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An isolated particle of mass \(m\) is moving in a horizontal plane \((x-y)\), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses \(m / 4\) and \(3 m / 4\). An instant later, the smaller fragment is at \(y=+15 \mathrm{~cm}\). The larger fragment at this instant is at (a) \(y=-5 \mathrm{~cm}\) (b) \(y=+20 \mathrm{~m}\) (c) \(y=+\overline{5 \mathrm{~cm}}\) (d) \(y=-20 \mathrm{~cm}\)

Short Answer

Expert verified
The larger fragment is at y = -5 cm.

Step by step solution

01

Understand the conservation of momentum

Since the explosion occurs in a closed system (no external forces assuming other gravity components orthogonal don't affect motion), momentum is conserved. Initially, the particle moves along the x-axis, so the total y-momentum is zero before the explosion. After the explosion, the sum of the y-components of momentum of both fragments remains zero.
02

Apply conservation of y-momentum

Let the velocity of the smaller fragment in the y-direction be denoted as \(v_{y1}\) and that of the larger fragment as \(v_{y2}\). The conservation of momentum in the y-direction is given by: \[\frac{m}{4}v_{y1} + \frac{3m}{4}v_{y2} = 0\] Rearrange this equation to find:\[v_{y2} = -\frac{1}{3}v_{y1}\]
03

Analyze the given positions

The smaller fragment is given at \(y = +15 \text{ cm}\). Assume this is due to velocity \(v_{y1}\). The larger fragment must have traveled in the opposite direction to maintain equilibrium in momentum. Using the relation from Step 2:\[v_{y2} = -\frac{1}{3}v_{y1}\] Thus, its distance traveled, due to its velocity and time elapsed being identical to the smaller, will be:\[y_2 = -\frac{1}{3} \times 15 \text{ cm} = -5 \text{ cm}\]
04

Verify the options

The possible y-coordinates of the larger fragment immediately after the explosion are compared:- (a) \(y=-5 \text{ cm}\) correct according to our calculation.- (b) \(y=+20 \text{ m}\) incorrect.- (c) \(y=+5 \text{ cm}\) incorrect.- (d) \(y=-20 \text{ cm}\) incorrect.

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Key Concepts

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

Explosion Fragments
When a particle explodes into fragments, the explosion creates segments of the particle that move separately in space. Each fragment has a portion of the original particle's mass, and their movements are influenced by the conservation of momentum. Understanding explosion fragments is crucial in solving problems related to the aftermath of an explosion and predicting fragment paths.

In this scenario, an isolated particle explodes into two fragments. The smaller fragment has a mass of \(\frac{m}{4}\), while the larger one has a mass of \(\frac{3m}{4}\). These fragments follow distinct trajectories, but they are linked through the momentum they share from the initial moving particle. By examining these fragments, we can calculate their post-explosion velocities and positions based on their mass and the initial conditions prior to the explosion.
  • The size relation between the fragments tells us about their mass distribution.
  • The explosion's effect is assumed instant without additional external forces affecting the system.
Momentum in Horizontal Plane
Momentum in physics refers to the quantity of motion an object possesses and is the product of an object's mass and velocity. In a horizontal plane, like the scenario of the explosion in our exercise, the investigation is often simplified as the focus is on motion along a defined plane without other complex factors such as vertical movement.

Prior to the explosion, the particle was moving along the x-axis in a horizontal plane, implying that only horizontal momentum was present. Due to this, the total momentum in the y-axis was initially zero since there was no vertical motion involved. Post-explosion, even though the fragments may move vertically, the sum of their momentum in that direction should still equal zero to satisfy the conservation principle.
  • Horizontal momentum remains constant unless acted on by external forces.
  • In solving, assume any y-components start from zero and must balance out among fragments post-event.
Conservation Laws in Physics
Conservation laws, such as the conservation of momentum, are fundamental principles in physics that dictate certain physical properties do not change within an isolated system. The law of conservation of momentum states that within an isolated system with no external forces, the total momentum remains constant over time.

In the problem presented, the conservation of momentum is applied specifically in the y-direction. Before the explosion, the y-momentum was zero, meaning the subsequent sum of the y-momentum of the fragments must also remain zero. This allows us to solve for unknown velocities and positions of the fragments following their explosion. To ensure accuracy:
  • Consider external forces: Here, ignoring gravity's orthogonal influence is key to simplifying the y-axis consideration.
  • Apply conservation in each plane independently when analyzing multi-directional motion.
The challenge in applying conservation laws effectively lies in accurately identifying the system boundaries and ensuring all potential influences are accounted for or justifiably ignored.

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