Fun with a Spring Gun

Part A =

- Mechanical energy is conserved because no dissipative forces perform work on the ball.
- The forces of gravity and the spring have potential energies associated with them.

Part B = 4.78m/s

Part C = 1.17m

Part D =

- increasing the spring constant k
- increasing the distance the spring is compressed
- decreasing the mass of the ball

Solutions Below:

_{max}(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.

Part A

Check all that apply:

Mechanical energy is conserved because no dissipative forces perform work on the ball.

The forces of gravity and the spring have potential energies associated with them.

No conservative forces act in this problem after the ball is released from the spring gun.

You should have checked #1: Mechanical energy is conserved because no dissipative forces perform work on the ball. and #2: The forces of gravity and the spring have potential energies associated with them.

Part B

_{m}the muzzle velocity of the ball (i.e., the velocity of the ball at the spring’s equilibrium position y=0).

When the spring is released, all of the spring potential energy will be converted to kinetic energy. So we can find the velocity of the ball by first figuring out the ball’s kinetic energy at the point of release, and then solving a kinematic equation to find the velocity at a given displacement.

Find the spring potential energy first. The spring is compressed 25.0cm which is 0.250m:

U = 1/2kx^{2}

U = 1/2(667)(0.250)^{2}

U = 20.84375J

Now find the ball’s velocity when it’s released – just solve for kinetic energy:

KE = 1/2mv^{2}

20.84375 = 1/2(1.50)v^{2}

20.84375 = 0.75v^{2}

27.79167 = v^{2}

v = 5.27m/s

Next, find the velocity at y = 0. Since the spring was compressed 0.250m, that means the initial velocity is at y = -0.250 and the ball will travel 0.250m upwards to reach y = 0. We can use the kinematic equation:

v_{f}^{2} = v_{0}^{2} + 2a(d)

a will be acceleration due to gravity, so:

v_{f}^{2} = 5.27^{2} + 2(-9.8)(0.250)

v_{f}^{2} = 27.77 – 19.6(0.250)

v_{f}^{2} = 22.87

v_{f} = 4.78m/s

Part C

_{max}of the ball.

We can solve this using the same kinematic equation:

v_{f}^{2} = v_{0}^{2} + 2a(d). At the peak (maximum height), we know that the velocity will be zero. So:

v_{f}^{2} = v_{0}^{2} + 2a(d)

0^{2} = 5.27^{2} – 19.6d

0 = 27.77 – 19.6d

19.6d = 27.77

d = 1.42m

And subtract 0.250m since the ball started from y = -0.250:

1.42 – 0.250 = 1.17

= 1.17m

Note that the initial velocity above was from the moment the ball was released (y = -0.250). We could have also solved using the velocity at y = 0 and then wouldn’t have had to subtract the 0.250m.

Part D

Check all that apply:

reducing the spring constant k

increasing the spring constant k

decreasing the distance the spring is compressed

increasing the distance the spring is compressed

decreasing the mass of the ball

increasing the mass of the ball

tilting the spring gun so that it is at an angle \theta; < 90 degrees from the horizontal

You should have checked #2: increasing the spring constant k, #4: increasing the distance the spring is compressed, and #5: decreasing the mass of the ball.