Question

A 584.5 g rubber ball is dropped from a height of 24.31 m and undergoes a perfectly elastic collision with the earth. What is the earth's speed after the collision? Assume the earth was at rest just before the collision. How many years would it take the earth to move 4.65 mm at this speed?

Answer 1

here,

mass of rubber ball , m1 = 584.5 g = 0.5845 kg

height , h = 24.31 m

the speed of ball before the collison , u1 = sqrt(2*g*h)

u1 = sqrt(2*9.81*24.31) m/s

u1 = 21.8 m/s

mass of earth , m2 = 5.98 * 10^24 kg

let the speed of earth and ball be v1 and v2

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.5845 * 21.8 = 0.5845 * v1 + 5.98 * 10^24 * v2 .....(1)

and

using conservation of energy

0.5 * m1 * u1^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

0.5845 * 21.8^2 = 0.5845 * v1^2 + 5.98 * 10^24 * v2^2 .....(2)

from (1) and (2)

v1 = - 21.8 m/s

v2 = 4.26 * 10^-24 m/s

s = 4.65 mm = 0.00465 m

the time taken to travel this distance , t = s /v2

t = 0.00465 /( 4.26 * 10^-24) s

t = 8.31 * 10^14 years