Question

A meteoroid of mass = 555 kg has a speed of 90.0 m/s when 700 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.49 m. (a) How much work does the force of gravity do on the meteoroid on the way to the surface? GJ (b) What is the speed of the meteoroid just before striking the sand? m/s (c) How much work does the force of gravity do on the meteoroid after it hits the surface? GJ (d) How much work does the sand do to stop the meteoroid? GJ (e) What is the average force exerted by the sand on the meteor? GN (f) How much thermal energy is produced? GJ

Answer 1

a)

Work done is given by : W = . = F S COS

Where,

W = Work done

F = Force

S = displacement

= angle between Force Vector and Displacement vector.

As the meteoroid falls ,the only force acting on it is the gravity,since air resistance has been neglected in the question.

The work done by gravity is possitive as mg (gravitational force) acts in downwards direction and displacement is also

in downward direction.

W_gravity = F S cos

F = mg = 555 x 9.8 = 5439 N

S = 700 km = 700000 metres

= 0 ; cos = 1

Hence , W = mgs = (5439 x 700000) joules = 3807300 kilo joules

b)To find the velocity before hitting the ground we cannot directly apply equation of constant acceleration,as the acceleration due to gravity varies with height if the height is as large as 700 km.

Apply energy conservation principle to find out the velocity before hitting.

Let us denote the initial position of 700 km height as point A and the ground as point B

(K.E + P.E) _A = (K.E +P.E)_B

To solve the question further,the question should dpecify the radius of earth and mass of earth, because the formula of potential energy for the earth - meteroid system is given by:

U = - G M_meteroid xM_earth /R_e +h

Where,

R_e = Radius of the earth

However,you can take radius of earth = 6400 km and solve the question.

c)

To solve this we need to know the answer of part (b)