Question

a space shuttle while travelling at a speed of 4000 km/h with respect to the earth disconnects and ejects the disconnected module at a speed of 100 km/h with respect to the state of the shuttle before the junction what is the final velocity of The Subtle weight of the ejected part is one sixth of the shuttle ​

Answer 1

Answer:

4020k(m/h)

Explanation:

Let the mass of the shuttle including the module is M, then the mass of the module is M/6.

The initial linear momentum before ejection

= M(4000km/h)

After the module is ejected

The velocity of the module with respect to earth = its velocity with respect to the shuttle + the velocity of shuttle with respect to earth

= -100k(m/h) + 4000k(m/h)=3900k(m/h)

If the final velocity of the shuttle is V the the final liner momentum

=(5M/6)V + M/6 * 3900k(m/h)

As per the principle of conservation of linear momentum

M(4000km/h)= (5M/6)V + M/6 * 3900k(m/h)

Or V = 4020k(m/h)

Answer 2

Answer:

thanks bro for this points(_^_)

Explanation:

plz folow me...