Conservation of Momentum

Section 7.6 Conservation of Momentum

We found above that the change in the total momentum of any system, whether it contains a single particle or several particles, is caused only by the external forces on the system, and the rate of that change at any instant is exactly equal to the net external force on all parts of the system.

\begin{equation*} \frac{d\vec P_\text{system}}{dt} = \vec F_\text{net}^\text{ext}. \end{equation*}

Therefore, if the net external force on a system is zero, then the total momentum cannot change. That is, the magnitude and direction of the momentum of a system that does not interact with anything else will remain fixed in time, i.e. conserved. We will call such systems isolated systems. This result is also called the principle of conservation of momentum .

By a system we usually mean any physical body of interest. A choice of a physical body for study leads to an artificial division of the entire universe into a system and the rest, called the surroundings or the external world, as illustrated in Figure 7.69.

Figure 7.69. Division of the universe into system and surroundings takes place when we focus on a part of the universe for study. The objects of interest make up the system and everything else is the surroundings or external to the system. We usually draw a dashed boundary line to indicate the objects that are within the system from those which are external to the system.

We study the overall motion of the system by examining the forces on the system by the external world. The surroundings is external to the system and influences the changes in momentum of the system. Similarly, the system is external to the surroundings and therefore, influences the changes in the momentum of the surroundings. The system and surroundings will obey their own equations of motion which can be written as:

\begin{align*} \amp \textrm{System:} \ \ \frac{d\vec P_{\textrm{system}}}{dt} = \vec F_{\textrm{net}}^{\textrm{on system}}\\ \amp \textrm{Surroundings:} \ \ \frac{d\vec P_{\textrm{surroundings}}}{dt} = \vec F_{\textrm{net}}^{\textrm{on surroundings}} \end{align*}

Now, when we add the two equations we obtain the equations of motion of the universe. Since the forces from the surroundings on the system are equal in magnitude but opposite in directions to the forces from the system on the surroundings, the net external force on the universe is zero.

\begin{equation*} \frac{d\vec P_{\textrm{Univese}}}{dt} = 0. \end{equation*}

This makes sense, since the larger system, which we have called the universe above, is an isolated system. Since there is nothing external to the larger system, the total momentum of the system and surroundings together would not change with time.

Principle of conservation of momentum:

The total momentum of a system and the rest of the universe does not change in time. We will call the rest of the universe ``the surroundings’’ or ``the environment’’. The force on the system by the objects in the environment changes the momentum of the system and the force on the objects of the environement by the system changes the momentum of the environment. But, since the forces are equal in magnitude and opposite in directions, the impulse on the system imparted by the environment in any interval must be equal in magnitude and opposite in direction to the impulse imparted to the environment by the system. Consequently, any change in the momentum of the system is accompanied by a change in the momentum of the environment of an equal magnitude and of opposite direction.

\begin{equation*} \Delta \vec P_{\textrm{System}} = -\Delta \vec P_{\textrm{Environment}}. \end{equation*}

Since, momentum is a vector quantity, the conservation of momentum applies independently for each direction in space. Decomposing these vectors in $x$, $y$ and $z$-components, we find that changes in different components of the momenta of the system are accompanied by equal changes of opposite sign in the changes in corresponding components of the momenta of the surroundings.

\begin{align*} \amp \Delta P_x^{\textrm{System}} = -\Delta P_x^{\textrm{Environment}}\\ \amp \Delta P_y^{\textrm{System}} = -\Delta P_y^{\textrm{Environment}}\\ \amp \Delta P_z^{\textrm{System}} = -\Delta P_z^{\textrm{Environment}} \end{align*}

When you bring the changes in momenta of the system and the environment on one side of the equation, we find that the net change in any component of the total momentum of the system and the environment together is always zero, i.e the total momentum is conserved component-by-component. The decomposition in three Cartesian components makes it clear that the momentum component in any direction is independently conserved from the momentum component in any other direction that is perpendicular to it.

For instance, when a football is kicked at an angle to the ground, the momentum in the horizontal direction is conserved independently of whatever happens in the momentum component in the vertical direction. Actually, although the horizontal component of the momentum of a football in free flight does not change, the absolute value of the vertical component changes with time, decreasing in the upward part of the flight, becoming momentarily zero at the top of the flight, and increasing afterwards until the ball hits the ground. The change in the vertical component of the momentum happens because of the external force on the ball in that direction, which is the gravitational force of the Earth on the ball.

