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Two disks sliding on a frictionless horizontal plane with opposite velocities of the same magnitude \(v_{0}\) hit each other squarely. Disk \(A\) is known to have a weight of 6 lb and is observed to have zero velocity after impact. Determine \((a)\) the weight of disk \(B,\) knowing that the coefficient of restitution between the two disks is \(0.5,\) (b) the range of possible values of the weight of disk \(B\) if the coefficient of restitution between the two disks is unknown.

Short Answer

Expert verified
(a) 12 lb. (b) Greater than 6 lb.

Step by step solution

01

Define the Givens and Target

We know that: - Disk A has a weight of 6 lb.- Initial velocity of both disks is \(v_0\) and they hit each other directly with opposite velocities.- After collision, velocity of Disk A is 0.- Coefficient of restitution \(e = 0.5\) (for part a only). We want to find the weight of Disk B for part (a) and determine the possible range for part (b). Convert weights to masses using \(W = mg\).
02

Momentum Conservation Equation

For both disks, write the momentum conservation equation before and after impact. Before collision:\[ \text{Total initial momentum} = m_A v_0 + m_B(-v_0) = 0 \]After collision, Disk A has a velocity of 0, so the momentum equation becomes: \[ 0 = m_A \cdot 0 + m_B v_B' \rightarrow m_B v_B' = 0 \] We need to analyze the collision to find \(v_B'\) and relate it to \(e\).
03

Use Coefficient of Restitution (Part a)

For part (a), use the equation for the coefficient of restitution:\[ e = \frac{v_B' - 0}{0 + v_0} = 0.5 \]This simplifies to \(v_B' = 0.5 v_0\). Substitute this into the momentum conservation equation:\[ m_B \times 0.5 v_0 = m_A \times v_0 \rightarrow m_B = 2m_A \]Convert to weight:\[ W_B = 2 \times 6 = 12 \text{ lb} \]
04

Calculate for Part (b) - Unknown Restitution

For part (b), without knowledge of \(e\), we only know Disk A's post-collision velocity is 0. Hence, momentum equation after impact:\[ m_B v_B' = m_A v_0 \rightarrow v_B' = \frac{m_A v_0}{m_B} \]For \(v_B'\) to be valid (i.e., positive), \(m_B\) must satisfy:\[ 0 < \frac{m_A v_0}{m_B} \leq v_0\]Thus, \(m_B > m_A = 6 \text{ lb} \). The weight of Disk B, \(W_B\), could be any value larger than 6 lb.

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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 conservation is a powerful concept that helps us analyze interactions such as collisions. When two objects collide, the total momentum of the system before the collision is equal to the total momentum after the collision. This principle is rooted in the law of conservation of momentum. In the current exercise, two disks collide and we use this principle to gain insights.

Here's a simplified view:
  • Momentum is the product of an object's mass and velocity.
  • In the absence of external forces, the momentum of a system remains unchanged.
  • For two colliding objects, the total momentum pre-collision equals the total post-collision momentum.
This exercise provides a practical application. Before the collision, the combined momentum is zero because the disks are moving with equal magnitude but opposite velocities. After the collision, the momentum must still sum to zero, guiding us in solving for the outcomes. This principle greatly simplifies our calculations, particularly when identifying unknowns like the weight of the second disk.
Coefficient of Restitution
The coefficient of restitution (commonly denoted as \(e\)) is a scalar value that represents how elastic a collision is.It measures the relative velocity of separation divided by the relative velocity of approach between two colliding bodies.
In this particular example, understanding \(e\) is crucial for part (a) of the problem.
  • If \(e = 1\), the collision is perfectly elastic, meaning no kinetic energy is lost.
  • If \(e = 0\), the collision is perfectly inelastic, and the bodies stick together post-collision.
  • In our exercise, \(e = 0.5\), indicating a partially elastic collision where some kinetic energy is lost.
We use \(e\) to determine how the velocities change post-impact.
The formula for the coefficient of restitution is given by:\[ e = \frac{v_{B}' - v_{A}'}{v_{A} - v_{B}} \]In this problem, we find the post-collision velocity of Disk B using this formula, which subsequently allows us to find its mass and weight.
Frictionless Plane
A frictionless plane is a hypothetical environment often used in physics to simplify problems. It allows us to study the pure effects of motion without complications like friction.
In the given exercise, the frictionless plane ensures that the only forces at work are those from the collision itself. This is why we can apply momentum conservation so straightforwardly.
  • Frictionless surfaces reduce external forces, allowing for more accurate momentum conservation analysis.
  • Without friction, once in motion, objects will continue indefinitely if uninhibited by other forces.
By ignoring friction, the exercise zeroes in on the dynamics of the collision, making it easier to understand how related principles like momentum conservation and the coefficient of restitution are applied. This setting is ideal for educational purposes, providing clarity and focus on the principles at hand.

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Most popular questions from this chapter

Observations show that a celestial body traveling at \(1.2 \times 10^{6} \mathrm{mi} / \mathrm{h}\) appears to be describing about point \(B\) circle of radius equal to 60 light years. Point \(B\) is suspected of being a very dense concentration of mass called a black hole. Determine the ratio \(M_{B} / M_{S}\) of the mass at \(B\) to the mass of the sun. (The mass of the sun is \(330,000\) times the mass of the earth, and a light year is the distance traveled by light in 1 year at \(186,300 \mathrm{mi} / \mathrm{s}\).)

Ball \(B\) is hanging from an inextensible cord. An identical ball \(A\) is released from rest when it is just touching the cord and drops through the vertical distance \(h_{A}=8\) in. before striking ball \(B .\) Assuming \(e=0.9\) and no friction, determine the resulting maximum vertical displacement \(h_{B}\) of the ball \(B .\)

A test machine that kicks soccer balls has a 5 -lb simulated foot attached to the end of a 6 -ft long pendulum arm of negligible mass. Knowing that the arm is released from the horizontal position and that the coefficient of restitution between the foot and the 1 -lb ball is \(0.8,\) determine the exit velocity of the ball \((a)\) if the ball is stationary, \((b)\) if the ball is struck when it is rolling towards the foot with a velocity of \(10 \mathrm{ft} / \mathrm{s}\).

A 2-kg sphere moving to the right with a velocity of 5 m/s strikes at A, which is on the surface of a 9-kg quarter cylinder that is initially at rest and in contact with a spring with a constant of 20 kN/m. The spring is held by cables, so it is initially compressed 50 mm. Neglecting friction and knowing that the coefficient of restitution is 0.6, determine (a) the velocity of the sphere immediately after impact, (b) the maximum compressive force in the spring.

A 180-lb man and a 120-lb woman stand at opposite ends of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.

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