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Chapter 9: Problem 25

A \(12.0-\mathrm{g}\) wad of sticky clay is hurled horizontally at a \(100-\mathrm{g}\) wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides \(7.50 \mathrm{m}\) before coming to rest. If the coefficient of friction between the block and the surface is \(0.650,\) what was the speed of the clay immediately before impact?

Short Answer

Expert verified
The initial speed of the clay immediately before impact is obtained by substituting the calculated value of \(v'\) from Step 2 into the equation from Step 1 and solving for \(v\).

Step by step solution

01

Use Conservation of Momentum

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Let's denote the speed of the clay immediately before impact as \(v\), the speed of clay-block system just after the impact as \(v'\). The momentum before the collision can be expressed as \(m_{clay} \cdot v\), and the momentum after as \((m_{clay} + m_{block}) \cdot v'\). Setting these two equalities to each other we get: \(m_{clay} \cdot v = (m_{clay} + m_{block}) \cdot v'\). Here, the masses are given but we need to find \(v\) and \(v'\). We will proceed to calculate \(v'\) in the next step.
02

Use Work-Energy Principle

The work-energy principle states that the work done on an object equals the change in its kinetic energy. In the given exercise, the block is slowing down due to friction which acts against the movement and this frictional force does work which leads to the object's kinetic energy turning into thermal energy. The kinetic energy just after the collision can be written as \(\frac{1}{2} \cdot (m_{clay} + m_{block}) \cdot {v'}^2\). This energy will then completely turn into thermal energy due to friction, which can also be calculated with the formula: \(F_{friction} \cdot d = m_{total} \cdot g \cdot \mu \cdot d\). By setting the energy transformations equal, we can solve for \(v'\).
03

Calculate Initial Speed

Now that we have obtained the value of \(v'\), we can substitute it into the equation derived in step 1. Solving the derived equation for \(v\) will yield the initial speed of the clay immediately before impact.

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Key Concepts

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

Work-Energy Principle
The Work-Energy Principle is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy. When you perform work on an object, you either increase or decrease its energy depending on the type of work. This principle is especially crucial in understanding collisions and motion.
This principle can be expressed as:
  • Work done = Change in kinetic energy
In the given problem, after the sticky clay collides with the wooden block, the combined object slides due to its kinetic energy being converted into thermal energy because of friction. This friction does work by opposing the motion. The work done by friction can be calculated by multiplying the frictional force with the distance the block slides. This also equals the loss in kinetic energy as the block comes to rest due to friction acting against it. By using this principle, you can determine the velocity of the clay-block system just after the collision, which is essential for further calculations.
Coefficient of Friction
The coefficient of friction, denoted by \( \mu \), is a dimensionless scalar that represents the frictional force resistance between two surfaces. It's a crucial factor in problems involving motion and collisions since it directly influences the amount of frictional force.
  • Static friction is the friction that keeps an object at rest, and is usually higher than the kinetic friction.
  • Kinetic friction involves moving objects, like our clay-block system sliding over the surface.
In this exercise, the coefficient of friction is given as 0.650. This number means that the friction force is 0.650 times the normal force exerted by the block on the surface. The frictional force, which slows down the block, is calculated by multiplying the normal force (equal to the object's weight, \(m_{total} \times g\)) by the coefficient of friction. This concept is vital for finding how far the block will move before stopping after the collision.
Kinetic Energy
Kinetic energy is the energy of motion. Any object in movement carries kinetic energy, which is calculated using the formula:\[ KE = \frac{1}{2} \times m \times v^2 \]where \(m\) is the mass and \(v\) is the velocity of the object. Kinetic energy is a key player in problems concerning motion and collisions, such as in this exercise.
  • Before the impact, the clay has kinetic energy while the block at rest has none.
  • After collision, both the clay and block move together, sharing kinetic energy.
In the problem, when the clay sticks to the block, their combined mass and velocity just after the impact allows us to compute their shared kinetic energy. This energy diminishes due to friction as the block and clay system slides along the surface until coming to a stop. Understanding how kinetic energy transforms is essential to solving the problem, as it helps you deduce the system's velocity just after the impact, hence determining the initial speed of the clay.

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Most popular questions from this chapter

An archer shoots an arrow toward a target that is sliding toward her with a speed of \(2.50 \mathrm{m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{m} / \mathrm{s}\) and passes through the 300 -g target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

A \(0.500-\mathrm{kg}\) sphere moving with a velocity \((2.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}+$$1.00 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s}\) strikes another sphere of mass \(1.50 \mathrm{kg}\) moving with a velocity \((-1.00 \hat{\mathbf{i}}+2.00 \hat{\mathbf{j}}-3.00 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s}\) (a) If the velocity of the \(0.500-\mathrm{kg}\) sphere after the collision is \((-1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}-8.00 \mathbf{k}) \mathrm{m} / \mathrm{s},\) find the final velocity of the 1.50-kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic). (b) If the velocity of the \(0.500-\mathrm{kg}\) sphere after the collision is \((-0.250 \hat{\mathbf{i}}+0.750 \hat{\mathbf{j}}-\) \(2.00 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s},\) find the final velocity of the \(1.50-\mathrm{kg}\) sphere and identify the kind of collision. (c) What If? If the velocity of the \(0.500-\mathrm{kg}\) sphere after the collision is \((-1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}+\) \(a \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s},\) find the value of \(a\) and the velocity of the \(1.50-\mathrm{kg}\) sphere after an elastic collision.

A golf ball \((m=46.0 \mathrm{g})\) is struck with a force that makes an angle of \(45.0^{\circ}\) with the horizontal. The ball lands \(200 \mathrm{m}\) away on a flat fairway. If the golf club and ball are in contact for \(7.00 \mathrm{ms},\) what is the average force of impact? (Neglect air resistance.)

George of the Jungle, with mass \(m\), swings on a light vine hanging from a stationary tree branch. A second vine of equal length hangs from the same point, and a gorilla of larger mass \(M\) swings in the opposite direction on it. Both vines are horizontal when the primates start from rest at the same moment. George and the gorilla meet at the lowest point of their swings. Each is afraid that one vine will break, so they grab each other and hang on. They swing upward together, reaching a point where the vines make an angle of \(35.0^{\circ}\) with the vertical. (a) Find the value of the ratio \(m / M\) (b) What If? Try this at home. Tie a small magnet and a steel screw to opposite ends of a string. Hold the center of the string fixed to represent the tree branch, and reproduce a model of the motions of George and the gorilla. What changes in your analysis will make it apply to this situation? What If? Assume the magnet is strong, so that it noticeably attracts the screw over a distance of a few centimeters. Then the screw will be moving faster just before it sticks to the magnet. Does this make a difference?

A 75.0 -kg firefighter slides down a pole while a constant friction force of \(300 \mathrm{N}\) retards her motion. A horizontal 20.0 -kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest \(4.00 \mathrm{m}\) above the platform, and the spring constant is \(4000 \mathrm{N} / \mathrm{m} .\) Find \((\mathrm{a})\) the firefighter's speed just before she collides with the platform and (b) the maximum distance the spring is compressed. (Assume the friction force acts during the entire motion.)
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