Let us look at an example of a system of two bodies of masses $m_1 $ and $m_2 \text{.}$ Let their initial velocties be $\vec v_1 $ and $\vec v_2 \text{.}$ If net impulse by external forces on these two bodies is zero, then their net momentum will not change. However, just because net momentum will not change, their individual momenta may change, e.g., as a result of collision between them or some other forces between them. Let their velocities after some time be $\vec v_1^{\,\prime} $ and $\vec v_2^{\,\prime} \text{.}$ Hence, in the absence of external impulse, we will find that

\begin{equation*} m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec v_1^{\,\prime} + m_2 \vec v_2^{\,\prime}, \end{equation*}

or equivalently,

\begin{equation*} \vec p_1 + \vec p_2 = \vec p_1^{\,\prime} + \vec p_2^{\,\prime}, \end{equation*}

That is, in the absence of any external impulses, the momenta of various parts of a system can get redistributed among the parts but can never be lost. Note that this is vector equation, and therefore, in most situations, we would analyze it in the component form.

The principle of conservation of momentum provides a powerful tool for solving problems by giving us a conservation equation for the direction in which there are no external forces. It is easy to peel off that component of the motion that will have conserved momentum and depending upon the question to be investigated, that may be enough for the problem.

Example 7.70. Momentum of Ice Skaters Pushing On Each Other.

Alan and Betsy are ice skating on a smooth surface. Alan’s mass is $60\text{ kg} $ and that of Betsy’s is $50\text{ kg} \text{.}$ They start from rest by pushing on each other with an average force of magnitude $30\text{ N}$ and directed horizontally for $1.5\text{ sec}\text{.}$ Assume the skating surface to be horizontal and frictionless.

Find the speeds of the two skaters at an instant after $t = 1.5\text{ sec} $ when they are no longer in contact.

Answer.

Betsy’s speed is $0.9\text{ m/s} $ and Alan’s speed is $0.75\text{ m/s} \text{.}$

Solution.

To deal with the vector nature of momentum, let us introduce a coordinate system. Since all motion is along the Alan-Betsy line, we just need the $x $ axis. Let positive $x $ direction be from Alan-to-Betsy direction.

Since the net force during the time of interest is that of the push on Alan or Betsy by the other, the momentum change of either of them can be obtained by using the momentum/impulse form of Newton’s second law.

Betsy’s velocity using impulse/momentum. Let’s work out the momentum change of Betsy, which we denote with letter $B\text{,}$ by computing the impulse on her.

\begin{equation} \Delta p_{B,x} = J_x^{\text{on B}} = 30\text{ N} \times 1.5\text{ sec}= 45\text{ N.s} = 45\text{ kg.m/s}.\tag{7.71} \end{equation}

Since initially momentum was zero, the momentum at $1.5\text{ sec}$ would be

\begin{equation*} p_{B,x} = p_{B,i,x} + \Delta p_{B,x} = 45\text{ kg.m/s}, \end{equation*}

which upon dividing by Betsy’s mass gives $x $ component of her velocity

\begin{equation*} v_{B,x} = \dfrac{p_{B,x}}{m_B} = 0.9\text{ m/s}. \end{equation*}

Since this is the only component of velocity, Betsy’s speed is $0.9\text{ m/s} \text{.}$

Using conservation of momentum to get change in Alan’s momentum. Next, we note that if we think of Alan and Betsy as a single system, the net external force horizontally on this system is actually zero along horizontal direction. Hence total $x$-momentum of Alan$+$Betsy system remains unchanged, although each clearly gains momentum due to the push on each other. But, their momenta are oppositely directed and hence cancel out. Therefore, the change in Alan’s momentum can be immediately obtained from Eq. (7.71). Let us attach letter $A $ for Alan.

\begin{equation*} \Delta p_{A,x} = - \Delta p_{B,x} = - 45\text{ kg.m/s}. \end{equation*}

Note the negative sign - this is due to Alan’s motion being directed towards negative $x $ axis. Now, we divide by Alan’s mass to obtain Alan’s $x $ velocity component.

\begin{equation*} v_{A,x} = \dfrac{p_{A,x}}{m_A} = - 0.75\text{ m/s}. \end{equation*}

Since this is the only component of velocity, Alan’s speed is $0.75\text{ m/s} \text{.}$

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