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. horizontal frictionless surface spring compressed

Explanation:

Step1: Calculate initial elastic - potential energy of the spring

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

Before the collision, the momentum of the $3.0 - kg$ block is $p_{1}=m_{1}v_{1}$, and we first find $v_{1}$ from the conservation of mechanical energy. The initial elastic - potential energy is converted to kinetic energy of the $3.0 - kg$ block just before the collision. So, $\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}kx^{2}$. Solving for $v_{1}$, we get $v_{1}=\sqrt{\frac{kx^{2}}{m_{1}}}=\sqrt{\frac{2600\times(0.10)^{2}}{3.0}}\approx2.94\ m/s$.
The momentum of the $3.0 - kg$ block before the collision is $p_{1}=m_{1}v_{1}=3.0\times2.94 = 8.82\ kg\cdot m/s$. After the collision, the two blocks ($m_{1}=3.0\ kg$ and $m_{2}=1.0\ kg$) stick together, and by conservation of momentum $p_{1}=(m_{1} + m_{2})v_{2}$, where $v_{2}$ is the velocity of the combined - mass system. So, $v_{2}=\frac{m_{1}v_{1}}{m_{1}+m_{2}}=\frac{8.82}{3.0 + 1.0}=2.205\ m/s$.

Step3: Calculate the kinetic energy of the combined - mass system

The kinetic energy of the combined - mass system ($m = m_{1}+m_{2}=4.0\ kg$) is $K=\frac{1}{2}(m_{1}+m_{2})v_{2}^{2}=\frac{1}{2}\times4.0\times(2.205)^{2}\approx9.72\ J$

Step4: Calculate the energy lost in the collision

The energy lost $\Delta E$ is the initial elastic - potential energy minus the kinetic energy of the combined - mass system. $\Delta E=13 - 9.72 = 3.28\ J$

Answer:

The velocity of the combined blocks after the collision is approximately $2.21\ m/s$ and the energy lost in the collision is approximately $3.28\ J$.