What is the Direction of the Third Piece After the Bomb Explosion?

[MrRottenTreats]

Advanced Momentum Question -- need help

The Question

A bomb initially at rest on a smooth, horizontal surface explodes into 3 pieces. Two pieces fly across the surface at a 60 degree angle to each other: a 2.0 kg at 20 m/s and a 3.0 kg piece at 12 m/s. The 3rd pieces flys across the surface as well with a vector velocity of 30 m/s.

--> make a hypothetical prediction based on the direction of the 3rd piece.

assuming 100% conservation of momentum in both horizontal and veritcal directions, solve for the direction of the 3rd mass.

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I made my cartesian plane, had the 2 bomb going [60 east of north] and [60 east of south] and the third somewhere in the NW direction.

from here i am kind of lost, i was thinking to break each into there compentents?

 

[nrqed]

The Question

A bomb initially at rest on a smooth, horizontal surface explodes into 3 pieces. Two pieces fly across the surface at a 60 degree angle to each other: a 2.0 kg at 20 m/s and a 3.0 kg piece at 12 m/s. The 3rd pieces flys across the surface as well with a vector velocity of 30 m/s.

--> make a hypothetical prediction based on the direction of the 3rd piece.

assuming 100% conservation of momentum in both horizontal and veritcal directions, solve for the direction of the 3rd mass.

-----------------------------------------------------------------------
I made my cartesian plane, had the 2 bomb going [60 east of north] and [60 east of south] and the third somewhere in the NW direction.

from here i am kind of lost, i was thinking to break each into there compentents?

Yes, the only way to proceed is to break the momenta into x and y components. Then impose that the sum of the x components of all 3 pieces is zero and same for the sum of the y components.
Then isolate the x and y components of the piece moving at 30 m/s. You will get something of the form

$$m v_x = A$$
$$m v_y = B$$
where A and B are some numbers. Then square both expressions and add them up. Using the fact that $v_x^2 + v_y^2 = (30 m/s)^2 = 900 m^2/s^2$ you will find the mass. Then go back to one of the two equations above and find theta (using, say $v_x = 30 m/s cos (\theta)$).

Pat

 

[MrRottenTreats]

okay thank you very much i have this solved , i broke the horizontal and vertical up, then isolated for the m3 , then equated them to get rid of the m3.

i got:

-38 / 30cos(Theta) = -65.8 / 30sin(theta)

cross multiplied and got 30 degress as my angle. :D

 

[nrqed]

okay thank you very much i have this solved , i broke the horizontal and vertical up, then isolated for the m3 , then equated them to get rid of the m3.

i got:

-38 / 30cos(Theta) = -65.8 / 30sin(theta)

cross multiplied and got 30 degress as my angle. :D

Ok..
I am not sure how you defined your theta. Using v_x = 30 cos(theta) and v_y = 30 sin(theta) for the third mass would have led to a theta larger than 90 degrees, obviously. But You may have defined your theta to be North of West in which case 30 degrees seems plausible.

I haven't checked your number but that may be right.
Notice that you should *definitely* double check your answers for theta and the mass by plugging them back in your initial equations for momentum conservation along x and y and see that the total momentum is zero.

Glad I could help.

Regards

Pat