4.3 Conservation of Linear Momentum

Syllabus
2024
Topic
4.3
Level

Learning objectives

4.3A—Describe the behavior of a system using conservation of linear momentumDescribe the behavior of a system using conservation of linear momentum.• A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.- i. For a collection of objects, the velocity of a system’s center of mass can be calculated using the equation- ii. The velocity of a system’s center of mass is constant in the absence of a net external force.• The total momentum of a system is the sum of the momenta of the system’s constituent parts.• In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings.- i. The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton’s third law.- ii. A system may be selected so that the total momentum of that system is constant.- iii. If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system. Relevant equation:• Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions. BOUNDARY STATEMENT AP Physics 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the AP Physics 1 Exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 includes a full treatment of conservation of momentum in two dimensions for problems that include one unknown final velocity. | AP Physics 1: Algebra-Based Course and Exam Description4.3B—Describe how the selection of a system determines whether the momentum of that system changesDescribe how the selection of a system determines whether the momentum of that system changes.• Momentum is conserved in all interactions.• If the net external force on the selected system is zero, the total momentum of the system is constant.• If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment. 84 Linear Momentum UNIT 4 TOPIC 4.4 Elastic and Inelastic Collisions

Conserving momentum in a system

Represent the system

Total system momentum is the vector sum of its parts. Dividing that total by total mass gives the center-of-mass velocity.

psys=ipi=imivivcm=ipiimi\begin{aligned}\vec p_{\text{sys}}&=\sum_i\vec p_i=\sum_i m_i\vec v_i\\[4pt]\vec v_{\text{cm}}&=\frac{\sum_i\vec p_i}{\sum_i m_i}\end{aligned}

Balance internal changes

If the net external force is zero, vcm\vec v_{\text{cm}} and psys\vec p_{\text{sys}} stay constant. Internal forces come in Newton's-third-law pairs, so their impulses are equal and opposite: one part's momentum gain is balanced by another part's loss.

Apply conservation

One-dimensional explosion: a 3.0kg3.0\,\text{kg} system starts at rest and separates into 1.0kg1.0\,\text{kg} and 2.0kg2.0\,\text{kg} parts. Choose right as positive. If the lighter part leaves at +6.0m/s+6.0\,\text{m/s}, then

0=(1.0kg)(+6.0m/s)+(2.0kg)v20=(1.0\,\text{kg})(+6.0\,\text{m/s})+(2.0\,\text{kg})v_2,

so v2=3.0m/sv_2=-3.0\,\text{m/s}. The total momentum remains zero, although each part now has nonzero momentum.

Keep the system and course boundary

Conservation applies to the system total, not separately to every object. In two dimensions, conserve momentum by components. AP Physics 1 expects semiquantitative reasoning and correct equation setup, not solving a full simultaneous-equation system for an unknown final velocity.

System choice controls momentum change

Define the boundary

Choose the system boundary first. Forces between included objects are internal; forces from outside are external. Only net external impulse changes the selected system's momentum.

Δpsys=JextFext,net=0Δpsys=0\begin{aligned}\Delta\vec p_{\text{sys}}&=\vec J_{\text{ext}}\\[4pt]\vec F_{\text{ext,net}}=\vec 0&\Rightarrow\Delta\vec p_{\text{sys}}=\vec 0\end{aligned}

Compare two selections

Selected system during a short cart collision Force between carts Momentum conclusion
Both carts Internal If other external impulse is negligible, total momentum of both carts is constant
Cart A only Force from cart B is external Cart A's momentum changes by the impulse from cart B

Track momentum transfer

Momentum is conserved in every interaction when the system and environment are considered together. If a chosen subsystem gains momentum, an equal-and-opposite amount is transferred elsewhere. A nonzero net external force therefore signals momentum crossing the selected boundary—not momentum being created or destroyed.

Test isolation

A collision does not automatically make the selected system isolated. Check the external impulse over the interaction interval. A force may be internal for a two-object system but external for either object considered alone.