A common scuba tank is an aluminum cylinder that weighs 31.7 pounds empty (Figure \(\PageIndex{4}\)). When full of compressed air, the internal pressure is between 2500 and 3000 psi (pounds per square inch). Suppose such a tank, which had been sitting motionless, suddenly explodes into three pieces. The first piece, weighing 10 pounds, shoots off horizontally at 235 miles per hour; the second piece (7 pounds) shoots off at 172 miles per hour, also in the horizontal plane, but at a 19° angle to the first piece. What is the mass and initial velocity of the third piece? (Do all work, and express your final answer, in SI units.)

To use conservation of momentum, we need a closed system. If we define the system to be the scuba tank, this is not a closed system, since gravity is an external force. However, the problem asks for just the initial velocity of the third piece, so we can neglect the effect of gravity and consider the tank by itself as a closed system. Notice that, for this system, the initial momentum vector is zero.

We choose a coordinate system where all the motion happens in the xy-plane. We then write down the equations for conservation of momentum in each direction, thus obtaining the x- and y-components of the momentum of the third piece, from which we obtain its magnitude (via the Pythagorean theorem) and its direction. Finally, dividing this momentum by the mass of the third piece gives us the velocity.

###### Solution

First, let’s get all the conversions to SI units out of the way:

\[31.7\; lb \times \frac{1\; kg}{2.2\; lb} \rightarrow 14.4\; kg\]

\[10\; lb \rightarrow 4.5\; kg\]

\[235\; \frac{miles}{hour} \times \frac{1\; hour}{3600\; s} \times \frac{1609\; m}{mile} = 105\; m/s\]

\[7\; lb \rightarrow 3.2\; kg\]

\[172 \frac{mile}{hour} = 77\; m/s\]

\[m_{3} = 14.4\; kg - (4.5\; kg + 3.2\; kg) = 6.7\; kg \ldotp\]

Now apply conservation of momentum in each direction.

x-direction:

\[\begin{split} p_{f,x} & = p_{0,x} \\ p_{1,x} + p_{2,x} + p_{3,x} & = 0 \\ m_{1} v_{1,x} + m_{2} v_{2,x} + p_{3,x} & = 0 \\ p_{3,x} & = -m_{1} v_{1,x} - m_{2} v_{2,x} \end{split}\]

y-direction:

\[\begin{split} p_{f,y} & = p_{0,y} \\ p_{1,y} + p_{2,y} + p_{3,y} & = 0 \\ m_{1} v_{1,y} + m_{2} v_{2,y} + p_{3,y} & = 0 \\ p_{3,y} & = -m_{1} v_{1,y} - m_{2} v_{2,y} \end{split}\]

From our chosen coordinate system, we write the x-components as

\[\begin{split} p_{3,x} & = - m_{1} v_{1} - m_{2} v_{2} \cos \theta \\ & = - (4.5\; kg)(105\; m/s) - (3.2\; kg)(77\; m/s) \cos (19^{o}) \\ & = -705\; kg\; \cdotp m/s \ldotp \end{split}\]

For the y-direction, we have

\[\begin{split} p_{3,y} & = 0 - m_{2} v_{2} \sin \theta \\ & = - (3.2\; kg)(77\; m/s) \sin (19^{o}) \\ & = -80.2\; kg\; \cdotp m/s \ldotp \end{split}\]

This gives the magnitude of p_{3}:

\[\begin{split} p_{3} & = \sqrt{p_{3,x}^{2} + p_{3,y}^{2}} \\ & = \sqrt{(-705\; kg\; \cdotp m/s)^{2} + (-80.2\; kg\; \cdotp m/s)} \\ & = 710\; kg\; \cdotp m/s \ldotp \end{split}\]

The velocity of the third piece is therefore

\[v_{3} = \frac{p_{3}}{m_{3}} = \frac{710\; kg\; \cdotp m/s}{6.7\; kg} = 106\; m/s \ldotp\]

The direction of its velocity vector is the same as the direction of its momentum vector:

\[\phi = \tan^{-1} \left(\dfrac{p_{3,y}}{p_{3,x}}\right) = \tan^{-1} \left(\dfrac{80.2\; kg\; \cdotp m/s}{705\; kg\; \cdotp m/s}\right) = 6.49^{o} \ldotp\]

Because \(\phi\) is below the −x -axis, the actual angle is 186.49° from the +x-direction.

**Significance**

The enormous velocities here are typical; an exploding tank of any compressed gas can easily punch through the wall of a house and cause significant injury, or death. Fortunately, such explosions are extremely rare, on a percentage basis.