Calculating the Velocity of a Catapulted Object

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    Catapult Velocity
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Homework Help Overview

The discussion revolves around calculating the velocity of an object catapulted by a force, specifically focusing on a scenario where a stone is launched from a catapult. The subject area includes concepts from mechanics, particularly energy conservation and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the kinetic energy formula and work done equations to find the velocity but expresses uncertainty in the application. Other participants question the assumptions regarding force, distance, and the mass of the catapult components. There is also a discussion about the relationship between work done and kinetic energy.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying concepts. Some guidance on equating work done to kinetic energy has been suggested, but there is no explicit consensus on the approach or assumptions being made.

Contextual Notes

Participants note constraints such as the assumption of zero mass for the catapult parts and the immovability of the catapult itself. There is also a mention of keeping the discussion at a GCSE standard, indicating a level of complexity expected in the responses.

I-Love-Maths2
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A catapult is pulled back 10cm with a force of 30N. How fast does the stone of mass 0.5kg got at maximum speed assuming no air resistance.

I have absolutely no idea how to do this...
But, I am assuming you use Ek=1/2mv^2, though I can't figure out how to apply it

To rearrange that:
V=sqrt[Ek/(1/2m)]

So, to make it clearer
- 10cm would be the displacement (s)
- 30N would be the force (F)
- 0.5kg would be the mass (m)
- 9.8m/s is the given unit for gravity in GCSE (g)

Other equations that might be useful:
- WD=Fs so WD=300J
- Ep=mgh
- W=mg

So from that we need an equation that uses WD and m

Can anybody help?
 
Last edited:
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How much force will be applied to the stone - and over what distance?
I think we are supposed to assume that the parts of the catapult have zero mass...
and that the catapult is immovable.
 
I-Love-Maths2 said:
WD=Fs so WD=300J
what becomes of WD once the projectile is away ?

By the way, you do have an error in there ...
 
Keep this down to GCSE standard please
Does WD=Ek according to the conservation of energy?
 
I think the exercise allows you to set the kinetic energy equal to the work done.
I-Love-Maths2 said:
Keep this down to GCSE standard please
I'm not from a different planet, but I'm not familiar with this standard. All I know is you don't get Joules if you multiply N with cm :rolleyes:
 
Oh, and a belated: hello Love, :welcome:
 

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