Question

use your answers to questions 71 through 75 on the information and diagram below. a spring - constant of 2600 newtons per meter is compressed 0.10 meter along a horizontal, frictionless surface. the spring is released, propelling a 3.0 - kilogram block. this block then collides with a stationary 1.0 - kilogram block. the blocks remain joined and move together as shown in the diagram. horizontal, frictionless surface spring compressed 3.0 kg 0.10 m 1.0 kg before collision show all work, including the equation and units.

Explanation:

Step1: Calculate 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 be $m_1 = 1.0\ kg$ and the mass of the second block be $m_2 = 3.0\ kg$. After the collision, they move together with a combined mass $m = m_1 + m_2=4.0\ kg$.
By conservation of energy, the initial elastic - potential energy $U$ is converted into kinetic energy of the combined blocks after the collision. So, $U=\frac{1}{2}mv^{2}$, where $v$ is the velocity of the combined blocks.
We can solve for $v$:
$v=\sqrt{\frac{2U}{m}}=\sqrt{\frac{2\times13}{4}}=\sqrt{6.5}\approx 2.55\ m/s$

Answer:

The velocity of the two - block system after the collision is approximately $2.55\ m/s$.