A 500.0 kg module is attached to a 400.0 kg shuttle craft, which moves at 1000ms relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0ms relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft increase because of the explosion?

Short Answer

Expert verified

Fraction of kinetic energy of the module and shuttlecraft increase is K.EK.Ei=2.5×10-3.

Step by step solution

01

Understanding the given information

  1. Mass of module,Mis500.0kg.
  2. Mass of shuttle craft,mis500.0kg.
  3. Speed of module and shuttle,Viis1000ms.
  4. Speed of module relative to shuttle, VRis100.0ms.
02

Concept and formula used in the given question

Consider by conservation of momentum and kinetic energy find the velocity of the shuttle. From the kinetic energies of the module and shuttle and find the fraction. The formula used are stated below.

m1vi1+m2vi2=m1vf1+m2vf2K.E=12mv2

03

Calculate the fraction for which the kinetic energy of the module and shuttle craft increase because of the explosion

Determine the speed of the shuttle:

mvi1+mvi2=m1vf1+m2vf2

Initially, both the module and the shuttle had the same speed solve as:

mvi1+mvi1=mVshuttle-100.0+mvshuttle500.0+400.01000=500.0Vshuttle-100.0+400.0Vshuttle900000=900.0vshuttle-50000900.0vshuttle=950000

Solve further as:

Vshuttle=950000900.0Vshuttle=1055.6ms

New speed for shuttle,

Vnew=Vshuttle-100.0=1055.6-100=955.6ms

Finally, the module and shuttle have different speeds.

Now the change in kinetic energy of the module and shuttle is determined as:

K.Ef-K.Ei=12500.0955.62+124001055.62-1290010002K.Ef-K.Ei=1151112

So, the fraction is,

K.Ef-K.EiK.Ei=11511124.5×108=2.5×10-3

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