Question

a spring with a spring constant of 2600 n/m from its unstretched position. the spring is compressed 0.10 meter along a horizontal, frictionless surface. this 3.0 - kg block is propelled, then collides with a stationary 1.0 - kg block and remains joined together as shown in the diagram. horizontal, frictionless surface. 3.0 kg. 1.0 kg. spring compressed. base your answers to questions below and on your knowledge of physics. show all work, including the equation and substitution with units.

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 $x=0.10\ m$.
$U=\frac{1}{2}\times2600\times(0.10)^{2}$
$U = 13\ J$

Step2: Use conservation of momentum and energy

The total mass of the two - block system is $m = 3.0\ kg+1.0\ kg=4.0\ kg$. Let the velocity of the combined blocks just after the spring is released be $v$.
By conservation of mechanical energy, the initial elastic - potential energy $U$ is converted into kinetic energy $K=\frac{1}{2}mv^{2}$ of the combined blocks.
$\frac{1}{2}mv^{2}=U$
$v=\sqrt{\frac{2U}{m}}$
Substitute $U = 13\ J$ and $m = 4.0\ kg$ into the formula:
$v=\sqrt{\frac{2\times13}{4}}=\sqrt{6.5}\approx 2.55\ m/s$

Answer:

The velocity of the combined blocks just after the spring is released is approximately $2.55\ m/s$.