A small spaceship with a mass of only 2.8 ✕ 103 kg (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a 30 kW laser beam, what speed will the ship attain in 1.0 day because of the momentum carried away by the beam?

962.14m/s

Explanation:

Data obtained from the question include:

m (mass) = 2.8x10^3 kg

P (power) = 30 kW = 30 x 1000 = 30000W

V (velocity) =?

t (time) = 1 day

There are 24 hours in a day.

t (time) = 1 day = 24 hours

We need to covert 24 hours to seconds. This is illustrated below:

There are 60 minutes in 1 hour and 60 seconds in 1 minutes.

Therefore, 24hours = 24 x 60 x 60 = 86400 seconds.

t (time) = 1 day = 24 hours = 86400 seconds

Power is related to velocity according to equation:

Power = force x Velocity

P = F x v (1)

Recall Force (F) = Mass (m) x a (acceleration) i.e F = ma

Substituting the value of F into equation 1, we have:

P = F x v

P = ma x v

P = m x a x v (2)

But: acceleration (a) = Velocity(v)/time(t) i.e a = v/t

Substituting the value of a into equation 2, have:

P = m x a x v

P = m x v/t x v

P = (m x v^2)/ t

Now, with this equation

P = (m x v^2)/ t, we can obtain the speed of the spaceship as follow:

P = (m x v^2)/ t

30000 = (2.8x10^3 x v^2) /86400

Cross multiply to express in linear form

2.8x10^3 x v^2 = 30000 x 86400

Divide both side by 2.8x10^3

v^2 = (30000 x 86400)/ 2.8x10^3

v^2 = 925714.2857

Take the square root of both side

v = √(925714.2857)

v = 962.14m/s

Therefore, the speed of the spaceship is 962.14m/s

found in the nucleus

has mass of one amu

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