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A dart gun suspended by strings hangs in equilibrium. The mass of the gun is 355 grams, not including a dart. The gun fires a 57.0 gram dart, causing it to swing backwards. The gun swings up to a height of 18.3 centimeters. What was the dart's speed in meters per second just after firing?

Respuesta :

Answer:

dart's speed is 11.77 m/s

Explanation:

given data

mass of the gun m = 355 gram = 0.355 kg

height = 18.3 centimeters = 0.183 m

dart = 57.0 gram = 0.057 kg

to find out

dart's speed

solution

we apply here law of conservation of energy that is express as

mgh = 0.5 × m × v²      ...........1

so speed of gun will be here as

V = [tex]\sqrt{2gh}[/tex]     ..................2

V = [tex]\sqrt{2*9.8*0.183}[/tex]

V =  1.89 m/s

and

now we find speed of dart by use law of conservation of momentum that is

M×V  = m×v         ...............3

so speed of the dart is

v = [tex]\frac{M*V}{m}[/tex]

v =  [tex]\frac{0.355*1.89}{0.057}[/tex]

v = 11.77 m/s

so dart's speed is 11.77 m/s

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