Question

A meteor whose mass was about 1.3×10⁸ kgstruck the Earth (mE=6.0×1024kg) with a speed of about 17 km/s and came to rest in the Earth.

a) What was the Earth's recoil speed?

b) What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth?

c) By how much did the Earth's kinetic energy change as a result of this collision?

Answer 1

Given,

m = 1.3 x 108 kg ; Me = 6 x 1024 kg ;

v = 17 km/s = 17000 m/s

(a)Let V be the recoil speed of earth.

As the collision is completely inelasti, only momentum of the system will be conserved.

Pi = Pf

mv = (m + Me) V

V = mv/(m+Me)

V = 1.3 x 10⁸ x 17000 / (1.3 x 10⁸ + 6 x 10²⁴) = 3.683 x 10^-13 m/s

Hence, V = 3.683 x 10^-13 m/s.

b)The kinetic energy of meteor is:

KEm = 1/2 m v² = 0.5 x 1.3 x 10⁸ x (17000)² = 1.88 x 10¹⁶ J

KEe = 1/2 Me V² = 0.5 x 6 x 10²⁴ x (3.683 x 10^-13)² = 0.407 J

The fraction will be:

KEe/KEm = 0.407J /  1.88 x 10¹⁶ J = 2.16 x 10^-17

Hence, fraction = 2.16 x 10^-17 gets transformed tp earth.

c)The initial KE of earth was 0 and later it became KEm = 0.407 J

KEm = 0.407 J.