11.2.3 A Strategy for Conservation of Momentum Problems

PROBLEM-SOLVING STRATEGY 11.1
Conservation of momentum

MODEL Clearly define the system.

If possible, choose a system that is isolated $(\vec{F}_{net} = \vec{0})$ or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved.
If it’s not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or conservation of energy.

VISUALIZE Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you’re trying to find.

SOLVE The mathematical representation is based on the law of conservation of momentum: $\vec{P}_{f} = \vec{P}_{i}$ . In component form, this is

\begin{matrix} (p_{f x})_{1} + (p_{f x})_{2} + (p_{f x})_{3} + \dots = (p_{i x})_{1} + (p_{i x})_{2} + (p_{i x})_{3} + \dots \\ (p_{f y})_{1} + (p_{f y})_{2} + (p_{f y})_{3} + \dots = (p_{i y})_{1} + (p_{i y})_{2} + (p_{i y})_{3} + \dots \end{matrix}

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 17

EXAMPLE 11.3 Rolling away

Bob sees a stationary cart 8.0 m in front of him. He decides to run to the cart as fast as he can, jump on, and roll down the street. Bob has a mass of 75 kg and the cart’s mass is 25 kg. If Bob accelerates at a steady $1.0 m / s^{2}$ , what is the cart’s speed just after Bob jumps on?

MODEL This is a two-part problem. First Bob accelerates across the ground. Then Bob lands on and sticks to the cart, a “collision” between Bob and the cart. The interaction forces between Bob and the cart (i.e., friction) act only over the fraction of a second it takes Bob’s feet to become stuck to the cart. Using the impulse approximation allows the system $Bob + cart$ to be treated as an isolated system during the brief interval of the “collision,” and thus the total momentum of $Bob + cart$ is conserved during this interaction. But the system $Bob + cart$ is not an isolated system for the entire problem because Bob’s initial acceleration has nothing to do with the cart.

VISUALIZE Our strategy is to divide the problem into an acceleration part, which we can analyze using kinematics, and a collision part, which we can analyze with momentum conservation. The pictorial representation of FIGURE 11.14 includes information about both parts. Notice that Bob’s velocity ${(v_{1 x})}_{B}$ at the end of his run is his “before” velocity for the collision.

The figure consists of three parts that illustrate a man jumping onto a cart. The figure also shows the acceleration and velocity associated with the action.

FIGURE 11.14 Full Alternative Text

SOLVE The first part of the mathematical representation is kinematics. We don’t know how long Bob accelerates, but we do know his acceleration and the distance. Thus

{(v_{1 x})}_{B}^{2} = {(v_{0 x})}_{B}^{2} + 2 a_{x}_{} Δ x = 2 a_{x} x_{1}

His velocity after accelerating for 8.0 m is

{(v_{1 x})}_{B} = \sqrt{2 a_{x} x_{1}} = 4.0 m / s

The second part of the problem, the collision, uses conservation of momentum: $P_{2 x} = P_{1 x}$ . Equation 11.26 is

m_{B} {(v_{2 x})}_{B} + m_{C} {(v_{2 x})}_{C} = m_{B} {(v_{1 x})}_{B} + m_{C} {(v_{1 x})}_{C} = m_{B} {(v_{1 x})}_{B}

where we’ve used ${(v_{1 x})}_{C} = 0 m / s$ because the cart starts at rest. In this problem, Bob and the cart move together at the end with a common velocity, so we can replace both ${(v_{2 x})}_{B}$ and ${(v_{2 x})}_{C}$ with simply $v_{2 x}$ . Solving for $v_{2 x}$ , we find

v_{2 x} = \frac{m_{B}}{m_{B} + m_{C}} {(v_{1 x})}_{B} = \frac{75 kg}{100 kg} \times 4 .0 m/s = 3.0 m / s

The cart’s speed is $3.0 m / s$ immediately after Bob jumps on.

Notice how easy this was! No forces, no acceleration constraints, no simultaneous equations. Why didn’t we think of this before? Conservation laws are indeed powerful, but they can answer only certain questions. Had we wanted to know how far Bob slid across the cart before sticking to it, how long the slide took, or what the cart’s acceleration was during the collision, we would not have been able to answer such questions on the basis of the conservation law. There is a price to pay for finding a simple connection between before and after, and that price is the loss of information about the details of the interaction. If we are satisfied with knowing only about before and after, then conservation laws are a simple and straightforward way to proceed. But many problems do require us to understand the interaction, and for these there is no avoiding Newton’s laws.