/*! 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 77 An 8.00 -kg block of wood sits a... [FREE SOLUTION] | 91Ó°ÊÓ

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An 8.00 -kg block of wood sits at the edge of a frictionless table, 2.20 \(\mathrm{m}\) above the floor. A \(0.500-\mathrm{kg}\) blob of clay slides along the length of the table with a speed of 24.0 \(\mathrm{m} / \mathrm{s}\), strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

Short Answer

Expert verified
The combined object travels approximately 0.94 meters horizontally.

Step by step solution

01

Calculate the total mass

First, we need to calculate the total mass of the block of wood and the clay after the collision. This is simply the sum of their individual masses: \\[ m_{\text{total}} = m_{\text{block}} + m_{\text{clay}} = 8.00 \text{ kg} + 0.500 \text{ kg} = 8.50 \text{ kg} \]
02

Apply conservation of momentum

Since the collision happens along a frictionless surface, the horizontal momentum is conserved. Set up the conservation of momentum equation: \\[ m_{\text{clay}} \cdot v_{\text{clay}} = m_{\text{total}} \cdot v_{\text{combo}} \] \where \( v_{\text{clay}} = 24.0 \text{ m/s} \) is the initial speed of the clay. Solving for \( v_{\text{combo}} \): \\[ (0.500 \text{ kg}) \cdot (24.0 \text{ m/s}) = (8.50 \text{ kg}) \cdot v_{\text{combo}} \] \\[ v_{\text{combo}} = \frac{12.0}{8.50} = 1.41 \text{ m/s} \]
03

Calculate time of fall

Next, calculate the time it takes for the combined object to fall to the floor, which is a vertical motion problem:\\[ h = \frac{1}{2} g t^2 \] \where \( h = 2.20 \text{ m} \) is the height and \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. Solving for \( t \):\\[ 2.20 = \frac{1}{2} \cdot 9.81 \cdot t^2 \] \\[ t^2 = \frac{2.20 \times 2}{9.81} \] \\[ t^2 = 0.448 \] \\[ t = \sqrt{0.448} = 0.669 \text{ s} \]
04

Calculate horizontal distance traveled

The horizontal distance traveled is given by the horizontal velocity times the time of flight. Use the formula: \\[ \text{distance} = v_{\text{combo}} \cdot t \] \\[ \text{distance} = 1.41 \text{ m/s} \cdot 0.669 \text{ s} = 0.943 \text{ m} \]

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Key Concepts

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

Projectile Motion
Projectile motion is a type of motion experienced by an object that is launched into the air with an initial velocity and then influenced only by gravity, ignoring air resistance. In this exercise, the block and clay fall under projectile motion once they travel horizontally off the table.

When analyzing projectile motion, it's crucial to separate the motion into vertical and horizontal components, as they are independent of each other.
  • The horizontal component involves the speed the object travels parallel to the ground. Here, the block and clay's horizontal velocity is the result from the collision.
  • The vertical component is controlled by gravity, dictating how long an object takes to fall a certain height, like the 2.20 m drop in this example.
Since the horizontal and vertical motions don't affect each other, the block and clay travel horizontally at a constant speed determined by the post-collision speed, while accelerating downward due to gravity until they reach the ground.
Frictionless Surface
A frictionless surface is a hypothetical surface where there is no resistance to sliding motion. This concept allows us to ignore the effects of friction which normally oppose motion. When considering the collision between the clay blob and the wooden block, the frictionless assumption greatly simplifies the analysis.

On a frictionless surface, conservation of momentum is straightforward because no external horizontal forces act on the system, i.e., friction does not lead to any energy loss or momentum gain. Thus, the horizontal momentum of the system before the collision (just the clay moving) equals the momentum after the collision (both clay and block moving together).

This allows us to directly relate the initial speed of the clay and the post-collision speed of the system using the equation:
  • Momentum before collision: \( m_{\text{clay}} \times v_{\text{clay}} \)
  • Momentum after collision: \( m_{\text{total}} \times v_{\text{combo}} \)
Without friction, the calculations efficiently account for momentum changes without additional factors complicating the physics.
Collision Physics
Collision physics involves studying interactions where objects strike each other, exchanging energies and momenta. In our exercise, the clay collides with the block and sticks to it – a classic inelastic collision.

In an inelastic collision like this:
  • Kinetic energy is not conserved. Some of it transforms into other energy forms, e.g., heat or deformation energy, because objects stick together.
  • Momentum is still conserved, which enables us to calculate the velocity of the combined clay and block system after their melding.
The properties of collisions dictate how we calculate post-collision speeds and are key to understanding how predictions are made in physics problems. The seamless merging of two objects denotes that mass sums while velocities are calculated from preserved momentum, forming the basis of collision physics and allowing for problem-solving even amidst energy transformations.

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Most popular questions from this chapter

Jack and Jill are standing on a crate at rest on the frictionless, horizontal surface of a frozen pond. Jack has mass 75.0 kg, Jill has mass \(45.0 \mathrm{kg},\) and the crate has mass 15.0 \(\mathrm{kg}\) . They remember that they must fetch a pail of water, so each jumps horizontally from the top of the crate. Just after each jumps, that person is moving away from the crate. with a speed of 4.00 \(\mathrm{m} / \mathrm{s}\) relative to the crate. (a) What is the final speed of the crate if both Jack and Jill jump simultaneously and in the same direction? (Hint: Use an inertial coordinate system attached to the ground.) (b) What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction? (c) What is the final speed of the crate if Jill jumps first and then Jack, again in the same direction?

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750 -kg car traveling to the right at 1.50 \(\mathrm{m} / \mathrm{s}\) collides with a \(1450-\mathrm{kg}\) car going to the left at 1.10 \(\mathrm{m} / \mathrm{s} .\) Measurements show that the heavier car's speed just after the collision was 0.250 \(\mathrm{m} / \mathrm{s}\) in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.0 \(\mathrm{m}\) long and has mass 20.0 \(\mathrm{kg} .\) Burt has mass 30.0 \(\mathrm{kg}\) and Ernie has mass 40.0 \(\mathrm{kg} .\) Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.

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