Problem 89 - University Physics with Modern Physics

Question

An \(8.00-\mathrm{kg}\) ball, hanging from the ceiling by a light wire 135 \(\mathrm{cm}\) long, is struck in an elastic collision by a 2.00 -kg ball moving horizontally at 5.00 \(\mathrm{m} / \mathrm{s}\) just before the collision. Find the tension in the wire just after the collision.

Step-by-Step Solution

Answer

The tension in the wire just after the collision is 102.18 N.

Step 1: Analyze the Collision

In an elastic collision, both momentum and kinetic energy are conserved. We can set up the conservation of momentum for the horizontal direction. Let \(v_1'\) be the velocity of the 2.00 kg ball after the collision, and \(v_2'\) be the velocity of the 8.00 kg ball after the collision. The equation for the conservation of momentum is: \[m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\] Given \(m_1 = 2.00 \text{ kg}\), \(v_1 = 5.00 \text{ m/s}\), \(m_2 = 8.00 \text{ kg}\), and the initial velocity of the 8.00 kg ball \(v_2 = 0\), we have:\[2.00 \times 5.00 + 8.00 \times 0 = 2.00v_1' + 8.00v_2'\]\[10.00 = 2.00v_1' + 8.00v_2'\]

Step 2: Apply Energy Conservation

Since the collision is elastic, the kinetic energy is also conserved. The equation for the conservation of kinetic energy is:\[\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2\]Substitute the known values, \(v_2 = 0\):\[\frac{1}{2} \times 2.00 \times 5.00^2 = \frac{1}{2} \times 2.00 \times v_1'^2 + \frac{1}{2} \times 8.00 \times v_2'^2\]\[25.00 = v_1'^2 + 4.00v_2'^2\]

Step 3: Solve the Equations

We have two equations:1. \(10.00 = 2.00v_1' + 8.00v_2'\)2. \(25.00 = v_1'^2 + 4.00v_2'^2\) Solve these equations simultaneously. From the first equation:\[v_1' = 5.00 - 4.00v_2'\]Substitute into the second equation:\[25.00 = (5.00 - 4.00v_2')^2 + 4.00v_2'^2\]Simplifying the above:\[25.00 = 25.00 - 40.00v_2' + 16.00v_2'^2 + 4.00v_2'^2\]\[0 = -40.00v_2' + 20.00v_2'^2\]Divide by 20, \[0 = -2.00v_2' + v_2'^2\]\[v_2'^2 = 2.00v_2'\]\[v_2'(v_2' - 2.00) = 0\]The solution is \(v_2' = 2.00 \text{ m/s}, v_1' = -3.00 \text{ m/s}\).

Step 4: Calculate Vertical Force

Just after the collision, the 8.00 kg ball has a horizontal velocity \(v_2' = 2.00 \text{ m/s}\). The horizontal motion will cause a deviation and tension in the wire. Apply the centripetal force concept:The tension \(T\) in the wire provides the centripetal force required to keep the ball moving in a circular path with radius \(L = 1.35 \text{ m}\):\[T - mg = m\frac{v_2'^2}{L}\]Solving for the tension \(T\):\[T = m\frac{v_2'^2}{L} + mg\]

Step 5: Calculate Tension

Substitute the known values into the expression for tension:\[m = 8.00 \text{ kg}, v_2' = 2.00 \text{ m/s}, L = 1.35 \text{ m}, g = 9.81 \text{ m/s}^2\] \[T = 8.00 \frac{(2.00)^2}{1.35} + 8.00 \times 9.81\]\[T = 23.70 \text{ N} + 78.48 \text{ N}\]\[T = 102.18 \text{ N}\]

Key Concepts

Momentum Conservation

In any elastic collision, one of the foundational principles is the conservation of momentum. Imagine momentum as the measure of motion, which needs to be maintained or conserved during a collision, transferring from one object to another.

In this context, before the collision, the total momentum of the system is solely due to the moving 2.00 kg ball. The 8.00 kg ball, initially at rest, receives some of this momentum after being struck. Even though the process is quick, the law of conservation of momentum ensures that the momentum before and after the collision remains the same, provided no external forces are at play.

• Momentum before = Momentum after
• Equation: \(m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\)

This equation helps us to solve for unknown velocities, assisting us to determine how fast each ball is moving after the collision.

Kinetic Energy Conservation

Kinetic energy is a measure of the energy that an object has due to its motion. In an elastic collision, not only is momentum conserved, but so is kinetic energy. This means that the sum of the kinetic energy of both objects remains the same before and after the collision.

The equation for the conservation of kinetic energy can be expressed as:

• \( \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2 \)
• This formula shows that the kinetic energy (half of mass times the square of velocity) must be equal before and after the impact.

These principles guide the calculation of how energy is shared between the two balls post-impact, serving as a cross-check by ensuring energy considerations align with the transfer of velocity between the balls.

Centripetal Force

Centripetal force is the invisible hand pushing an object towards the center of a circular path. Post-collision, when the 8.00 kg ball swings, it grips to a circular arc due to this force, courtesy of the tension in the wire.

This force

• acts perpendicular to the ball's velocity,
• is essential for maintaining circular motion,
• and is provided by the tension in the string.

The equation linking the centripetal force \(F_c\) to the tension \(T\) is given by \(T - mg = m \frac{v^2}{L} \), ensuring the collision's momentum turns the horizontal force into a circular challenge.
Tension supports the ball's weight and the extra force needed to curve its path.

Tension in a Wire

Tension can be understood as the pulling force transmitted through the wire when it is swung post-collision. It's the combination of forces maintaining the equilibrium of the hanging ball.

There are two forces that play a role in determining the tension in the wire:

• The ball's weight (\(mg\)) acts downwards due to gravity.
• The centripetal force, needed for circular motion, acts horizontally.

These forces add up to give us the total tension exerted by the wire. Mathematically, the tension can be computed by:

• \( T = m \frac{v^2}{L} + mg \)
• This accounts for the gravitational pull and the force required to keep the ball on its new path.
• The tension is crucial to ensure that the ball remains on its circular trajectory and doesn’t fall down."

By understanding these forces, one can calculate the exact tension, facilitating better design and safety in practical applications.