Question

A block of mass \(5.0 \mathrm{~kg}\) slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a mass of \(4.0 \mathrm{~kg}\) attached to a horizontal spring (with spring constant of \(k=2000.0 \mathrm{~N} / \mathrm{m}\) ), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest? a) \(0.40 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\)

Short answer

Answer: The maximum compression in the spring is 0.67 m.

Step-by-step solution

Law of Conservation of Momentum

Initially, only the 5kg block is moving. Calculate the total momentum before the collision. Let's denote the initial velocity of the 5kg block as \(v_1 = 8 \mathrm{~m/s}\) and the initial velocity of the 4kg block as \(v_2 = 0 \mathrm{~m/s}\) (since it is at rest). Total initial momentum = \(m_1v_1 + m_2v_2 = (5 \mathrm{~kg})(8 \mathrm{~m/s}) + (4 \mathrm{~kg})(0 \mathrm{~m/s}) = 40 \mathrm{~kg \cdot m/s}\).

Post-Collision Velocity

As the blocks stick together after the collision, they move together. Find the post-collision velocity. Let \(v\) be the post-collision velocity of both the blocks. Using conservation of momentum, we have total initial momentum = total final momentum. \((5 \mathrm{~kg} + 4 \mathrm{~kg})v = 40 \mathrm{~kg \cdot m/s}\). Solving for \(v\), we get \(v = \frac{40 \mathrm{~kg \cdot m/s}}{9 \mathrm{~kg}} = \frac{40}{9} \mathrm{~m/s}\).

Conservation of Energy

Now, we apply the conservation of energy. The initial kinetic energy of the system is converted into the potential energy stored in the spring. Initially, the spring is assumed to be in the uncompressed position. The initial kinetic energy can be calculated using the formula \(KE = \frac{1}{2}(m_1 + m_2)v^2\). \(KE = \frac{1}{2}(9 \mathrm{~kg})(\frac{40}{9} \mathrm{~m/s})^2 = \frac{1}{2}(9 \mathrm{~kg})(\frac{1600}{81} \mathrm{~m^2/s^2}) = \frac{7200}{81} \mathrm{~J}\).

Find maximum spring compression

Now, we set this kinetic energy equal to the potential energy stored in the spring at maximum compression. If \(x\) is the spring compression, the potential energy stored in the spring can be calculated using the formula \(PE = \frac{1}{2}kx^2\). Equating kinetic energy to potential energy, we get: \(\frac{7200}{81} \mathrm{~J} = \frac{1}{2}(2000 \mathrm{~N/m})x^2\). Solving for \(x\), we get \(x^2 = \frac{7200}{81 \cdot 1000} \mathrm{~m^2} \implies x = \sqrt{\frac{72}{81}} \mathrm{~m} = \frac{6}{9} \mathrm{~m} = 0.67 \mathrm{~m}\). Thus, the maximum spring compression is \(0.67 \mathrm{~m}\), which corresponds to option (e).

Key concepts

Conservation of Momentum

The principle of conservation of momentum is a fundamental concept in physics, stating that the total momentum of a closed system of objects is constant, provided no external forces are acting on the system. This principle is rooted in Newton's Third Law, which states that for every action, there is an equal and opposite reaction.

When applying this to collision problems, such as a sliding block hitting another block connected to a spring, we can say that the total momentum before the collision is equal to the total momentum after the collision. In our example, the 5 kg block sliding at 8 m/s has an initial momentum (product of mass and velocity) which is conserved even after it sticks to the 4 kg mass. This enables us to mathematically solve for the velocity of the two masses after the collision by setting the initial and final momentum equal to each other and solving for the unknown velocity.

Conservation of Energy

The Conservation of Energy is another pivotal concept in physics, asserting that the total energy in an isolated system remains constant over time. It implies that energy cannot be created or destroyed but can only be transformed from one form to another.

When the two blocks in our example collide and stick together, their collective kinetic energy (energy of motion) eventually converts into potential energy within the spring as it compresses. Initially, the 5 kg block possesses kinetic energy due to its motion. After impact, this energy transforms into the potential energy of the compressed spring - a perfect illustration of kinetic to potential energy conversion. We use this principle to equate the initial kinetic energy of the blocks to the potential energy of the spring at maximum compression, allowing us to solve for the extent of spring compression.

Kinetic to Potential Energy Conversion

The conversion from Kinetic to Potential Energy occurs when an object in motion begins to slow down and store energy in a different form. In the context of our spring compression problem, the moving blocks convert their kinetic energy entirely into the potential energy of the spring as they come to rest.

When in motion, an object’s kinetic energy can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. This energy is then transferred into the spring as it compresses, storing energy in the form of elastic potential energy, calculated as \( PE = \frac{1}{2}kx^2 \) where \( k \) is the spring constant and \( x \) is the displacement of the spring from its equilibrium position. By equating kinetic energy to potential energy, we find the maximum displacement of the spring which corresponds to the maximum potential energy.