Question

base your answers to questions 71 through 75 on the information and diagram below and on your knowledge of physics. 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 blocks remain joined and move together as shown in the diagram below. calculate the speed, v, of the two blocks after the collision.

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

Before the collision, all the energy is elastic - potential energy of the spring. After the collision, the two - block system has kinetic energy. Let the mass of the first block $m_1 = 3.0\ kg$ and the mass of the second block $m_2 = 1.0\ kg$, so the total mass $M=m_1 + m_2=4.0\ kg$.
By conservation of energy, the initial elastic - potential energy $U$ is converted into kinetic energy $K=\frac{1}{2}Mv^{2}$ of the combined blocks after the collision.
We set $U = K$, so $13=\frac{1}{2}\times4\times v^{2}$

Step3: Solve for the velocity $v$

First, rewrite the equation $13=\frac{1}{2}\times4\times v^{2}$ as $13 = 2v^{2}$.
Then, $v^{2}=\frac{13}{2}=6.5$.
Taking the square root of both sides, $v=\sqrt{6.5}\approx 2.55\ m/s$

Answer:

$2.55\ m/s$