Three particles break off after an explosion

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SUMMARY

The discussion focuses on a physics problem involving the conservation of momentum after an explosion of a 20 kg object into three pieces. The first piece (5 kg) moves NW at 30 m/s, and the second piece (4 kg) moves SE at 25 m/s. The third piece, weighing 11 kg, has its x and y components of velocity calculated as 98.692 m/s and -39.32 m/s, respectively. Additionally, the change in kinetic energy of the pieces post-explosion is derived using the equations for momentum and kinetic energy.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with vector components and trigonometry
  • Knowledge of momentum conservation principles
  • Ability to calculate kinetic energy using KE=0.5*m*v^2
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  • Study vector decomposition in physics
  • Learn about conservation laws in collisions
  • Explore kinetic energy transformations in explosions
  • Investigate advanced momentum problems involving multiple bodies
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Students studying physics, particularly those focusing on mechanics and momentum conservation, as well as educators looking for practical examples of explosive interactions in physical systems.

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Homework Statement


An object with total mass mtotal = 20 kg is sitting at rest when it explodes into three pieces.
The first piece with mass m1 = 5 kg moves NW at an angle of θ1 = 20° above the –x axis with a speed of v1 = 30 m/s.
A second piece with mass m2 = 4 kg moves SE at an angle of θ2 = 25° to the right of the -y axis at a speed of v2 = 25 m/s.
The mass of the third piece is 11kg.
What is the x-component and y component of the velocities of the third piece?
What is the change in kinetic energies of the pieces after the collision.
Note that momentum is conserved and Pf=0.


Homework Equations



p=mv
pi=pf=0
KE=0.5*m*v^2
KE=0.5*p^2/m

The Attempt at a Solution



In x direction--> calculating the momentum
m1v1cos20=m2v2sin25+Px where Px is the momentum of the third particle in x direction
5*30*cos20=4*25*sin25+Px
Px=98.692

In x direction--> calculating the momentum
m1v1sin20=m2v2cos25+Py where Py is the momentum of the third particle in x direction
5*30*sin20=4*25*cos25+Px
Py=-39.32
 
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Did you use Pf=0?
 

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