Question

base your answers to questions 71 through 75 on the information and diagram below. a spring with a spring - constant of 2600 newtons per meter is compressed 0.10 meter from its unstretched position. the spring is released, propelling a 3.0 - kilogram block along a horizontal, frictionless surface. this block then collides with a stationary 1.0 - kilogram block. the two blocks remain joined and move together as shown in the diagram.

Explanation:

Step1: Calculate initial elastic - potential energy

The formula for elastic - potential energy is $U = \frac{1}{2}kx^{2}$, where $k = 2600\ N/m$ and $x = 0.10\ m$.
$U=\frac{1}{2}\times2600\times(0.10)^{2}=13\ J$

Step2: Apply conservation of momentum and energy

Let the mass of the first block $m_1 = 1.0\ kg$ and the mass of the second block $m_2 = 3.0\ kg$. After the collision, they move together with mass $m = m_1 + m_2=4.0\ kg$. According to the conservation of energy, the initial elastic - potential energy is converted into kinetic energy of the combined blocks. The formula for kinetic energy is $K=\frac{1}{2}mv^{2}$.
We set $U = K$, so $\frac{1}{2}kx^{2}=\frac{1}{2}(m_1 + m_2)v^{2}$
$v=\sqrt{\frac{kx^{2}}{m_1 + m_2}}$
Substitute $k = 2600\ N/m$, $x = 0.10\ m$, $m_1 = 1.0\ kg$, and $m_2 = 3.0\ kg$ into the formula.
$v=\sqrt{\frac{2600\times(0.10)^{2}}{1.0 + 3.0}}=\sqrt{\frac{2600\times0.01}{4}}=\sqrt{6.5}\approx 2.55\ m/s$

Answer:

$2.55\ m/s$