In Figure 9-62, block 2 (mass 1.3 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 160 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 2.1 kg), traveling at speed v1 = 4.0 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

To find the distance by which the spring is compressed, we need to use the principle of conservation of energy. When the two blocks stick together and come to a stop, the initial kinetic energy of block 1 will be converted into potential energy stored in the compressed spring.

First, let's find the initial kinetic energy of block 1. The formula for kinetic energy is:

KE = 1/2 * m * v^2

where KE is the kinetic energy, m is the mass, and v is the velocity.

Given that block 1 has a mass of 2.1 kg and is traveling at a speed of 4.0 m/s, we can calculate its initial kinetic energy:

KE1 = 1/2 * 2.1 kg * (4.0 m/s)^2
= 1/2 * 2.1 kg * 16.0 m^2/s^2
= 1.68 kg·m^2/s^2
= 1.68 J

Since there is no friction, the total mechanical energy of the system is conserved. Before the collision, the mechanical energy is equal to the kinetic energy of block 1, and after the collision, it is equal to the potential energy stored in the compressed spring.

The formula for the potential energy stored in a spring is:

PE = 1/2 * k * x^2

where PE is the potential energy, k is the spring constant, and x is the compression or extension of the spring.

We can rearrange the formula to solve for x:

x = sqrt(2 * PE / k)

Since the initial kinetic energy is equal to the potential energy, we can substitute the values and solve for x:

x = sqrt(2 * 1.68 J / 160 N/m)
= sqrt(0.021 J/m)
≈ 0.145 m

Therefore, the spring is compressed by approximately 0.145 meters.