As per the question, the mass of meteorite [ m]= 50 kg
The velocity of the meteorite [v] = 1000 m/s
When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.
We are asked to calculate the gain in kinetic energy of earth.
The kinetic energy of meteorite is calculated as -
![Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2](https://tex.z-dn.net/?f=Kinetic%5C%20energy%5C%20%5BK.E%5D%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![=\frac{1}{2}50kg*[1000\ m/s]^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D50kg%2A%5B1000%5C%20m%2Fs%5D%5E2)

Here, J stands for Joule which is the S.I unit of energy.
Answer: This is the orbit (of the moon around Earth).
An orbit is a circular/oval path that planets, moons, comets, etc follow with a "subject" in the middle. In this case, the circle is the orbit of the moon around Earth.
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>