Calculating the Velocity of a Catapulted Object

  • Thread starter Thread starter I-Love-Maths2
  • Start date Start date
  • Tags Tags
    Catapult Velocity
AI Thread Summary
To calculate the maximum speed of a stone launched by a catapult, the work done (WD) can be equated to kinetic energy (Ek) using the formula Ek=1/2mv^2. The work done, calculated as WD=Fs, equals 300J when a force of 30N is applied over a displacement of 10cm. The kinetic energy is then set equal to the work done, leading to the equation 300J = 1/2(0.5kg)v^2. Rearranging this gives v=sqrt(1200), resulting in a maximum speed of approximately 34.64 m/s. The discussion emphasizes the importance of understanding energy conservation principles in this context.
I-Love-Maths2
Messages
18
Reaction score
10
Member advised to use the formatting template for all homework help requests
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:
Physics news on Phys.org
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:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top