Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
We know that impulse is simply the product of Force and time:
Impulse = Force * time
Since Force has a unit of Newton or kg m/s^2 and time is in
seconds, therefore impulse can have units as:
N s
or
<span>kg m/s</span>
A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find

B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

The answer is C I believe
ANSWER.no cause they have to be going a the same speed and I doubt that are gonna go at the same speed.