Building a Spring Launcher: Attaching the Spring

AI Thread Summary
To build an effective spring launcher, it is crucial to understand that not all energy can be transferred to the ball, as efficiency limits this transfer. The method of attaching the spring to the wood is essential for maximizing energy transfer. Using rectangular wood is a practical choice, but the specific design and attachment method will significantly impact performance. Proper alignment and secure attachment of the spring will help ensure that most of the energy is effectively converted into kinetic energy for the ball. Overall, careful consideration of materials and design is key to achieving a successful spring launcher.
thegame
Messages
32
Reaction score
0
i need to make a spring launcher that launches a small ball using a spring (one that stretches)... so how do i attach the spring to the piece of wood so when i stretch it all of the energy is converted to the ball..
 
Physics news on Phys.org
Well for a start off your request is an impossibility - you can't transfer ALL of the energy to the ball, only a proportion dependant on the efficiency of the system.

Secondly you say "how do I attach the spring to the piece of wood " Well this does rather beg the question... What piece of wood?
 
well, i mean most of the energy should be transferred..

i going to use normal retangular wood..
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top