Comic-strip hero Superman meets an asteroid in outer space, and hurls it at 760 m/s, as fast as a bullet. The asteroid is a 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

Juliana Hatfield

Physics

30 December, 22:34

Comic-strip hero Superman meets an asteroid in outer space, and hurls it at 760 m/s, as fast as a bullet. The asteroid is a 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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Aryanna Raymond · Answer 1

Answer: 1700071.6miles/hr

Explanation:

1. (Initial momentum) = (final momentum)

Therefore,

0 = (Superman's recoil momentum) + (Asteroid's momentum)

0 = {m * (recoil velocity) } + (1000m * 760m/s)

0 = (recoil velocity) + (1000 * 760m/s)

Recoil velocity = - 760,000m/s

Superman is holding an asteroid in outer space, but he hurls it at 760m/s. According to the law of conservation of momentum. Superman should have an equal and opposite recoil momentum in the opposite direction of the asteroid otherwise, the net momentum of the system of the system would not be conserved.

Superman and asteroid are at rest initially. The initial momentum is 0. We are given that the asteroid is 1000 times more massive than the superman, with 'M' as the mass of Superman. The asteroid would have mass of 1000m.

Notice the recoil velocity is solved to be negative: this is because the superman will move in the direction opposite to the asteroid, as expected in miles

In miles = 760,000/sec * 3600sec/hr * 1mile/1609.344m

= 1700071.6miles/hr