Please help Projectile breaks into fragments question

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Homework Statement


A projectile of mass 22kg is fired at an angle of 55 degrees to horizontal with a speed of 350ms-1. At the highest point of trajectory the projectile explodes into two equal fragments, one lands directly beneath the explosion and the other flies off.

I have done the parts of the question, worked out the first piece lands 5867m and the second piece lands 17600m from the cannon. I have also worked out the speeds of each fragment at the moment of the explosion

The last part says what is the energy released in the explosion?

I can't formulate it in my head, I can only think of KE[itex]_{initial}[/itex] + Potential energy [itex]_{explosion}[/itex]= GPE[itex]_{max height}[/itex] + KE[itex]_{left over}[/itex] (because its still moving in x direction at max height)+ KE[itex]_{explosion}[/itex]


I'm not sure of how I am going to find this energy.

Thanks
 
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Do you need to worry about the PE? How about looking only at the KE just before and just after after the explosion...

KEafter = KEbefore + KEexplosion

Edit: Humm. I'm having second thoughts about that. Do you know the mass of each bit after the explosion?
 
CWatters said:
Do you need to worry about the PE? How about looking only at the KE just before and just after after the explosion...

KEafter = KEbefore + KEexplosion

Edit: Humm. I'm having second thoughts about that. Do you know the mass of each bit after the explosion?

no no wait it gives the right answer if I do KE[itex]_{before}[/itex] = KE[itex]_{AFTER}[/itex] + KE [itex]_{explosion}[/itex]

but how did you think of this? why can I ignore potential energy?

also the mass of each piece is 10kg as it breaks into equal fragments.
 
DunWorry said:
no no wait it gives the right answer if I do KE[itex]_{before}[/itex] = KE[itex]_{AFTER}[/itex] + KE [itex]_{explosion}[/itex]

but how did you think of this? why can I ignore potential energy?

also the mass of each piece is 10kg as it breaks into equal fragments.

The change in gravitational potential energy is the same, but the change in kinetic energy gives us the work done by the projectile, which is what we are interested in.
 
DunWorry said:
no no wait it gives the right answer if I do KE[itex]_{before}[/itex] = KE[itex]_{AFTER}[/itex] + KE [itex]_{explosion}[/itex]

but how did you think of this? why can I ignore potential energy?

also the mass of each piece is 10kg as it breaks into equal fragments.

Immediately after the explosion, what is the potential energy?
 
DunWorry said:
but how did you think of this? why can I ignore potential energy?

What sammy said. Just after the explosion the two bits of shell have different velocity but they start from the same height as the shell.
 
Last edited:

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