11.4 Explosions

An explosion, where the particles of the system move apart from each other after a brief, intense interaction, is the opposite of a collision. The explosive forces, which could be from an expanding spring or from expanding hot gases, are internal forces. If the system is isolated, its total momentum during the explosion will be conserved. One of the most well-known consequences of an explosion is that a larger object recoils after launching a small but high-speed projectile.

EXAMPLE 11.6 Recoil from a javelin throw

The women’s javelin used in the Olympics has a mass of 600 g. An elite athlete can throw a javelin with a speed in excess of $25 m/s$ . Suppose a 55 kg athlete, standing on a 2.0 kg skateboard, throws a javelin horizontally at $25 m/s$ . What is her recoil speed as she rolls backward after the throw?

MODEL Define the system to be $athlete + skateboard + javelin$ . The throwing force, an action/reaction pair of forces between the athlete and the javelin, is an internal force. There’s no net vertical force. Rolling friction will eventually stop the skateboard, but—using the impulse approximation—it’s reasonable to assume that the friction force is much smaller than the throwing force and that it will not influence her motion during the fraction of a second it takes to throw the javelin. Thus this is an isolated system, and the law of conservation of momentum applies.

VISUALIZE FIGURE 11.24 shows a before-and-after pictorial representation. We’ve chosen to let the javelin travel to the right, which means that the athlete and the skateboard will have a negative recoil velocity.

The figure illustrates the before-and-after pictorial representations of a Javelin throw by a person standing on a skateboard.

FIGURE 11.24 Full Alternative Text

SOLVE The x-component of the total momentum is $P_{x} = (p_{x})_{A} + (p_{x})_{S} + (p_{x})_{J}$ . Everything is at rest before the throw, so the initial total momentum is zero. After the throw, the athlete and the skateboard move together with a common final velocity $(v_{f x})_{A}$ . The law of conservation of momentum is thus

P_{f x} = m_{J} (v_{f x})_{J} + (m_{A} + m_{S}) (v_{f x})_{A} = P_{i x} = 0

Solving for the recoil velocity, we find

(v_{f x})_{A} = - \frac{m_{J}}{m_{A} + m_{S}} (v_{f x})_{J} = - \frac{0.60 kg}{57 kg} \times 25 m/s = - 0.26 m / s

The minus sign indicates that the recoil is to the left. The recoil speed is $0.26 m / s$ .

We would not know where to begin to solve a problem such as this using Newton’s laws. But Example 11.6 is a simple problem when approached from the before-and-after perspective of a conservation law. The selection of $athlete + skateboard + javelin$ as “the system” was the critical step. For momentum conservation to be a useful principle, we had to select a system in which the complicated forces between the athlete’s hand and the javelin and between her feet and the skateboard were all internal forces. The athlete by herself is not an isolated system, so her momentum is not conserved.

EXAMPLE 11.7 Radioactivity

A $^{238} U$ uranium nucleus is radioactive. It spontaneously disintegrates into a small fragment that is ejected with a measured speed of $1.50 \times 10^{7} m / s$ and a “daughter nucleus” that recoils with a measured speed of $2.56 \times 10^{5} m / s$ . What are the atomic masses of the ejected fragment and the daughter nucleus?

MODEL The notation $^{238} U$ indicates the isotope of uranium with an atomic mass of 238 u, where u is the abbreviation for the atomic mass unit. The nucleus contains 92 protons (uranium is atomic number 92) and 146 neutrons. The disintegration of a nucleus is, in essence, an explosion. Only internal nuclear forces are involved, so the total momentum is conserved in the decay.

VISUALIZE FIGURE 11.25 shows the pictorial representation. The mass of the daughter nucleus is $m_{1}$ and that of the ejected fragment is $m_{2}$ . Notice that we converted the speed information to velocity information, giving $(v_{f x})_{1}$ and $(v_{f x})_{2}$ opposite signs.

The figure illustrates the before-and-after pictorial representations of the decay of the Uranium nucleus.

FIGURE 11.25 Full Alternative Text

SOLVE The nucleus was initially at rest, hence the total momentum is zero. The momentum after the decay is still zero if the two pieces fly apart in opposite directions with momenta equal in magnitude but opposite in sign. That is,

P_{f x} = m_{1} (v_{f x})_{1} + m_{2} (v_{f x})_{2} = P_{i x} = 0

Although we know both final velocities, this is not enough information to find the two unknown masses. However, we also have another conservation law, conservation of mass, that requires

m_{1} + m_{2} = 238 u

Combining these two conservation laws gives

m_{1} (v_{f x})_{1} + (238 u - m_{1}) (v_{f x})_{2} = 0

The mass of the daughter nucleus is

\begin{array}{l} m_{1} & = \frac{{(v_{f x})}_{2}}{{(v_{f x})}_{2} - {(v_{f x})}_{1}} \times 238 u \\ = \frac{1.50 \times 10^{7} m/s}{(1.50 \times 10^{7} - (- 2.56 \times 10^{5})) m/s} \times 238 u = 234 u \end{array}

With $m_{1}$ known, the mass of the ejected fragment is $m_{2} =$ $238 - m_{1} = 4 u$ .

REVIEW All we learn from a momentum analysis is the masses. Chemical analysis shows that the daughter nucleus is the element thorium, atomic number 90, with two fewer protons than uranium. The ejected fragment carried away two protons as part of its mass of 4 u, so it must be a particle with two protons and two neutrons. This is the nucleus of a helium atom, $^{4} He$ , which in nuclear physics is called an alpha particle $α$ . Thus the radioactive decay of $^{238} U$ can be written as $^{238} U \to^{234} Th + α$ .

Similar reasoning explains how a rocket or jet aircraft accelerates. FIGURE 11.26 shows a rocket with a parcel of fuel on board. Burning converts the fuel to hot gases that are expelled from the rocket motor. If we choose $rocket + gases$ to be the system, the burning and expulsion are both internal forces. There are no other forces, so the total momentum of the $rocket + gases$ system must be conserved. The rocket gains forward velocity and momentum as the exhaust gases are shot out the back, but the total momentum of the system remains zero.

The figure is the before-and-after pictorial representations of rocket propulsion.

FIGURE 11.26 Full Alternative Text

Section 11.6 looks at rocket propulsion in more detail, but even without the details you should be able to understand that jet and rocket propulsion is a consequence of momentum conservation.

11.4 Explosions

EXAMPLE 11.6 Recoil from a javelin throw

FIGURE 11.24 Before-and-after pictorial representation of a javelin throw.

EXAMPLE 11.7 Radioactivity

FIGURE 11.25 Before-and-after pictorial representation of the decay of a $^{238} U$ nucleus.

FIGURE 11.26 Rocket propulsion is an example of conservation of momentum.