Collisions with Springs: Videos & Practice Problems

In a collision involving two crates, conservation of momentum and energy principles are applied to determine the maximum compression of a spring. The initial momentum before the collision is equal to the combined momentum after, leading to a final velocity. The kinetic energy at the moment of collision transforms into potential energy stored in the spring at maximum compression. The equation for maximum compression, $\sqrt{\frac{M}{k}}v$, illustrates this relationship, where M is the total mass and k is the spring constant.

Collisions with Springs

In this scenario, we analyze a collision between two crates followed by their interaction with a spring. The process involves two key principles: conservation of momentum during the collision and conservation of energy as the crates compress the spring. Initially, we identify the states of the system at three critical points: before the collision (point A), immediately after the collision (point B), and at maximum compression of the spring (point C).

To begin, we apply the conservation of momentum to determine the velocity of the combined crates after the collision. The equation for momentum conservation is expressed as:

$$m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2 v_{2b}$$

In this case, since the two crates stick together after the collision, we can denote their combined mass as \(M = m_1 + m_2\). The final velocity \(v_b\) can be calculated using the initial velocities and masses of the crates. For example, if crate 1 has a mass of 10 kg and an initial velocity of 20 m/s, and crate 2 has a mass of 30 kg at rest, the equation simplifies to:

$$10 \times 20 + 30 \times 0 = (10 + 30) v_b$$

Solving this gives us \(v_b = 5 \, \text{m/s}\).

Next, we transition to the energy conservation aspect to find the maximum compression of the spring. The conservation of energy equation is given by:

$$K_b + U_b + W_{nc} = K_c + U_c$$

At point B, the crates have kinetic energy due to their combined motion, while the spring is uncompressed, meaning there is no potential energy stored in the spring. At point C, when the spring is maximally compressed, the kinetic energy is zero, and all energy is stored as elastic potential energy in the spring. The kinetic energy at point B can be expressed as:

$$K_b = \frac{1}{2} M v_b^2$$

And the potential energy at maximum compression is:

$$U_c = \frac{1}{2} k x_c^2$$

Setting these equal gives:

$$\frac{1}{2} M v_b^2 = \frac{1}{2} k x_c^2$$

By canceling the \( \frac{1}{2} \) and rearranging for \(x_c\), we find:

$$x_c = \sqrt{\frac{M}{k}} v_b$$

Substituting the values, where \(M = 40 \, \text{kg}\) (the combined mass of the crates) and \(k = 500 \, \text{N/m}\), we can calculate the maximum compression distance:

$$x_c = \sqrt{\frac{40}{500}} \times 5 = 1.41 \, \text{m}$$

This result indicates the maximum compression of the spring after the collision and subsequent motion of the crates. By applying the principles of conservation of momentum and energy, we effectively analyze the dynamics of the system and determine the desired compression distance.

An 8g piece of sticky clay strikes and embeds itself in a 0.992kg block at rest on a frictionless, horizontal surface. The block is attached to a spring with a spring constant of 5N/m. The impact compresses the spring 75.0 cm. What was the initial speed of the clay?

• A. 210 m/s
• B. 1.68 m/s
• C. 187 m/s
• D. 242.5 m/s

Answer: (A)

Here’s what students ask on this topic:

How do you calculate the maximum compression distance of a spring in a collision problem?

To calculate the maximum compression distance of a spring in a collision problem, you need to use the principles of conservation of momentum and energy. First, determine the final velocity after the collision using the conservation of momentum equation:

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v_b$

Next, use the conservation of energy equation to find the maximum compression distance:

$\frac{m_{\mathrm{total}}{v_b}^2}{2}=\frac{k{x}^2}{2}$

Solve for $x$ to get the maximum compression distance.

What is the role of conservation of momentum in collisions involving springs?

In collisions involving springs, the conservation of momentum principle is used to determine the final velocity of the combined masses after the collision. This principle states that the total momentum before the collision is equal to the total momentum after the collision. The equation is:

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v_b$

Here, $m_1$ and $m_2$ are the masses, and $v_1$ and $v_2$ are their velocities before the collision. $v_b$ is the velocity after the collision. This final velocity is then used in the conservation of energy equation to find the maximum compression of the spring.

How does the conservation of energy apply to collisions with springs?

The conservation of energy principle is crucial in collisions involving springs to determine the maximum compression of the spring. After the collision, the kinetic energy of the combined masses is converted into the potential energy stored in the spring. The equation used is:

$\frac{m_{\mathrm{total}}{v_b}^2}{2}=\frac{k{x}^2}{2}$

Here, $m_{\mathrm{total}}$ is the total mass, $v_b$ is the velocity after the collision, $k$ is the spring constant, and $x$ is the maximum compression distance. By solving this equation, you can find the maximum compression of the spring.

What is the difference between elastic and inelastic collisions in the context of spring compression?

In the context of spring compression, elastic and inelastic collisions differ in how they conserve kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved. The objects bounce off each other, and the spring compression involves only the elastic potential energy.

In an inelastic collision, momentum is conserved, but kinetic energy is not. The objects stick together, and some kinetic energy is converted into other forms of energy, such as heat. The spring compression in this case involves the combined mass of the objects and the remaining kinetic energy converting into the spring's potential energy.

How do you use the conservation of momentum and energy to solve collision problems involving springs?

To solve collision problems involving springs, follow these steps:

1. Use the conservation of momentum to find the final velocity after the collision:

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v_b$

2. Use the conservation of energy to find the maximum compression distance of the spring:

$\frac{m_{\mathrm{total}}{v_b}^2}{2}=\frac{k{x}^2}{2}$

3. Solve for $x$ to get the maximum compression distance. This approach ensures you account for both momentum and energy transformations in the system.