Help with another momentum question

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An explosion breaks a rock into three pieces, with two pieces moving at right angles: a 1.0 kg piece at 12 m/s and a 2.0 kg piece at 8 m/s. To find the direction of the third piece, the conservation of momentum principle is applied, revealing it must move in the opposite direction to balance the total momentum. The total momentum of the first two pieces is 28 kg*m/s, so the third piece must have a momentum of 40 kg*m/s to maintain equilibrium. Using the momentum formula p = mv, the mass of the third piece is calculated to be 1 kg. The discussion emphasizes the importance of vector diagrams and momentum conservation in solving such problems.
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An explosion blows a rock into 3 parts. Two peices go off at right angles to each other, a 1.0 kg piece at 12 m/s and a 2.0 kg piece at 8 m/s. The third piece flies off at 40 m/s.

a) Draw a vector diagram to show the direction in which the third piece goes.

I don't know how I can determine which direction it goes

b) What is its mass?

I don't know how I can get this if I am unable to determine the total momentum.
 
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Set up a coordinate system. Assume the 1 kg piece goes off in the positive x-direction, at 12 m/s, and that the 2 kg piece goes of in the positive y-direction (so the motions are at right angles) at 8 m/s. It should be east to calculate the total momentum of the two pieces. Since the mass was originally at rest and there was no external force (the explosion is an internal force) the total momentum of all three pieces must also be 0: the momentum of the last piece must be in exactly the opposite direction to the total momentum of the first two and of exactly the same length. Divide that length by the speed, 40 m/s, to find the mass of the third piece (sounds like it's going to be small).
 


a) To determine the direction of the third piece, we can use the principle of conservation of momentum. This means that the total momentum before the explosion is equal to the total momentum after the explosion. We know that the momentum of the first two pieces is 1.0 kg * 12 m/s = 12 kg*m/s and 2.0 kg * 8 m/s = 16 kg*m/s respectively. So, the total momentum before the explosion is 12 kg*m/s + 16 kg*m/s = 28 kg*m/s. After the explosion, the third piece has a momentum of 40 kg*m/s. This means that the total momentum after the explosion is also 40 kg*m/s.

To draw the vector diagram, we can represent the momentum of the first two pieces as vectors in different directions. Since the first two pieces are moving at right angles to each other, we can draw one vector horizontally and the other vertically. Then, we can draw the third vector starting from the end of the first two vectors and pointing in the direction of the third piece's momentum (40 kg*m/s).

b) To determine the mass of the third piece, we can use the equation for momentum: p = mv, where p is momentum, m is mass, and v is velocity. We know that the momentum of the third piece is 40 kg*m/s and the velocity is 40 m/s. Therefore, we can rearrange the equation to solve for mass: m = p/v = 40 kg*m/s / 40 m/s = 1 kg. So, the mass of the third piece is 1 kg.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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