Question

One end of a massless, 30-cm-long spring with a spring constant of 15 N/m is attached to a 250 g stationary air-track glider; the other end is attached to the track. A 600 g glider hits and sticks to the 250 g glider, compressing the spring to a minimum length of 22 cm . What was the speed of the 600g glider just before impact?

Answer 1

The potential energy stored in the block-spring support system when the block is just released is 0.72 J. The speed of the block when it crosses the point where the spring is neither compressed nor stretched is 1.632 m/s. The speed of the block when it has traveled a distance of 20 cm from the release point is also 1.632 m/s.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of energy. Initially, the system has only potential energy in the form of the compressed spring. The potential energy stored in the spring can be calculated using the formula U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the compression or extension of the spring. In this case, the compression of the spring is 12 cm or 0.12 m. Substituting these values into the formula, we get U = (1/2)(100 N/m)(0.12 m)^2 = 0.72 J.

When the block is released, the potential energy is converted into kinetic energy. At the point when the spring is neither compressed nor stretched, all the potential energy would have been converted into kinetic energy. We can calculate the speed of the block using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the block, and v is the speed of the block. Substituting the values, we get 0.72 J = (1/2)(0.3 kg)v^2, solving for v we get v = sqrt((2)(0.72 J)/(0.3 kg)) = 1.632 m/s.

If the block travels a distance of 20 cm or 0.2 m from the release point, its speed can be calculated using the formula KE = (1/2)mv^2 as well. Substituting the values, we get (1/2)(0.3 kg)(v^2) = 0.72 J, solving for v we get v = sqrt((2)(0.72 J)/(0.3 kg)) = 1.632 m/s.