At a time when mining asteroids has become feasible,astronauts have connected a

Question

At a time when mining asteroids has become feasible,astronauts have connected a line between their 3580-kg space tugand a 6150-kg asteroid. Using their ship's engine, they pull on theasteroid with a force of 490 N. Initially the tug and the asteroidare at rest, 420 m apart. How much time does it take for the shipand the asteroid to meet?
Thanks! At a time when mining asteroids has become feasible,astronauts have connected a line between their 3580-kg space tugand a 6150-kg asteroid. Using their ship's engine, they pull on theasteroid with a force of 490 N. Initially the tug and the asteroidare at rest, 420 m apart. How much time does it take for the shipand the asteroid to meet?
Thanks!

Explanation / Answer

At a time when mining asteroids has becomefeasible, astronauts have connected a line between their 3760-kgspace tug and a 6250-kg asteroid. Using their ship's engine, theypull on the asteroid with a force of 490 N. Initially the tug andthe asteroid are at rest, 420 m apart. How much time does it takefor the ship and the asteroid to meet in seconds? m = 6,250 kg
M= 3,760 kg
F = 490 N
L = 420 m

Lets denote
A - acceleration of asteroid,
a - acceleration of the ship.

The ship pulls asteroid with force F:
A = F/M

According to the third law of Newton, as re-action asteroid appliesforce -F to the ship:
a = -F/m

Relative acceleration of asteroid with respect too the ship:
a_relative = A-a
a_relative = F(1/M + 1/m)
a_relative = F (M + m)/(Mm)

The distance between the ship and asteroid obeys equation ofaccelerated motion:
d = L - a_relative t²/2

When the ship meets asteroid distance d becomes zero:
d = 0
L - a_relative t²/2 = 0
a_relative t²/2 = L
F (M + m)/(Mm) t²/2 = L


Answer:
time t = [2L/F Mm/(M+m)] = 63.44 second m = 6,250 kg
M= 3,760 kg
F = 490 N
L = 420 m

Lets denote
A - acceleration of asteroid,
a - acceleration of the ship.

The ship pulls asteroid with force F:
A = F/M

According to the third law of Newton, as re-action asteroid appliesforce -F to the ship:
a = -F/m

Relative acceleration of asteroid with respect too the ship:
a_relative = A-a
a_relative = F(1/M + 1/m)
a_relative = F (M + m)/(Mm)

The distance between the ship and asteroid obeys equation ofaccelerated motion:
d = L - a_relative t²/2

When the ship meets asteroid distance d becomes zero:
d = 0
L - a_relative t²/2 = 0
a_relative t²/2 = L
F (M + m)/(Mm) t²/2 = L


Answer:
time t = [2L/F Mm/(M+m)] = 63.44 second m = 6,250 kg
M= 3,760 kg
F = 490 N
L = 420 m

Lets denote
A - acceleration of asteroid,
a - acceleration of the ship.

The ship pulls asteroid with force F:
A = F/M

According to the third law of Newton, as re-action asteroid appliesforce -F to the ship:
a = -F/m

Relative acceleration of asteroid with respect too the ship:
a_relative = A-a
a_relative = F(1/M + 1/m)
a_relative = F (M + m)/(Mm)

The distance between the ship and asteroid obeys equation ofaccelerated motion:
d = L - a_relative t²/2

When the ship meets asteroid distance d becomes zero:
d = 0
L - a_relative t²/2 = 0
a_relative t²/2 = L
F (M + m)/(Mm) t²/2 = L


Answer:
time t = [2L/F Mm/(M+m)] = 63.44 second

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