/*! 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 45 A vase of mass \(m\) falls onto ... [FREE SOLUTION] | 91Ó°ÊÓ

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A vase of mass \(m\) falls onto a floor and breaks into three pieces that then slide across the frictionless floor. One piece of mass \(0.25 m\) moves at speed \(v\) along an \(x\) axis. The second piece of the same mass and speed moves along the \(y\) axis. Find the speed of the third piece.

Short Answer

Expert verified
The speed of the third piece is \(\frac{v}{\sqrt{2}}\).

Step by step solution

01

Conceptual Understanding

The problem involves conservation of momentum. Since the vase falls and breaks on a frictionless surface, the momentum of the system before breaking should equal the momentum of the system after breaking.
02

Setup of Initial Conditions

Before the vase breaks, it is at rest on the floor. This means the initial momentum of the system is zero. Accordingly, the total momentum in each direction (x and y axis) after breaking must also equal zero.
03

Express Momentum After Breaking

The two pieces of mass \(0.25m\) move along the x and y axes with speed \(v\). For the third piece of mass \(0.5m\) (since the total mass \(m = 0.25m + 0.25m + 0.5m\)), its momentum will be along some angle with components along both x and y axes.
04

Momentum Conservation in the X-axis

Conserve momentum in the x-axis: the momentum contributed by the first piece is \(0.25mv\), and \(P_{x,3rd}\) is the component of the third piece's momentum in the x direction. Thus, \(0.25mv + P_{x,3rd} = 0\).
05

Momentum Conservation in the Y-axis

Conserve momentum in the y-axis: the momentum contributed by the second piece is \(0.25mv\), and \(P_{y,3rd}\) is the component of the third piece's momentum in the y direction. Thus, \(0.25mv + P_{y,3rd} = 0\).
06

Solve for Momentum Components of the Third Piece

From Step 4, \(P_{x,3rd} = -0.25mv\) and from Step 5, \(P_{y,3rd} = -0.25mv\).
07

Calculate Speed from Momentum Components

The third piece's momentum is comprised of x and y components: \(P_{3} = \sqrt{P_{x,3rd}^2 + P_{y,3rd}^2}\). Substitute \(P_{x,3rd}\) and \(P_{y,3rd}\), and the speed \(v_{3}\) of the third piece is given by \(v_{3} = \frac{P_{3}}{0.5m}\). Evaluate \(v_{3} = \frac{\sqrt{(-0.25mv)^2 + (-0.25mv)^2}}{0.5m}\).
08

Simplify and Solve

Simplify the square root to find that \(v_{3} = \sqrt{0.5}v\). Thus, the speed of the third piece is \(v_{3} = \frac{v}{\sqrt{2}}\).

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Key Concepts

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

Momentum
Momentum is a key concept in physics that describes the amount of motion an object has. It is a product of the object's mass and velocity and is usually represented as \( p = mv \).
Momentum is a vector quantity, meaning it has both magnitude and direction. This is important because in any physical process, both the magnitude and the direction of momentum can affect the outcome.

When an object experiences a force and changes its velocity, its momentum changes. In problems involving multiple objects, such as the one with the vase, the total momentum of the system before an interaction must equal the total momentum after the interaction, provided no external forces act on the system. This principle is known as the conservation of momentum.
Collision
In the context of physics, a collision occurs when two or more objects exert forces on each other for a relatively short period of time.
Collisions can be perfectly elastic, perfectly inelastic, or somewhere in between. In elastic collisions, both kinetic energy and momentum are conserved. In inelastic collisions, only momentum is conserved, which often results in some deformation or generation of heat.

For this problem, involving a vase shattering, it's important to understand how the fragments move post-collision. Even though it wasn't a traditional collision between objects, the fragments behave similarly post-breakage. They separate with different velocities but still respect the laws of momentum conservation.
Vector Components
Vector components break down forces or motion into different directions, typically using an x, y, and sometimes z-axis in physics problems.
In the problem of the vase breaking, you're dealing with two main vector components: along the x-axis and y-axis. The pieces of the vase move independently in their respective directions.

To fully describe the motion of each piece, you calculate the momentum in these two directions separately. This involves breaking vectors into horizontal (\( x \)) and vertical (\( y \)) components, ensuring each component's momentum sums to zero along its direction per the conservation laws. Later, these components are combined again to find the resultant motion or speed of an object, like the third piece here.
Frictionless Surface
A frictionless surface is an idealized concept used to simplify physics problems. It means that there is no resistance to motion, allowing objects to move freely without any energy loss due to friction.
This simplification is crucial in momentum conservation problems because it eliminates external forces that could alter momentum. Without friction, any changes in an object's speed or path must result from interactions with other objects, not from energy loss to the surroundings.

For the falling and breaking vase, assuming a frictionless surface means that the total momentum before the vase hits the floor is conserved after breaking. It allows the problem to be analyzed using only internal interactions, which simplifies calculations and focuses solely on momentum principles.

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Most popular questions from this chapter

Basilisk lizards can run across the top of a water surface (Fig. 9-44). With each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. To avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before water can flow into the air cavity. If the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. Suppose the mass of a basilisk lizard is \(90.0 \mathrm{~g}\), the mass of each foot is \(3.00 \mathrm{~g}\), the speed of a foot as it slaps the water is \(1.50\) \(\mathrm{m} / \mathrm{s}\), and the time for a single step is \(0.600 \mathrm{~s}\). (a) What is the magnitude of the impulse on the lizard during the slap? (Assume this impulse is directly upward.) (b) During the \(0.600 \mathrm{~s}\) duration of a step, what is the downward impulse on the lizard due to the gravitational force? (c) Which action, the slap or the push, provides the primary support for the lizard, or are they approximately equal in their support?

A steel ball of mass \(0.600 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 9-57). At the bottom of its path, the ball strikes a \(2.80 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0\), a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \hat{i}\), with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.750 \mathrm{~s}\) and \(t=1.25 \mathrm{~s}\) ? (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0\) ?

A \(1000 \mathrm{~kg}\) automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of \(3.0 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant a \(2000 \mathrm{~kg}\) truck, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far is the com of the automobiletruck system from the traffic light at \(t=\) \(5.0 \mathrm{~s} ?\) (b) What is the speed of the com then?

Figure 9-47 shows a two-ended "rocket" that is initially stationary on a frictionless floor, with its center at the origin of an \(x\) axis. The rocket consists of a central block \(C\) (of mass \(M=6.00 \mathrm{~kg}\) ) and blocks \(L\) and \(R\) (each of mass \(m=2.00 \mathrm{~kg}\) ) on the left and right sides. Small explosions can shoot either of the side blocks away from block \(C\) and along the \(x\) axis. Here is the sequence: (1) At time \(t=0\), block \(L\) is shot to the left with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time \(t=0.80 \mathrm{~s}\), block \(R\) is shot to the right with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that block \(C\) then has. At \(t=2.80 \mathrm{~s}\), what are (a) the velocity of block \(C\) and (b) the position of its center?

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