How Far Does the Lighter Fragment Slide After an Explosion?

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Homework Help Overview

The problem involves an explosion of an object at rest on a flat surface into two fragments with differing masses. The heavier fragment, which is seven times the mass of the lighter fragment, slides a certain distance before stopping. The task is to determine how far the lighter fragment slides, given that both fragments experience the same coefficient of kinetic friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial momentum being zero and explore the relationship between the masses and their velocities post-explosion. There are attempts to incorporate distance into the calculations, with some participants suggesting the use of kinematic equations to find acceleration and distance.

Discussion Status

The discussion is ongoing, with participants providing various approaches to relate the distances and forces involved. Some guidance has been offered regarding the use of energy concepts and momentum conservation, but there is no explicit consensus on a single method to solve the problem.

Contextual Notes

Participants are navigating the complexities of the problem, including the need to account for frictional forces and the relationship between the two fragments. There is an acknowledgment of the challenge posed by having multiple unknowns in the equations being considered.

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Homework Statement



An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. the heavier fragment slides 8.2m before stopping. how far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction.



Homework Equations



0 = mv1 + mv2


The Attempt at a Solution



The initial momentum is 0 in this question because there is no movement initially.
using the equation 0 = mv1 + 7mv2

-mv1 = 7mv2
-v1 = 7v2

but... the problem I am having is how do i incorporate the distance ?
help!
 
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you can use vi power 2 -vi 0power 2=2ax and calculate accelerate then use again this formulla to find the distance of lighter mass.
 
what do you mean. how do you find the acceleration?
that equation vf^2 = vi^2 + 2ad has 2 unknowns, therefore it cannot be solved to find the acceleration...

could anyone explain how i can solve the problem ?
 
I believe you can solve this by recognizing the relationship

E = \int _a ^b \vec F \cdot \vec ds

But don't worry, you don't have to really do any calculus. Since the frictional forces can be approximated as constants, the above equation reduces to

E = \vec F \cdot \vec s

where E is the energy, \vec F is the frictional force, and \vec s is the displacement (i.e. distance) traveled.

Start with the heavier fragment. Calculate its normal force and multiply it by the coefficient of friction to get the frictional force. Use that, and the equation above, to determine the energy converted to frictional heat, associated with the heavier fragment (hint: this is where the 8.2 m comes into play, together with the frictional force which you just calculated).

Now you can use (1/2)mv2 to get the initial velocity of the heavier fragment. Once you know the heavier fragment's velocity you can express its momentum.

Now you can move on to the lighter fragment. Conservation of momentum means that the lighter fragment will have the same initial momentum magnitude as the heavier fragment, but in the opposite direction. If you know its initial momentum, you can calculate its initial velocity (you already know that the mass is 1/7 that of the heavier object, and you know its momentum, so you can solve for velocity). You can also calculate its normal force, and thus its frictional force, and also its initial kinetic energy. That's all the information you need to use in the above equations to calculate its sliding distance.
 
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