Exploding Pumpkin: Solving for Third Fragment's Velocity, Direction, and Mass

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Homework Help Overview

The problem involves a pumpkin of mass 3.30 kg that explodes into three pieces on a frictionless plane. Two pieces are known: a 0.50 kg piece moving north at 3.2 m/s and a 1.4 kg piece moving southwest at 4.1 m/s. The task is to determine the speed, direction, and mass of the third piece.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and the need to consider vector quantities. There are suggestions to draw diagrams to visualize forces and directions. Some participants question the assumptions made regarding the initial momentum and the setup of the equations.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem using vector equations and momentum conservation. There is an emphasis on clarifying the initial conditions and the need for a diagram to aid understanding.

Contextual Notes

Participants note that the initial momentum is zero, which is a critical aspect of the problem. There is also mention of the angles involved in the directions of the fragments, which may not be presented in a conventional manner.

Forerunner546
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Homework Statement


A pumpkin of mass 3.30kg at rest on a smooth frictionless plane, explodes into 3 pieces in the horizontal plane. A .50kg piece flies to the North at 3.2m/s. A 1.4kg piece flies off (SW) at 4.1m/s.

What is the speed, direction and mass of the third piece?

Homework Equations


M1V1o + M2V2o = M1V1f + M2V2f

The Attempt at a Solution


I assumed the piece was 1.4kg as I subtracted the original pumpkins mass by the known fragmentary masses. I used the conservation of momentum to subtract the original pumpkins mass and velocity by the unknown fragment and equaled it to the two known pieces. M1V1o - M3V3f = M1V1f + M2V2f

This yielded -5.2 as a velocity which seems wrong, and I have no idea how to find the direction of the third piece and no angles were included.
 
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The FIRST thing you should always do in working on such problems is to draw a force diagram. For unknown forces you estimate at direction and magnitude from the known forces/directions. Do that and present it here and then we can help you work on the problem.
 
They do give angles, just not in the sense you may be used to. Get a Compass Rose and find direction of North and South West, then figure out what angles these are.
 
Here’s a diagram
 

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Forerunner546 said:
This yielded -5.2 as a velocity which seems wrong
Can't tell where you are going wrong if you do not post your working.
 
Forerunner546 said:

Homework Statement


A pumpkin of mass 3.30kg at rest on a smooth frictionless plane, explodes into 3 pieces in the horizontal plane. A .50kg piece flies to the North at 3.2m/s. A 1.4kg piece flies off (SW) at 4.1m/s.

What is the speed, direction and mass of the third piece?

Homework Equations


M1V1o + M2V2o = M1V1f + M2V2f

The Attempt at a Solution


I assumed the piece was 1.4kg as I subtracted the original pumpkins mass by the known fragmentary masses. I used the conservation of momentum to subtract the original pumpkins mass and velocity by the unknown fragment and equaled it to the two known pieces. M1V1o - M3V3f = M1V1f + M2V2f

This yielded -5.2 as a velocity which seems wrong, and I have no idea how to find the direction of the third piece and no angles were included.
Forerunner546 said:
Here’s a diagram
6f89413f-811d-487b-be47-b9f7cdc691cb-jpeg.jpg


Hello @Forerunner546 . :welcome:

Your equations are vector equations, because momentum is a vector quantity. So is velocity.

What is the initial momentum ?
 

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SammyS said:
View attachment 234263

Hello @Forerunner546 . :welcome:

Your equations are vector equations, because momentum is a vector quantity. So is velocity.

What is the initial momentum ?
There is no initial momentum
 
Forerunner546 said:
There is no initial momentum
Right. So write out the momentum conservation equations for two orthogonal directions.
 
Forerunner546 said:
There is no initial momentum
By this I assume that you mean: the initial momentum is zero.

That's correct.

You can also solve this by arranging the momentum vectors in a triangle and using Law of Cosines and/or Law of Sines.
 

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