Question

A space shuttle, while travelling at a speed of 4000km/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 100km/h with respect to the state of the shuttle before the ejection, find the velocity of the shuttle.

Answer 1

We have been given the velocity of the module before the ejection as 100km/h and the velocity of the combined shuttle relative to the earth as 4000km/h. To find the velocity of the module after ejection subtract the velocity of the module before ejection from the combined velocity of the shuttle. Apply conservation of linear momentum and equate the initial and the final momentum together.

Complete step by step solution:
Find the velocity of the shuttle:
Here the mass of the shuttle including the module is M, the mass of the module is given as one-fifth of the actual mass $\dfrac{M}{5}$.
Here we apply the conservation of linear momentum, since there is no net external force applied on the system, so conservation of linear momentum can be applied.
Here the initial linear momentum including the mass of the module is:
${P_i} = MV$;
$\Rightarrow {P_i} = M\left( {4000km/h} \right)$;
Now, after the ejection of the module:
Velocity of the module = Velocity of the module with respect to shuttle before ejection + Velocity of the shuttle with respect to earth.
${V_{after}} = {V_{before}} + {V_e}$;
Put the given values in the above equation:
$\Rightarrow {V_{after}} = - 100 + 4000$;
$\Rightarrow {V_{after}} = 3900km/h$;
The final velocity of the shuttle is V, then the final linear momentum of the shuttle would be:
${P_f} = \left( {4M/5} \right)V + \left( {M/6} \right) \times 3900$;
As per the conservation of linear momentum ${P_i} = {P_f}$.$M\left( {4000km/h} \right) = \left( {4M/5} \right)V + \left( {M/6} \right) \times 3900$
Solve the above equation:
$\Rightarrow M\left( {4000km/h} \right) = M\left[ {\left( {4/5} \right)V + \left( {1/6} \right) \times 3900} \right]$;
Cancel out the common:
$\Rightarrow \left( {4000km/h} \right) = \left[ {\left( {4/5} \right)V + \left( {1/6} \right) \times 3900} \right]$;
$\Rightarrow \left( {4000km/h} \right) = \left( {4/5} \right)V + 650$;
Take the value which is not associated with the velocity on the LHS:
$\Rightarrow \left( {4000km/h} \right) - 650 = \left( {4/5} \right)V$;
$\Rightarrow 3350 \times \dfrac{5}{4} = V$;
The final velocity of the shuttle is:
$\Rightarrow V \simeq 4020km/h$;

Final Answer: If the shuttle ejects the disconnected module at a speed of 100km/h with respect to the state of the shuttle before the ejection, the velocity of the shuttle is 4020km/h.

Note: Here, first we need to calculate the initial linear momentum when the shuttle and the module are attached together and then we have to find the final linear momentum of the shuttle i.e. when the shuttle is detached from the module and equate them together by applying linear momentum together.