +5) V
            22.5 (kg-m)/s =   5.05 kg V
                 4.46 m/s   = V

b) delta P1 = 22.5 - .050 (4.46) = 22.3 (kg-m)/s  lost by bullet

c) delta P2 = m*V = 5(4.46) = 22.3 (kg-m)/s  gain by block

d) a = -9.8 m/s ^2, h= d, Si = 4.46 m/s
    d = (Sf^2 - Si^2) / 2a  = (0 - 4.46 ^2) / (2* -9.8) = 1.01 m

Question: A 500 kg cannon at rest fires a cannon ball of mas 2 kg. The velocity of the cannon ball is 200 m/s. a) What is the recoil velocity of the cannon ball? b) If the cannon is fired horizontally from the top of a tower 60 m tall, how far from the base of the tower does the cannon ball land? (This is an example of Newton's third law of motion, action= reaction).

a)        Before               After
m1*v1 + m2*v2 = m1*v1' + m2*v2' 
(500)(0) + 2 (0) = 500 *v1' + 2 (200)
                      0 = 500 * v1' + 400
              -.8 m/s = v1'
b) d = 1/2 * a * t^2
    60 = 1/2 (9.8) t^2
    3.5 s = t

    dH = vH * t = (200 m/s) 3.5s = 700 m


Question: Identical twins Kate and Karen each have a mass of 45 kg. They are rowing their boat on a hot summer afternoon when they decide to go for a swim. Kate jumps off the front of the boat at a speed of 3.00 m/s. Karen jumps off the back at a speed of 4.00 m/s. If the 70.-kilogram rowboat is moving at 1.00 m/s when the girls jump, what is the speed of the rowboat after the girls jump?

Use conservation of momentum:

momentum before collision = momentum after collision

m1*v + m2*v  + m3*v = m1*vb + m2*v1 + m3 *v2

( combine mass of boat and twins) * v = (70) * vb + (45)*3 + 45 * (-4),
the minus speed for Karen because she jump off to the left in the opposite direction of the moving boat

(70 + 45 + 45 kg) 1 m/s = 70 * vb + 135 + -180
                                160  = 70 * vb + -45
                                 205 = 70 * vb
                        2.93 m/s = vb, boat moving to the right

Note: Check the answer by plug vb into the original equation.

Pendulum Motion

• the time it take to swing back and forth is called the period (T), measured in seconds
• the length (L) of the string affects the period
• the amplitude and mass has no affect on the period and gravitational acceleration (g) 
• the numbers of swing in a second is call the frequency (f), measure in hertz
• period is the reciprocal of frequency, T = 1/f

The equation of the period for small angle (< 15°) is given by T = 2pi √(L/g), g = 9.8 m/s^2

If L = 1 meter, T = 2 * 3.14 * √(1/9.8) =  2 seconds

If gravity is unknown, and given T and L, rearrange the equation above to solve for g:

g = 4 pi^2 (L/T^2)

The equation of the frequency is f = [√(L/g)] / 2pi 

f = [√(1/9.8)] / (2*3.14) = .05 Hertz (Hz)

Find the velocity at the bottom of the swing, apply potential energy =  kinetic energy: mgh = (1/2)mv^2, solve for v =  √(2gh), g = 9.8, h = vertical drop distance

Uniform Circular Motion

Question: A satellite rotate around the earth once time a day at a distance of 4.21 x 10^7 meters from the earth's center. Find the magnitude of acceleration felt by the satellite.

• The centripetal acceleration is given by ac = v^2 / r
• The speed is given by the circumference of the period, v= 2pi*r / T

r = 4.21 x 10^7, T =  one day = 86400 seconds

V = [2pi * (4.21 x 10^7) ] / 86400 = 3062 m/s

a = (3062)^2 / (4.21 x 10^7) = .223 m/s^2

Question: A race car weight 1500 kg is running at a speed of 45 m/s (100 mph) through a circular corner of the track. The radius of the circle path is 180 m. What is centripetal force exert toward the center of the track?

• Centripetal force is given by Fc = mac = (m*v^2 ) / r

     Fc = [1500 kg (45 m/s)^2] / 180 m

     = 16875 N