The launching mechanism of a toy gun consists a spring of unknown constant. If t

Question

The launching mechanism of a toy gun consists a spring of unknown constant. If the spring is compressed a distance of 0.120-m and the gun fired vertically, the gun can launch a 20.0-g projectile from rest to a maximum height of 20.0-m above the starting point of the projectile. Neglecting all reistive forces,

Quesions: (A) Determine the spring constant. (B) Find the speed of the projectile as it moves through the equilibrium position of the spring (where x=0).

This picture shows both the initial situation (left) and the final situation (right). THIS IMPLIES THE CONSERVATIVE FORCE FORMULA.

 PlEaSe: Show your solutions in a step-by-step procedure, and explain it in a simple way. Thank you.

Explanation / Answer

a) We know (from equations of projectile motion) that Ymax=V^2 * sin()^2 / 2g (If you want to know how I get this, simply go back to the simple equations of motion...simple derivation).

Therefore, since we know Ymax=20m, and =90 degrees (since it's shooting straight up), We can solve for V^2:

20=V^2*sin(90)^2/2g -----> 40g=V^2.

Now, using conservation of energy:

(1/2)mv^2+(1/2)kx^2=0. Simply plug in 40g for V^2 and solve for K:

(1/2)(.002)(40g)+(1/2)k(.12)^2=0

so K= 54.4

b) (mgh)i+((1/2)mv^2)i=(mgh)f+((1/2)mv^2)f <------ again, conservation of energy
Simple plug and chug:
(.002*9.8*.12)+((1/2)(.002*0^2))=(.002*9.8*0)+((1/2)(.002*v^2))

.002352=.001v^2
V=2.352 m/s

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