9.5 Collisions in multiple dimensions (Page 3/4)

University physics volume 1 Page 3 / 4

A drawing of a scuba tank exploding, and the resulting three pieces of different sizes. — A scuba tank explodes into three pieces.

Strategy

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 the 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:

\begin{matrix} 31.7 lb \times \frac{1 kg}{2.2 lb} \to 14.4 kg \\ 10 lb \to 4.5 kg \\ 235 \frac{miles}{hour} \times \frac{1 hour}{3600 s} \times \frac{1609 m}{mile} = 105 \frac{m}{s} \\ 7 lb \to 3.2 kg \\ 172 \frac{mile}{hour} = 77 \frac{m}{s} \\ m_{3} = 14.4 kg - (4.5 kg + 3.2 kg) = 6.7 kg. \end{matrix}

Now apply conservation of momentum in each direction.

The three pieces of the scuba tank are shown on an x y coordinate system. The medium size piece is on the positive x axis and has momentum p 1 in the plus x direction. The smallest piece is at an angle theta above the positive x axis and has momentum p 2. The largest piece is at an angle phi below the negative x axis and has momentum p 3.

x -direction:

\begin{matrix} 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{matrix}

y -direction:

\begin{matrix} 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{matrix}

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

\begin{matrix} p_{3, x} & = − m_{1} v_{1} - m_{2} v_{2} cos θ \\ = − (14.5 kg) (105 \frac{m}{s}) - (4.5 kg) (77 \frac{m}{s}) cos (19 °) \\ = −1850 \frac{kg \cdot m}{s} . \end{matrix}

For the y -direction, we have

\begin{matrix} p_{3 y} & = 0 - m_{2} v_{2} sin θ \\ = − (4.5 kg) (77 \frac{m}{s}) sin (19 °) \\ = −113 \frac{kg \cdot m}{s} . \end{matrix}

This gives the magnitude of $p_{3}$ :

\begin{matrix} p_{3} & = \sqrt{p_{3, x}^{2} + p_{3, y}^{2}} \\ = \sqrt{{(−1850 \frac{kg \cdot m}{s})}^{2} + (−113 \frac{kg \cdot m}{s})} \\ = 1854 \frac{kg \cdot m}{s} . \end{matrix}

The velocity of the third piece is therefore

v_{3} = \frac{p_{3}}{m_{3}} = \frac{1854 \frac{kg \cdot m}{s}}{6.7 kg} = 277 \frac{m}{s} .

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

ϕ = {tan}^{−1} (\frac{p_{3, y}}{p_{3, x}}) = {tan}^{−1} (\frac{113 \frac{kg \cdot m}{s}}{1850 \frac{kg \cdot m}{s}}) = 3.5 ° .

Because $ϕ$ is below the $− x$ -axis, the actual angle is $183.5 °$ 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.

Check Your Understanding Notice that the mass of the air in the tank was neglected in the analysis and solution. How would the solution method changed if the air was included? How large a difference do you think it would make in the final answer?

The volume of a scuba tank is about 11 L. Assuming air is an ideal gas, the number of gas molecules in the tank is
$\begin{matrix} P V & = & N R T \\ N & = & \frac{P V}{R T} = \frac{(2500 psi) (0.011 m^{3})}{(8.31 J/mol \cdot K) (300 K)} (\frac{6894.8 Pa}{1 psi}) \\ = & 7.59 \times 10^{1} mol \end{matrix}$
The average molecular mass of air is 29 g/mol, so the mass of air contained in the tank is about 2.2 kg. This is about 10 times less than the mass of the tank, so it is safe to neglect it. Also, the initial force of the air pressure is roughly proportional to the surface area of each piece, which is in turn proportional to the mass of each piece (assuming uniform thickness). Thus, the initial acceleration of each piece would change very little if we explicitly consider the air.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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