A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest. What is the common velocty of the two-car train after the collisions?

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,

V is the common velocity after the collisions

So, the two car train will move with a common velocity of 6.79 m/s.

Let m =  mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.

Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)

The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
     = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
    = 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)

1.1x10¹⁶ kg