/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 A completely inclastic collision... [FREE SOLUTION] | 91Ó°ÊÓ

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A completely inclastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass \(3.0 \mathrm{~kg},\) is moving upward at \(20 \mathrm{~m} / \mathrm{s}\) and the other ball, of mass \(2.0 \mathrm{~kg}\), is moving downward at \(12 \mathrm{~m} / \mathrm{s}\). How high do the combined two balls of putty risc above the collision point? (Ncglect air drag.)

Short Answer

Expert verified
The combined balls rise approximately 2.645 meters above the collision point.

Step by step solution

01

Determine Momentum Before Collision

Calculate the total momentum of the system before the collision using the formula for momentum, which is \( p = mv \) where \( m \) is mass and \( v \) is velocity.The momentum of the 3 kg ball moving upwards: \[ p_1 = 3.0 \, \text{kg} \times 20 \, \text{m/s} = 60 \, \text{kg} \cdot \text{m/s} \]The momentum of the 2 kg ball moving downwards:\[ p_2 = 2.0 \, \text{kg} \times (-12) \, \text{m/s} = -24 \, \text{kg} \cdot \text{m/s} \]The total momentum of the system:\[ p_{total} = p_1 + p_2 = 60 - 24 = 36 \, \text{kg} \cdot \text{m/s} \]
02

Calculate the Velocity After Collision

Since the collision is completely inelastic, both balls move together with a single velocity \( v' \) after the collision. Use the conservation of momentum:\[ m_1v_1 + m_2v_2 = (m_1 + m_2)v' \]Substitute the known values:\[ 36 = (3 + 2)v' \]Solve for \( v' \):\[ v' = \frac{36}{5} = 7.2 \text{ m/s} \]
03

Calculate Maximum Height Achieved

Use the conservation of energy to find how high the mass rises. The kinetic energy just after the collision is completely converted into gravitational potential energy at maximum height:\[ \frac{1}{2} (m_1 + m_2) v'^2 = (m_1 + m_2) g h \]Cancel \((m_1 + m_2)\) from both sides:\[ \frac{1}{2} v'^2 = g h \]Solving for height \( h \):\[ h = \frac{v'^2}{2g} = \frac{(7.2)^2}{2 \times 9.8} \approx \frac{51.84}{19.6} \approx 2.645 \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Conservation
In the realm of physics, momentum is a foundational concept when analyzing collisions, like the one involving our two balls of putty. Momentum, which is the product of an object's mass and velocity (\( p = mv \) ), is conserved in isolated systems, meaning the total momentum before and after a collision remains constant.
This principle applies perfectly to inelastic collisions, where colliding objects stick together post-impact. Although kinetic energy might not be conserved here, momentum still is.
In the provided problem, we calculated the total momentum of each ball before collision and found it to be:\[ p_{total} = 60 - 24 = 36 \, \text{kg} \cdot \text{m/s} \]
  • The upward-moving ball had a momentum of 60 \( \text{kg} \cdot \text{m/s} \).
  • The downward-moving ball had a momentum of -24 \( \text{kg} \cdot \text{m/s} \), indicating the opposite direction.
The sum of these momenta gives us the system's total momentum. After the collision, this total momentum allows us to calculate the shared velocity of the combined mass, ensuring momentum's conservation through the entire event.
Kinetic Energy
Kinetic energy represents the energy an object possesses due to its motion, given by the formula:\( KE = \frac{1}{2}mv^2 \) . It's a scalar quantity and depends on both mass and the square of velocity.
Unlike momentum, kinetic energy isn't necessarily conserved in an inelastic collision. When the balls collide and stick together, mechanical energy is lost to internal forces, like deformation or heat.
In our scenario:
  • The initial kinetic energy is concentrated in the movement of the two separate balls.
  • Post-collision, the combined mass's kinetic energy is less than the sum of both initial energies.
Despite this loss, the key outcome of kinetic energy in this setup is how it transforms. Right after the impact, the kinetic energy of the now singular body paves the way for evaluating how high this body will rise, by converting into gravitational potential energy.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy stored in an object due to its height above the Earth's surface. It's calculated with the formula:\( GPE = mgh \) , where \( m \) is mass, \( g \) is gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and \( h \) is height.
After an inelastic collision, when all kinetic energy is converted into gravitational potential energy, we can estimate the maximum height reached using the relation between the two.
In this exercise:
  • The kinetic energy right after collision transitions completely into gravitational potential energy at the peak of the rise.
  • This transition is given by:\[ \frac{1}{2} v'^2 = gh \] where each term represents the energy type at different phases.
By solving for \( h \), we find that the new mass system rises up to a certain height, in this case, around 2.645 meters. This demonstrates how potential energy reflects the influence of gravity working against the motion resulting from the collision.

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Most popular questions from this chapter

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