A space shuttle, while travelling at a speed of 4000 km/h with respect to the earth, disconnects and ejects a module backward, Weighing one fifth of the residual part. If the shuttle ejects the disconnected module at a speed of 100 km/h with respect to the state of the shuttle before the ejection, find the final velocity of the shuttle. (4020 km/h)

Open in App

Solution

If the final velocity of the shuttle is $V$ then the total final linear momentum

$= \frac{5 M}{6} V + \frac{M}{6} \times (4000 - 100)$

$\Rightarrow 4000 M = \frac{5 M}{6} V + \frac{M}{6} \times (4000 - 100)$

$V = 4020 k m / h$

Suggest Corrections

Similar questions

Q. A space shuttle, while travelling at a speed of

4000 k m / h

with respect to the earth, disconnects and ejects a module backward, weighing one fifth residual part. If the shuttle ejects the disconnected module at a speed of

100 k m / h R

with respect to the state of the shuttle before the ejection, find the final velocity of the shuttle.

Q. The

105, 000 k g

space shuttle is doing a training exercise in outer space.
How much energy would it take to bump up the space shuttle from an altitude of

500 k m

1000 k m

above the surface of the earth?
Assume the radius of the Earth is

6.37 \times 10^{6} m

and the mass of the Earth is

5.98 \times 10^{24} k g

Imagine you are in a space shuttle and went to space but due to any reason you came out of the space shuttle in the space and now the space shuttle is about to go back then how will you enter inside the space shuttle back so that you can come along with it?

Q. A space shuttle of mass 86400 kg is revolving in a circular orbit of radius

6.66 \times 10^{6}

m around the earth. It takes 90.1 minutes for the shuttle to complete one revolution. On a repair mission it moves 1 m closer to a disabled satellite every 3.0 s. Find the kinetic energy of shuttle relative to the satellite

Q. A space shuttle is launched in a circular orbit near the earth's surface. The additional velocity be given to the space - shuttle to get free from the influence of gravitational force, will be.

Join BYJU'S Learning Program

If the final velocity of the shuttle is V then the total final linear momentum=5M6V+M6×(4000−100)⇒4000M=5M6V+M6×(4000−100)V=4020km/h

If the final velocity of the shuttle is $V$ then the total final linear momentum

$= \frac{5 M}{6} V + \frac{M}{6} \times (4000 - 100)$

$\Rightarrow 4000 M = \frac{5 M}{6} V + \frac{M}{6} \times (4000 - 100)$

$V = 4020 k m / h$