Question

6) A meteor of mass 575 kg has speed 90.0 m/s when it is 800 km above the surface of the Earth. It is falling vertically and strikes a bed of sand in which it is brought to rest in 3.25 m. Ignore the effects of air resistance. a) What is its speed just before striking the sand? b) How much work does the sand do to stop the meteor? c) What is the average force exerted by the sand on the meteor? d) How much thermal energy is produced?

Answer 1

Part A.

Using energy conservation on meteor at initial and final positions:

KEi + PEi = KEf + PEf

KEi = (1/2)*m*Vi^2

Vi = Initial speed of meteor = 90.0 m/s

PEi = initial gravitational potential energy = G*m*Me/Ri

Ri = initial distance of meteor from center of earth = Re + h = 6370 + 800 = 7170 km = 7.17*10^6 m

KEf = (1/2)*m*Vf^2

PEf = -G*m*Me/Rf

Rf = final height of meteor = 6.37*10^6 m

Me = mass of earth = 5.98*10^24 kg

G = Gravitational Constant = 6.674*10^-11

So, Using these values

0.5*m*Vi^2 - G*m*Me/Ri = 0.5*m*Vf^2 - G*m*Me/Rf

Vf^2 = Vi^2 + 2*G*Me*(1/Rf - 1/Ri)

Vf = sqrt [Vi^2 + 2*G*Me*(1/Rf - 1/Ri)]

Vf = sqrt [90.0^2 + 2*6.674*10^-11*5.98*10^24*(1/(6.37*10^6) - 1/(7.17*10^6))]

Vf = final speed of meteor = 3740 m/sec = 3.74*10^3 m/sec

Part B.

Using Work-energy theorem:

W = dKE = KEf - KEi

W = (1/2)*m*Vf^2 - (1/2)*m*Vi^2

W = (1/2)*575*3740^2 - (1/2)*575*90.0^2

W = 4.02*10^9 J

Since above value is work done on the sand by meteor, while we need Work-done by sand, So

Work-done by sand on meteor = -4.02*10^9 J

Part C.

Since We know that:

Work-done = Force*displacement

W = F*d

F = avg force = W/d

d = displacement = -3.25 m

So,

F = -4.02*10^9/(-3.25)

F = 1.24*10^9 N = Average force exerted

Part D.

Thermal energy produced will be equal to work-done by meteor, So from part B

Thermal energy produced = 4.02*10^9 J

Let me know if you've any query.