/*! 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 101 a \(3.2 \mathrm{~kg}\) box of ru... [FREE SOLUTION] | 91Ó°ÊÓ

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a \(3.2 \mathrm{~kg}\) box of running shoes slides on a horizontal frictionless table and collides with a \(2.0 \mathrm{~kg}\) box of ballet slippers initially at rest on the edge of the table, at height \(h=0.40 \mathrm{~m}\). The speed of the \(3.2 \mathrm{~kg}\) box is \(3.0 \mathrm{~m} / \mathrm{s}\) just before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor?

Short Answer

Expert verified
The kinetic energy is approximately 8.87 J before they strike the floor.

Step by step solution

01

Calculate Initial Momentum

Use the conservation of momentum for the system. The initial momentum is the product of mass and velocity of the moving box: \[ p_{initial} = m_1 \times v_1 = 3.2 \, \text{kg} \times 3.0 \, \text{m/s} = 9.6 \, \text{kg m/s} \]Since the second box is at rest initially, its momentum is zero.
02

Calculate Combined Mass After Collision

The two boxes stick together after the collision, so their combined mass is:\[ m_{total} = m_1 + m_2 = 3.2 \, \text{kg} + 2.0 \, \text{kg} = 5.2 \, \text{kg} \]
03

Calculate Velocity After Collision

Apply the conservation of momentum to find the velocity after the collision:\[ p_{initial} = p_{final} \]\[ 9.6 \, \text{kg m/s} = 5.2 \, \text{kg} \times v_{final} \]Solving for \( v_{final} \), we get:\[ v_{final} = \frac{9.6}{5.2} = 1.846 \, \text{m/s} \]
04

Calculate Kinetic Energy Before Impact

Use the kinetic energy formula for the combined mass just before they strike the floor:\[ KE = \frac{1}{2} m_{total} v_{final}^2 \]Plugging in the values, we get:\[ KE = \frac{1}{2} \times 5.2 \, \text{kg} \times (1.846 \, \text{m/s})^2 \]\[ KE = \frac{1}{2} \times 5.2 \times 3.408 \approx 8.8712 \, \text{J} \]Thus, the kinetic energy just before they strike the floor is approximately 8.87 J.

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Key Concepts

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

Elastic and Inelastic Collisions
In physics, collisions are often classified into two main types: elastic and inelastic. Understanding these concepts is critical in solving momentum and energy problems.
- **Elastic Collisions:** These conserve both momentum and kinetic energy. The objects bounce off each other without deformation or heat generation. Commonly seen in ideal gas particles, where after the collision, they remain unchanged.
- **Inelastic Collisions:** Only momentum is conserved, not kinetic energy. The objects may stick together, deform, or heat up. In our problem, since the boxes stick together after colliding, it is classified as a perfectly inelastic collision, where the maximum amount of kinetic energy is lost, converting to other forms like sound or heat.
Understanding collision types helps predict the behavior of colliding objects and the energy transformation post-collision. This forms the basis to approach and solve collision-related physics problems effectively.
Kinetic Energy Calculation
Kinetic energy represents the energy that an object possesses due to its motion. The formula to calculate kinetic energy is:\[ KE = \frac{1}{2} m v^2 \]where \(m\) is mass, and \(v\) is velocity. In the provided problem, once the collision has occurred and the velocities of the combined masses (the two boxes) have been determined, calculating their kinetic energy helps understand how much energy they possess before hitting the ground.
- **Initial Step:** Combine the masses (3.2 kg + 2.0 kg = 5.2 kg) since they stick together.
- **Determine Velocity after Collision:** Use the conservation of momentum to solve for the velocity of the combined mass.
Subsequently, applying these values in the kinetic energy formula gives the energy just before impact, found to be approximately 8.87 Joules. Knowing how to calculate kinetic energy is essential for analyzing motion efficiency and energy transformations in physics.
Physics Problem Solving
Solving physics problems, like the one described, often requires a structured approach. Here's a simple breakdown to follow:
1. **Understand the Problem:** Carefully read the details to comprehend what is being asked. Identify the physical principles such as conservation of momentum or energy involved.
2. **Identify Known and Unknown Variables:** List what values are given (mass, velocity) and what needs solving (e.g., kinetic energy).
3. **Apply Relevant Physics Principles:** Use equations and formulas pertinent to the case, such as momentum (\[p = mv\]), and kinetic energy (\[KE = \frac{1}{2}mv^2\]). These establish relationships between the variables.
4. **Solve & Validate:** Do the math to find the unknowns. Recheck steps to ensure consistency and correctness, especially for unit conversions.
5. **Interpret Results:** Consider what the solution means in the context of the problem.
This systematic approach helps in tackling complex problems effectively, ensuring one does not overlook crucial aspects that impact the solution. Mastering these strategies will greatly enhance problem-solving skills in physics.

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Most popular questions from this chapter

block 1 of mass \(m_{1}\) slides from rest along a frictionless ramp from height \(h=2.50 \mathrm{~m}\) and then collides with stationary block \(2,\) which has mass \(m_{2}=2.00 m_{1}\). After the collision, block 2 slides into a region where the coefficient of kinetic friction \(\mu_{k}\) is 0.500 and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

Suppose a gangster sprays Superman's chest with 3 g bullets at the rate of 100 bullets/min, and the speed of cach bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest?

block 1 of mass \(m_{1}\) slides along an \(x\) axis on a frictionless floor with a speed of \(v_{1 j}=4.00 \mathrm{~m} / \mathrm{s}\). Then it undergoes a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=0.500 m_{1}\). Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass \(m_{3}=0.500 \mathrm{~m}_{2}\). (a) What then is the speed of block 3 ? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block \(1 ?\)

Shows a slab with dimensions \(d_{1}=11.0 \mathrm{~cm}\), \(d_{2}=2.80 \mathrm{~cm}, \quad\) and \(\quad d_{3}=13.0 \mathrm{~cm} .\) Half the slab consists of aluminum (density \(=2.70 \mathrm{~g} / \mathrm{cm}^{3}\) ) and half consists of iron (density = \(7.85 \mathrm{~g} / \mathrm{cm}^{3}\) ). What are (a) the \(x\) coordinate, (b) the \(y\) coordinate, and (c) the \(z\) coordinate of the slab's center of mass?

In a common but dangcrous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\). and the collision on the floor lasts \(0.082 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

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