If superman in space throws a boulder with 10 times as much mass as himself, and the boulder leaves his hands at 100 m/s, how fast does superman recoil?

A) 1000 m/s.
B) The same speed, 100 m/s.
C) 10 m/s.
D) None of these are correct.

1000 m = m v

v = 1000

To determine how fast Superman recoils, we can apply the conservation of momentum. According to this principle, the total momentum before the event (Superman and the boulder) should be equal to the total momentum after the event (Superman and the boulder).

Let's denote the mass of Superman as M and the mass of the boulder as m. Given that the boulder has 10 times the mass of Superman, it can be expressed as m = 10M.

We can also denote the initial velocity of the boulder as v1 = 100 m/s and the recoil velocity of Superman as v2.

Using the conservation of momentum, we have:
Total momentum before = Total momentum after

(M + m) * 0 = M * v2 + m * (-v1)

Substituting the known values, we have:
(M + 10M) * 0 = M * v2 + 10M * (-100)

Simplifying the equation:
11M * 0 = M * v2 - 1000M
0 = M * v2 - 1000M

Since this equation is valid for any non-zero value of M, we can cancel M on both sides:
0 = v2 - 1000

Therefore, v2 = 1000 m/s

Based on these calculations, the correct answer is option A) 1000 m/s. Superman recoils at a speed of 1000 m/s.