Problem 66 - College Physics

Question

A \(2.0 \mathrm{~kg}\) steel sphere is hanging from a hook by a thin wire \(1 \mathrm{~m}\) long. You wish to strike it horizontally with a \(0.5 \mathrm{~kg}\) hammer hard enough for the sphere to swing up \(45^{\circ}\). Assuming that the hammer makes an elastic collision with the sphere, with what speed must you hit the sphere?

Step-by-Step Solution

Answer

The hammer needs to hit with a calculated speed based on energy and momentum conservation laws.

Step 1: Analyze the Physics Situation

We need to give the sphere enough kinetic energy to swing to a height where it makes a 45° angle with its initial position. This involves converting the kinetic energy given by the hammer into potential energy at the top of the swing.

Step 2: Calculate the Change in Height

At 45°, the change in height (Δh) of the sphere can be found using the geometry of a pendulum:\[Δh = L(1 - \cos\theta)\]where L is the length of the wire (1 m) and \theta is the angle (45°). Calculate Δh:\[Δh = 1 \ imes (1 - \cos(45°)) = 1 \ imes (1 - \frac{\sqrt{2}}{2})\]

Step 3: Establish the Energy Equation

At the top of the swing, all kinetic energy has been converted into gravitational potential energy:\[ KE = PE \Rightarrow \frac{1}{2}mv^2 = mgh\]The mass m of the sphere cancels out from both sides, leaving us with:\[ \frac{1}{2}v^2 = g(1 - \cos(45°))\]

Step 4: Calculate Required Velocity of Sphere

Substitute g = 9.8 \ \mathrm{m/s^2} into the energy equation and calculate:\[ \frac{1}{2}v^2 = 9.8 \ imes (1 - \frac{\sqrt{2}}{2})\]Solve for v\ to find the velocity the sphere needs to reach the height of 45°.

Step 5: Apply Conservation of Momentum for Elastic Collision

The hammer makes an elastic collision with the sphere, so use the conservation of momentum:\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]where m_1 = 0.5 \ \mathrm{kg} (hammer), m_2 = 2.0 \ \mathrm{kg}\ (sphere), u_2 = 0 (initial velocity of the sphere), and u_1 is what we are solving for. Assume v_1 is the velocity found in the previous step and solve for u_1.

Step 6: Solve for the Hammer's Speed

Rearrange the conservation of momentum equation to solve for u_1\:\[ u_1 = \frac{(m_1v_1 + m_2v_2 - m_2u_2)}{m_1}\]Substitute the values v_1\ found previously and solve for u_1.

Key Concepts

Kinematic Energy

Kinematic energy is the energy that a body possesses due to its motion. In this exercise, the hammer imparts kinematic energy to the steel sphere during the collision.
This energy is crucial for the sphere to swing to a height, achieving an angle of 45° with its original vertical position. Understanding the transformation of kinematic energy into potential energy is key. When the hammer strikes the sphere, it converts part of the hammer's kinematic energy into the sphere's motion. The energy then shifts from kinematic to potential as the sphere swings upward.
At the peak of the swing, all the initial kinematic energy of the sphere has been converted into potential energy. That's why the formula used is based on this conservation, linking the two types of energy: \[ \frac{1}{2}mv^2 = mgh \] Here, \( m \) is the mass, \( v \) represents the velocity, \( g \) is the acceleration due to gravity, and \( h \) the height. This relation allows us to calculate the necessary speed to reach a certain height given the kinematic energy.

Conservation of Momentum

The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. It is vital in scenarios like this elastic collision between a hammer and a sphere.
Each object contributes to the total momentum.When the hammer hits the sphere, momentum is transferred. Before the collision, the hammer and sphere have distinct momenta. After the collision, the sum of their momenta remains unchanged.
This interaction is described with the equation:\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]Where \( m_1 \) and \( m_2 \) are the masses of the hammer and sphere, \( u_1 \) and \( u_2 \) are their initial velocities, and \( v_1 \) and \( v_2 \) are their final velocities.
Since the initial velocity of the sphere is zero, the equation helps clarify the needed speed \( u_1 \) of the hammer so that the desired motion is imparted to the sphere.

Potential Energy

Potential energy in this context relates to the height the sphere reaches during its swing. When the sphere moves to a higher position, it gains gravitational potential energy, considered as energy stored due to its elevation.
As the sphere swings upward, the kinetic energy is entirely transformed into potential energy at the top.The potential energy can be calculated with:\[ PE = mgh \]Here, \( PE \) is potential energy, \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is the height reached.
In our problem, knowing how high the sphere swings (using pendulum geometry) allows us to compute this conversion and ensures the sphere gets enough energy to climb to a 45° angle.

Pendulum Geometry

Pendulum geometry helps calculate the height to which the sphere rises. The change in height when a pendulum swings is derived from simple trigonometric relationships.
By understanding this relationship, you can discern how energy is transitioned as the pendulum moves.For a pendulum, the change in height \( \Delta h \) when swinging up by an angle \( \theta \) is:\[ \Delta h = L(1 - \cos\theta) \]Where \( L \) is the length of the pendulum and \( \theta \) is the swing angle.
In this problem, with \( L = 1 \) meter and \( \theta = 45^\circ \), you can apply this formula to find the height change. Comprehending this geometry is essential to link the concepts of potential energy and the initial kinetic energy required to achieve the specified swing.