General Conservation of Momentum Problem

[pinkerpikachu]

A meteor whose mass was about 10^8 kg struck the Earth (m = 6.0 X 10^24) with a speed of about 15 km/s and came to rest in the earth. A) what was the Earth's recoil speed? B) what fraction of the meteor's KE was transformed into KE of the earth? C) By how much did the Earth's KE change as a result of this collison.


I know how to figure out part A. m1v1/m2 = v2 = 2.5 X 10^-12

A little guidance on B and C please?

 

[JoshuaR]

For part B: You know the meteor's initial KE. What is that of the earth's? You know mass of earth, and velocity of earth-meteor, so you should be able to find KE of earth. There you can find the fraction.

An additional question: Does this answer make sense? If yes, why, if not, why? What type of collision is this (elastic, inelastic, etc.)?

 

[thomate1]

B) To calculate the fraction of the meteor's kinetic energy (KE) that was transformed into the Earth's KE, we can use the conservation of momentum equation: m1v1 + m2v2 = (m1 + m2)v3, where m1 and v1 are the mass and velocity of the meteor before the collision, m2 and v2 are the mass and velocity of the Earth before the collision, and v3 is the final velocity of the combined system after the collision.

We can rearrange this equation to solve for the fraction of the meteor's KE that was transformed into the Earth's KE: KE(Earth)/KE(meteor) = (m1v1 - m3v3)/(m1v1). Plugging in the values given in the problem, we get KE(Earth)/KE(meteor) = (10^8 kg x 15 km/s - 6.0 x 10^24 kg x 2.5 x 10^-12 km/s) / (10^8 kg x 15 km/s) = 2.4999 x 10^-12, or approximately 0.0000000000025.

C) The change in the Earth's KE as a result of this collision can be calculated by subtracting the Earth's KE before the collision from the Earth's KE after the collision. Using the equation KE = 1/2mv^2, we can calculate the Earth's KE before the collision as 1/2(6.0 x 10^24 kg)(2.5 x 10^-12 km/s)^2 = 9.375 x 10^16 J. After the collision, the Earth's KE would be 1/2(6.0 x 10^24 kg)(2.5 x 10^-12 km/s)^2 + 1/2(10^8 kg)(2.5 x 10^-12 km/s)^2 = 9.37500000000001 x 10^16 J. Therefore, the change in the Earth's KE is 9.37500000000001 x 10^16 J - 9.375 x 10^16 J = 0.00000000000001 x 10^16 J, or approximately 0.00000000000001 joules.