Physics Spring Constant

Question

How would I determine the spring constant of a marble launcher? It shoots a 0.0051 kg marble a height of 2.3 above its muzzle, assuming it is angled at 62 degrees above the horizontal? The marble compressed the spring 0.080 m before being released.

Answer 1

Here is a sketch of the situation.

In order to reach a vertical height of 2.3 m, we can calculate the initial velocity in the y-direction from the kinematic equation:

v_f-y² = v_i-y² + 2ay where v_f-y = 0 (vertical velocity at the top of the marbles flight), a = g = -9.8 m/s², and y is the vertical height 2.3 (I'm assuming you meant to say "meters").

0 = v_i-y² + 2(-9.8)(2.3)

v_i-y = √45.08 = 6.714 m/s

Using trig, we can say that sin(62°) = opp/hyp = v_i-y/v_i so vi = v_i-y/sin(62°) = 6.714/0.8829 = 7.604 m/s

(Note: we could have also used energy to solve for v_i)

The kinetic energy that the marble had at the moment it left the muzzle is 1/2mv² = (1/2)(0.0051)(7.604²) = 0.1475 joules

The potential energy the marble had at the moment it left the muzzle is mgh = (0.0051)(9.81)(0.080sin(62°) = 0.0035 joules

The total energy the marble had at the moment it left the muzzle was KE + PE = 0.1475 joules + 0.0035 joules = 0.1510 joules

We will assume friction is small so that no work was lost in shooting the marble meaning the potential energy prior to launch was also 0.1510 joules. The potential energy of a spring is given from U_S = 1/2kx² so:

0.1510 = (1/2)k(0.08²) or k = 47 N/m

Note: It's possible your teacher did not intend you to add in the potential energy at the end of the muzzle since the muzzle length was not included in your givens. I chose to use it (I assumed it was just the length of the spring compression) because it did make a difference. Without it, the k = 46 N/m