A 500g rubber ball is dropped from a height of 10m and undergoes a perfectly elastic collision with the earth.
(a) What is the Earth’s velocity after the collision, assuming the Earth is at rest just before the collision.
(b) How many years would it take the Earth to move 1.0mm at this speed?
Well, the second assumption in statement (a) is incorrect, but let’s do the math anyway:
(a)First calculate time for ball to drop 10m
s=1/2 g t^2
s=10m
g=9.8 m/s^2 (the accelaration due to gravity)
solve for t
t= sqrt(2s/g) = sqrt (2.04) or about 1.43 seconds
(2)Next calculate velocity of ball at impact = t multiplied by g
va= 14.01 m/sec
(3)Calculate velocity of Earth after impact
Use equation at top of page of following link
Formulas for elastic collisions
with ma=0.5kg mb=5.9742 × 10^24 kilograms
vb=0
va=velocity calculated above
Vb’ = velocity of earth after impact in meters/second
vb’ = 2 * ma * va / ( ma+mb)
= 2*14.01*0.5 /(0.5+5.97*10^24)
Mass of ball can be ignored in the denominator part of this calculation
= 14.01 / 5.97*10^24
= 2.34 * 10^-24 m/s
(4)Calculate time to move 1.0 mm = s=1×10^-3 meters
T= s/vb’ = 1×10^-3 divided by 2.34 *10^-24
= 0.42 * 10^21 = 4.2 * 10^20
Result will be in seconds
Seconds in one year= 60*60*24*365= 3.15 x 10^7 seconds
Divide T by the number of seconds in one year to get the answer in years.
Answer = 4.2 x 10^20 divided by 3.15 x 10^7
= 1.33 x 10^ 13 years
Also look here for more information:
Elastic collision
Frame of reference
Now let’s redo the calculation but not ASSUME the Earth is at rest when the ball hits:
This is actually a far deeper question than it first appears because we are also dealing with gravity as well as elastic collisions.
(1)At the moment the ball is dropped, the momentum of the ball (and the earth) can be considered zero.
(2)At the moment the ball hits the ground, the ball is moving at:
(a)First calculate time for ball to drop 10m
s=1/2 g t^2
s=10m
g=9.8 m/s^2
solve for t
t= sqrt(2s/g) = sqrt (2.04) or about 1.43 second
(b)Next calculate velocity of ball at impact = t multiplied by g=
va= 14.01 m/sec
(3)However, gravity is a two way street, and the ball is also attracting the earth
The gravitational force of the Earth on the ball= Force of ball on the earth
F=mg = 0.5kg x 9.8 m/s^2 = 4.9 N
Mearth = 5.9742 x 10^24 Kg
Therefore, acceleration of the earth =
F/Mearth = 4.9N divided by 5.9742 x 10^24 Kg = 8.2 x 10 ^-25 m/s^2
Before impact, the earth is moving toward the ball at
Vearth = Aearth x t = 8.2 x 10 ^-25 m/s^2 x 1.43 seconds
Vearth = -1.17 x 10 ^-24 m/s (toward the ball)
(4) To calculate the Velocity after impact use the formula on this page, but first we need
to change reference frame for Mb (the earth) so it it’s velocity is zero.
Since we are interested in the velocity at the time of impact, we can ignore the fact that we have
an accelerating reference frame (due to gravity).
To correct the reference frame we simply subtract Vearth from both objects (Ball and the earth)
http://ac.wwu.edu/~vawter/PhysicsNet/Top…
with ma=0.5kg mb=5.9742 × 10^24 kilograms
vb=0 va=velocity of ball calculated above – Vearth
Vb’ = velocity of earth in meters/second
vb’ = 2 * ma * va / ( ma+mb)
= 2*14.01*0.5 /(0.5+5.97*10^24)
Mass of ball and reference frame correct can be ignored in calculation
= 14.01 / 5.97*10^24
= 2.34 * 10^-24 m/s
now we correct for the change in reference frame by adding
Vearth(before collision) = -1.17 x 10 ^-24 m/s (toward the ball)
Therefore
Vearth(after collision) = 2.34 * 10^-24 m/s -1.17 x 10 ^-24 m/s
= 1.17 x 10 ^-24 m/s (away from the ball)
(5)Calculate time to move 1.0 mm = s=1×10^-3 meters
T= s/vb’ = 1×10^-3 divided by 1.17 *10^-24
= 0.854 * 10^21 = 8.54 * 10^20
Result will be in seconds
Seconds in one year= 60*60*24*365= 3.15 x 10^7 seconds
Divide T by the number of seconds in one year to get your answer.
Answer = 8.54 x 10^20 divided by 3.15 x 10^7
= 2.71 x 10^ 13 years
However, the Earth will NOT keep moving for this period!
It will actually stop as soon as the ball reaches zero velocity at the top of its next bounce.