A toy gun has a barrel 7.25 cm long and a spring with a spring constant of 82.3

Question

A toy gun has a barrel 7.25 cm long and a spring with a spring constant of 82.3 N/m. When ready to fire, the spring is compressed the length of the barrel, and when the 25.5 gram dart leaves the gun, the spring is at its natural length. a. If the gun is fired perfectly horizontally, what is the speed of the dart as it leaves the gun? b. If we fired the gun perfectly vertically, what additional type of energy would we have to worry about? c. What would be the speed of the dart as it leaves the gun if it is fired vertically? A good diagram really helps here! d. How high above the gun does the dart get if fired vertically?

Explanation / Answer

A) if horizontally fired then

using law of conservation of energy

energy at the compreesed position = energy in the relaxed position

0.5*k*x^2 = 0.5*m*v^2

v = Sqrt(k/m)*x = sqrt(82.3/0.0255)*0.0725 = 4.12 m/s

B) additional energy is the gravitational potential energy

C) cosidering compressed position as the reference point

0.5*k*x^2 = m*g*h + 0.5*m*v^2

0.5*m*v^2 = (0.5*82.3*0.0725*0.0725)-(0.0255*9.81*0.0725) = 0.198

v = 3.94 m/s

d) height reached is Hmax = v^2/(2*g) = 3.94*3.94/(2*9.81) = 0.792 m

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