Question

You and your crew must dock your 2.55 × 10⁴ kg spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha’s control tower has requested that you ram another vessel, a freight ship of mass 1.60 × 10⁴ kg, latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of 1.80 × 10⁷ kg. Alpha’s automated system that guides incoming spacecraft into dock requires that the incoming speed is less than 2.0 m/s.

(a) Assuming a perfectly linear alignment of your ship’s velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha’s docking port, what must be your ship’s speed (before colliding with the freight ship) in order that the combination of the freight ship and your ship arrive at Alpha’s docking port with a speed of 1.40 m/s?
(b) What will be the velocity of Spaceport Alpha when the combination of your vessel and the freight ship successfully docks with it?
(c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of freight ship in that case (assuming your ship had the same initial velocity as you had calculated in part (a))?

Answer 1

A) we are just required to conserve the momentum of the system consisting our ship and freight ship.

Given initial velocity of frieght ship = 0 = V​​​​​​2

Final combined velocity = 1.40m/s = V

Mass of our ship = 2.55*104kg = m​​​​​​1

Mass of freight ship = 1.60*104kg = m​​​​​​2

Assumed initial velocity of our ship = V​​​​​​1

momentum conservation,

m​​​​​​1*V​​​​​​1 + m​​​​​​2*V​​​​​​2 = (m​​​​​​1 + m​​​​​​2)*V

2.55*104*V​​​​​​1 + 1.60*104*0 = (2.55+1.60)*104*1.40

Solving this we get,

V​​​​​​1 = 2.278m/s, approximately (2.28 m/s)

B) if it is still going and combining with the spacecraft, then, again applying momentum conservation,

m​​​​​​s = mass of spacecraft = 1.80*107 kg

V' = final velocity of the system,

(m​​​​​​1 + m​​​​​​2)*V = (m​​​​​​1 + m​​​​​​2 + m​​​​​​s)*V'

(2.55+1.60)*104 * 2.278 = (2.55*104 + 1.6*104 +1.80*107)*V'

Solving this we get,

V' = 0.00322m/s = 3.22mm/s

C) if collision is elastic then,

Let the velocity of your ship after collision be '+x' and the velocity of freight ship after collision be 'y'

So now applying momentum conservation,

m​​​​​​1*V​​​​​​1 + m​​​​​​2*V​​​​​​2 = m​​​​​​1*x + m​​​​​​2*y

2.55*104*2.28 + 1.60*104*0 = 2.55*104*x + 1.60*104*y (i)

Secondly we know the collision is elastic hence coefficient of restitution is unity.

That is 'e' = 1

Which is the ratio of velocity of separation is to velocity of approach.

Velocity of approach = V​​​​​​1​​​​​ - V​​​​​​2 = 2.28m/s

Velocity of separation = y - x

And since e = 1

We get ,

(y - x)/2.28 = 1

y - x = 2.28 (ii)

Solving (i) & (ii),

We get,

x = 0.522 m/s

y = 2.802 m/s

So the velocity of freight ship after collision is 2.80m/s approximately.