Question

A mass, \(m_{1}=8.00 \mathrm{~kg},\) is at rest on a frictionless horizontal surface and connected to a wall by a spring with \(k=70.0 \mathrm{~N} / \mathrm{m},\) as shown in the figure. A second mass, \(m_{2}=5.00 \mathrm{~kg}\), is moving to the right at \(v_{0}=17.0 \mathrm{~m} / \mathrm{s}\). The two masses collide and stick together. a) What is the maximum compression of the spring? b) How long will it take after the collision to reach this maximum compression?

Short answer

Answer: The maximum compression of the spring is 0.601 m, and the time needed to reach that maximum compression is 0.8535 s.

Step-by-step solution

Conservation of Momentum for the Collision

Since there is no outside force acting on the system, the total momentum before the collision is equal to the total momentum after the collision. Let's write the equation for conservation of linear momentum: $$m_1v_1+m_2v_2=(m_1+m_2)V_{f}$$ where \(v_1\) and \(v_2\) are the initial velocities of masses \(m_1\) and \(m_2\), respectively, and \(V_f\) is the final velocity after the collision.

Calculate the final velocity after collision

Plug in the given values and solve for \(V_f\): $$8.00 \text{kg} \times 0\text{m/s} + 5.00\text{kg} \times 17.0\text{m/s} = (8.00\text{kg} + 5.00\text{kg})V_{f}$$ Solving for \(V_f\) gives us: $$V_f = \frac{5.00\text{kg} \times 17.0\text{m/s}}{8.00\text{kg} + 5.00\text{kg}} = \frac{85.0\text{ kg m/s}}{13.00\text{kg}} = 6.54 \text{ m/s}$$

Conservation of Mechanical Energy

The total mechanical energy of the system after the collision is stored as kinetic energy. As the compressed spring reaches its maximum compression, the mass comes momentarily to rest, and the kinetic energy is completely converted to potential energy stored in the spring. We can write the conservation of mechanical energy as: $$ \frac{1}{2}(m_1+m_2)V_{f}^2 = \frac{1}{2}kx_{\text{max}}^2$$ where \(x_{\text{max}}\) is the maximum compression of the spring.

Calculate the maximum compression of the spring

Now, we can plug in the known values and solve for \(x_{\text{max}}\): $$ \frac{1}{2}(13.00\text{kg})(6.54\text{m/s})^2 = \frac{1}{2}(70.0\text{N/m})x_{\text{max}}^2$$ Solving for \(x_{\text{max}}\) gives us: $$x_{\text{max}} = \sqrt{\frac{13\text{kg}\times(6.54\text{m/s})^2}{70\text{N/m}}} = 0.601\text{m}$$ So, the maximum compression of the spring is 0.601 m.

Use the simple harmonic motion properties to find the time

Now we need to find the time required to reach maximum compression. We can use the properties of simple harmonic motion to find the period \(T\) of oscillation: $$ T = 2\pi\sqrt{\frac{m_{\text{eq}}}{k}}$$ where \(m_{\text{eq}}\) is the equivalent mass (sum of \(m_1\) and \(m_2\)) and \(k\) is the spring constant.

Calculate the period of oscillation

Plug in the known values to find \(T\): $$T = 2\pi\sqrt{\frac{13.00\text{kg}}{70.0\text{N/m}}} = 1.707\text{s}$$

Finding the time to reach maximum compression

Since the time required to reach maximum compression is half of the period (as during the other half, it returns to its original position), we can find the time \(t\) as: $$t = \frac{T}{2} = \frac{1.707\text{s}}{2} = 0.8535\text{s}$$ This gives us the answers to the problem: a) The maximum compression of the spring is 0.601 m. b) The time to reach the maximum compression is 0.8535 s.

Key concepts

Collision and Momentum

Understanding how objects interact during a collision is fundamental in physics, particularly in analyzing systems where two or more objects collide and combine to move together. The principle guiding this behavior is known as the conservation of momentum. The total momentum of a closed system—that is, one without external forces—remains constant.

In our exercise, we see two masses colliding on a frictionless surface. Since external forces like friction are absent, the momentum before and after the collision remains the same. This allows us to calculate the final velocity after the objects stick together. The equation \(m_1v_1 + m_2v_2 = (m_1 + m_2)V_f\) encapsulates this concept, where \(m_1\) and \(m_2\) are the masses of the objects, \(v_1\) and \(v_2\) are their respective velocities before the collision, and \(V_f\) is their combined velocity after the collision.

Remember, during a perfectly inelastic collision like the one posed in our problem, objects stick together, and some kinetic energy is typically converted into other forms of energy.

Conservation of Mechanical Energy

The conservation of mechanical energy is a powerful tool that allows us to equate the energy in different forms when a system is isolated from external non-conservative forces. Mechanical energy consists of two primary components: potential energy and kinetic energy. If no non-conservative forces (like friction) are doing work, the mechanical energy of a system is conserved.

In our collision scenario, we find that the mechanical energy before and after the collision must be equal, since the surface is frictionless. After the masses collide and stick together, they compress the spring. At the point of maximum compression, all the kinetic energy has been temporarily transformed into elastic potential energy within the spring. We describe this transformation with the equation \(\frac{1}{2}(m_1+m_2)V_f^2 = \frac{1}{2}kx_{\text{max}}^2\), where \(k\) is the spring constant and \(x_{\text{max}}\) is the maximum compression of the spring. Unraveling this formula grants us the ability to find the maximum compression of the spring post-collision.

Simple Harmonic Motion

Another pivotal concept from our exercise is simple harmonic motion (SHM), which describes the back-and-forth oscillation of objects like springs and pendulums. SHM is characterized by a restoring force that is proportional to the displacement and acts in the opposite direction.

For a spring-mass system, this is depicted by Hooke's law, where the force is calculated as \(-kx\). When the spring is compressed or stretched, it exerts a force that tries to bring the mass back to its equilibrium position, resulting in an oscillatory motion.

The time it takes for one complete cycle of this motion is called the period (\(T\)), given by the equation \(T = 2\text{π}\sqrt{\frac{m_{\text{eq}}}{k}}\), where \(m_{\text{eq}}\) is the total mass and \(k\) the spring constant. By understanding SHM, we can determine how long it will take the joined masses in our exercise to reach maximum compression after collision.

Spring Compression Calculation

Calculating spring compression involves understanding how much a spring shortens or lengthens from its resting position. This calculation is essential when examining systems where springs store potential energy, such as in our collision exercise. The spring compression can be determined using the energy concepts we have discussed.

Once we have the final speed of the mass-spring system from the conservation of momentum, we use the conservation of mechanical energy to find the extent of the spring's compression. The formula \(\frac{1}{2}kx_{\text{max}}^2 = \frac{1}{2}(m_1+m_2)V_f^2\) equates the kinetic energy at the moment just after collision to the potential energy at maximum compression. By rearranging the equation and solving for \(x_{\text{max}}\), we determine the extent to which the spring is compressed.

It is crucial to recognize that the compression of the spring directly relates to the energy transferred from the moving masses to the spring system in that perfect energy-conserving situation.