Comic-strip hero superman meets an asteroid in outer space, and hurls it at 850 m/s, as fast as a bullet. the asteroid is three thousand times more massive than superman. in the strip, superman is seen at rest after the throw. taking physics into account, what would be his recoil velocity?

Question

Maxim Burton

Physics

29 December, 17:22

Comic-strip hero superman meets an asteroid in outer space, and hurls it at 850 m/s, as fast as a bullet. the asteroid is three thousand times more massive than superman. in the strip, superman is seen at rest after the throw. taking physics into account, what would be his recoil velocity?

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Answers (1)

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Keith Garner · Answer 1

Before the asteroid is threw, the total momentum is zero, since neither Superman nor the asteroid are moving.

Conservation of momentum commands the total momentum after the astronaut is threw must be zero too. This means that Superman's backward momentum afterward throwing the asteroid is equivalent to the asteroid forwards momentum, in size.

Momentum is mass times velocity. We know the mass of the asteroid is 1000M and its velocity is 850 m/s, so its momentum is

(1000M) (850 m/s) = 850,000M m/s.

So to get the answer: dividing by Superman's mass, M, gives his recoil velocity o 850,000 m/s.