Grade 12 Physics Bomb explosion

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The discussion revolves around calculating the velocity and direction of Piece 1 from a bomb explosion, using momentum conservation principles. The initial momentum is zero since the bomb was at rest, leading to equations for both horizontal and vertical momentum. The horizontal equation simplifies to 0 = 2kg (V1F) cos(x) - 60 kg/m/s, while the vertical equation simplifies to 0 = 2kg (V1F) sin(x) - 30 kg/m/s. The ratio of the equations leads to the conclusion that tan(x) = 1/2, indicating the angle of Piece 1's trajectory. Participants clarify that V1F cancels out in the calculations, allowing for the determination of direction without needing its specific value.
Shin-
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Question:
A bomb explodes and 3 pieces are flying in different directions.
Piece 1 : weighs 2kg and is traveling at the North East direction
Piece 2 : weighs 1kg and is traveling in the south direction ( exactly downards )
Piece 3 : weighs 3 kg and is traveling in the west direction ( exactly to the left )
Find the velocity and direction that Piece 1 is traveling at.

I know to use the momentum equation twice , one with the horizontal direction and one on the vertical direction.

Horizontal:
M1V1(initial)+M2V2(initial)+M3V3(initial) = M1V1(Final)+M2V2(Final)+M3V3(final)
Since all the initial velocities are 0 ( because bomb was at rest before explosion , the entire left side is equal to zero)

0 = 2kg ( V1F ) cosx + 1kg ( 30 m/s) cos 90 + 3kg ( 20 m/s ) Cos 180
= 2kg (V1F) cosx - 60 Kg/m/s
30 m/s = V1F cosx
Above , i canceled M2V2(FINAL) because i set that as the reference direction and angle , so it counts as 0. and i did minus 60 kg/m/s because its to the negative direction.

Vertical :
M1V1(initial)+M2V2(initial)+M3V3(initial) = M1V1(Final)+M2V2(Final)+M3V3(final)
Since all the initial velocities are 0 ( because bomb was at rest before explosion , the entire left side is equal to zero)

0 = 2kg ( V1F ) sinx + 1kg ( 30 m/s) sin 90 + 3kg ( 20 m/s ) sin 180

= 2kg ( V1F) sinx - 30 Kg/m/s
15 m/s = V1F sinx

Then divided the final equations to get

Sinx V1F 15 m/s
--------- = -------
Cosx V1F 30 m/s



Tanx V1F = .5 m/s

now i have no idea wat to do to find the direction and speed :( can anyone help please?
 
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You neglected to cancel out V1F in your final equations.
 
No , i just rewrote it as Tanx V1f

Sinx/Cosx = Tanx ( trig identity )
 
any ideas?
 
Hi Shin,

I think you misunderstood Phlogistonian's post.

You had:

Sinx V1F 15 m/s
--------- = -------
Cosx V1F 30 m/s

this simplifies to

tan x = (1/2)

The V1F is not inside the trig function and so will cancel since it is in both numerator and denominator.
 
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