Cons. Momentum/Energy of rubber block blown up by explosive

[strangeeyes]

Homework Statement

A small explosive charge is placed in a rubber block resting on a smooth (frictionless) surface. When the charge is detonated, the block breaks into three pieces. A 200-g piece travels at 1.4 m/s, and 300g piece travels at 0.90m/s. The third piece flies off at a speed of 1.8m/s. If the angle between the first two piece is 80 degrees, calculate the mass and direction of the third piece.

 

[gneill]

That's nice. What have you tried so far? What's your procedure?

 

[strangeeyes]

well that's just it, i don't know where to begin. beyond the fact that momentum and energy is conserved, I'm at a loss with this one. i tried throwing together a momentum equation where 0=m1v1'+m2v2'+m3v3' just to see what came of it and got something like 305g which obviously isn't correct. do you have any ideas? help is greatly appreciated.

 

[gneill]

While total momentum is conserved, it is also true that momentum in any given direction is also conserved. That means you should be able to write separate expressions for the total momentum in the X and Y directions. Does that help?

 

[strangeeyes]

not sure. are you suggesting something like px=m1v1x'+m2v2x'+m3v3x'? wouldn't i then have two variables m3 and its angle? can the conservation of energy formula be used to substitute m3 into that equation or is energy not conserved in the explosion? I'm in grade twelve, so i don't really understand what the formula for energy would be if it was conserved seeing as the block initially has no potential kinetic energy and i don't know of any formula for energy transferred by an explosion

 

[gneill]

not sure. are you suggesting something like px=m1v1x'+m2v2x'+m3v3x'? wouldn't i then have two variables m3 and its angle? can the conservation of energy formula be used to substitute m3 into that equation or is energy not conserved in the explosion? I'm in grade twelve, so i don't really understand what the formula for energy would be if it was conserved seeing as the block initially has no potential kinetic energy and i don't know of any formula for energy transferred by an explosion

Momentum is always conserved. Even if something decomposes explosively and moving energy around the system, momentum will be conserved.

You are looking to find two unknown values: m3 and the angle of the third piece with respect to the others. If you have two unknowns, you need (at least) two simultaneous equations. Since momentum is conserved separately in the X and Y directions, you can immediately write two independent equations in your two unknowns. Stir, bake, solve.

Oh, and beware of quadrant issues with the angle; if your solution for the mass turns out negative, that just means that the mass is really traveling in the opposite direction along the velocity vector (flip the angle by 180 degrees). Always check your answer by plugging the results back into your original equations.

 

[strangeeyes]

so I've been stirring & baking but this wrong answer is the best i could find:
0=m1v1x'+m2v2x'+m3v3x'
=0.2(1.4)cos0+0.3(0.9)cos80+m3(1.80cos@)
m3=-0.1816cos@

then plugged that value into the y momentum equation:
0=m1v1y'+m2v2y'+m3v3y'
=0+0.3(0.9)sin80+(-0.1816cos@)1.8sin@
0.32688(cos@sin@)=0.265898
0.5sin(2@)=0.728665
@=46.77 degrees

then to find m3:
m3=-0.1816cos@
=-0.1257 kg

i was hopeful here but then i tried putting into the unsimplified equations and got:
=0.2(1.4)cos0+0.3(0.9)cos80+m3(1.80cos46.77)
=-0.2623 kg

-or-

using the y:
=0+0.3(0.9)sin80+m3(1.8)sin@
=-.20467 kg

arg!
i don't understand where i am going wrong?

 

[gneill]

so I've been stirring & baking but this wrong answer is the best i could find:
0=m1v1x'+m2v2x'+m3v3x'
=0.2(1.4)cos0+0.3(0.9)cos80+m3(1.80cos@)
m3=-0.1816cos@

Check that last line. Since the starting equation began with the product m3*cos(@), I think you should find that m3 and cos(@) are inversely proportionate. That is, you should have:

m3 = k/cos(@)

for some value of k that grinds out of all the other known stuff.

You might find it easier to derive the relationship symbolically first, plugging in the numerical values only at the end.

then plugged that value into the y momentum equation:
0=m1v1y'+m2v2y'+m3v3y'
=0+0.3(0.9)sin80+(-0.1816cos@)1.8sin@
0.32688(cos@sin@)=0.265898
0.5sin(2@)=0.728665
@=46.77 degrees

then to find m3:
m3=-0.1816cos@
=-0.1257 kg

i was hopeful here but then i tried putting into the unsimplified equations and got:
=0.2(1.4)cos0+0.3(0.9)cos80+m3(1.80cos46.77)
=-0.2623 kg

-or-

using the y:
=0+0.3(0.9)sin80+m3(1.8)sin@
=-.20467 kg

arg!
i don't understand where i am going wrong?

Your methodology looks sound. Fix up that little problem above and see where it leads.