Comic-strip hero Superman meets an asteroid in outer space, and hurls it at 750 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?

His momentum would have to equal that of the asteroid, but in the opposite direction. That would require that

V(superman) = 3000*750 m/s = 2.25*10^6 m/s

which is about 1% of the speed of light.

Where do you get 10^6 m/s?

So is the answer 22.5 or 225,000,000

No, it is neither of those two numbers you wrote.

It is what it is.
Do the multiplication.

Comic-strip hero Superman meets an asteroid in outer space and hurls it at 800 m/s, as fast as a bullet. The asteroid is 1000 times more massive that Superman. In the strip, Superman is seen at rest after the throw. Taking physics into account, what would be his recoil velocity? Show your work?...

Whats the answer?

To determine Superman's recoil velocity, we can use the principle of conservation of momentum. According to this principle, in the absence of external forces, the total momentum of a system remains constant.

In this case, the system consists of Superman and the asteroid. Let's assume that Superman's initial velocity is zero (since he is seen at rest after the throw) and denote the final velocity of the asteroid as v_af and Superman's recoil velocity as v_sr.

The momentum of the asteroid before the throw can be calculated using the equation:

momentum_initial = mass_asteroid * velocity_initial

Since the mass of the asteroid is given as three thousand times more than Superman's mass, we can write:

momentum_initial = (3000 * mass_Superman) * velocity_initial

The momentum of the system after the throw can be calculated as the sum of the momentum of the asteroid and Superman's momentum:

momentum_final = (3000 * mass_Superman) * v_af + mass_Superman * v_sr

According to the principle of conservation of momentum, momentum_initial should be equal to momentum_final:

momentum_initial = momentum_final

Working with the given values and rearranging the equation, we get:

(3000 * mass_Superman) * velocity_initial = (3000 * mass_Superman) * v_af + mass_Superman * v_sr

Since mass_Superman is a common factor, it cancels out on both sides of the equation:

(3000 * velocity_initial) = (3000 * v_af) + v_sr

Now, we can substitute the values provided in the question. The velocity_initial is given as 750 m/s:

(3000 * 750) = (3000 * v_af) + v_sr

2250000 = 3000 * v_af + v_sr

Lastly, from the information given in the question, we know that Superman hurls the asteroid at the same speed as a bullet, which is approximately 750 m/s. Therefore, the final velocity of the asteroid (v_af) would also be approximately 750 m/s.

Substituting this into the previous equation, we can now solve for Superman's recoil velocity (v_sr):

2250000 = 3000 * 750 + v_sr
2250000 = 2250000 + v_sr
v_sr = 0

Therefore, the recoil velocity of Superman after throwing the asteroid is 0 m/s (at rest).

Note: This answer assumes an idealized scenario without considering factors such as air resistance or the effects of Superman's superhuman strength.