Question

A spring-loaded toy gun is used to shoot a ball of mass 1.5 kg straight up in the air, as shown. The spring has spring constant 777 N/m. If the spring is compressed a distance of 20.0 centimeters from its equilibrium position and thern released, the ball reaches a maximum height (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis. (a) How much energy is stored in the spring when it is compressed 20.0 cm? (b) Find the maximum height of the ball. (C) Find the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position)

Answer 1

Answer:

Part a)

U = 15.5 J

Part b)

h = 1.05 m

Part c)

v = 4.55 m/s

Explanation:

Part a)

Initial potential energy stored in the spring is given by

$U = \frac{1}{2}kx^2$

now we have

$k = 777 N/m$

x = 20.0 cm

now from above formula we will have

$U = \frac{1}{2}(777)(0.20)^2$

$U = 15.5 J$

Part b)

By mechanical energy conservation law we can say that initial spring potential energy stored = final gravitational potential energy at the top position

$\frac{1}{2}kx^2 = mgh$

$15.5 = 1.5(9.8)h$

$h = 1.05 m$

So maximum height from gun is 1.05 m

Part c)

For muzzle velocity of the ball we can use the energy conservation again

According to which the final kinetic energy of the ball = initial spring energy stored in it

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$(\frac{1}{2})(1.5 v^2) = 15.5$

$v = v = 4.55 m/s$