Question

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?

Answer 1

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 = $\sqrt{2gh}$     ..................2

V = $\sqrt{2*9.8*0.183}$

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 = $\frac{M*V}{m}$

v =  $\frac{0.355*1.89}{0.057}$

v = 11.77 m/s

so dart's speed is 11.77 m/s