Question

A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high will the ball go if the gun is aimed vertically? The acceleration of gravity is 9.81 m/s 2 . Assume that at the end of the barrel, the spring’s displacement and the ball’s height are each zero

Answer 1

The answer to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Step-by-step explanation:

The energy in the spring → Energy, E =
`(1)/(2)kx^2`

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E =
`(1)/(2) 579*0.01^2 = 0.02895 J`

The spring energy is transferred to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE =
`(1)/(2)mv^2`

From which v =
`\sqrt{(2*KE)/(m) }` =
`\sqrt{(2*0.02895)/(0.021) }` = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

u = initial velocity = 1.66 m/s

a = g = Acceleration due to gravity

S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m