Calculating the Speed of a Ball Shot from a Spring Gun

[Jtappan]

Homework Statement

A spring gun (k = 22 N/m) is used to shoot a 56 g ball horizontally. Initially the spring is compressed by 19 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground, 1.5 m below the spring gun?
____ m/s

Homework Equations

kinetic energy, potential energy and spring constant equation

The Attempt at a Solution

I have attempted to do this problem with the kinetic energy, potential energy and spring constant equation. But I must be messing it up some how with the two sping compression lengths or the 1.5 m above ground. Can anyone shed some light on how to do this one?

 

[Astronuc]

Well there are two parts to this problem - horizontal velocity (speed) and vertical velocity (speed). To get the 'speed' of the ball 1.5 m below, one needs to find the resultant from both velocities.

The spring provides the energy to achieve a horizontal motion (velocity and KE).

The springs stored mechanical energy is transformed as the spring extends from 19 cm to 12 cm, so find the energy from that change. The spring energy becomes the kinetic energy of the ball.

When the ball leaves the gun, it begins a free fall under gravity, so use equation of motion for free fall under constant acceleration of gravity.

Here is a good reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

For elastic or spring potential energy - http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

 

[ogb p]

I would approach this problem by first identifying the relevant equations and variables. The given information includes the spring constant (k = 22 N/m), the mass of the ball (m = 56 g = 0.056 kg), and the initial and final compression lengths of the spring (x_i = 19 cm and x_f = 12 cm). The unknown variable is the speed of the ball (v).

To solve for the speed, we can use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the initial potential energy stored in the compressed spring will be converted into kinetic energy as the ball is launched.

We can set up the following equation:

Potential energy at initial compression = Kinetic energy at final compression + Potential energy at final compression + Gravitational potential energy at final position

Since the spring gun is horizontal, there is no change in gravitational potential energy. Therefore, we can ignore the last term in the equation.

Now, we can plug in the values for potential and kinetic energy:

(1/2)kx_i^2 = (1/2)mv^2 + (1/2)kx_f^2

Solving for v, we get:

v = √[(k/m)(x_i^2 - x_f^2)]

Plugging in the given values, we get:

v = √[(22 N/m)/(0.056 kg)][(0.19 m)^2 - (0.12 m)^2] = 3.30 m/s

Therefore, the speed of the ball when it hits the ground is 3.30 m/s.