AP®︎ Physics 1: Topic 4.3 Practice Test

Prepare for your quiz, test, or the AP exam with focused practice questions on Topic 4.3 of AP Physics 1 – Conservation of Linear Momentum.


Questions List

Topic 4.3

Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16

Question 1 Easy

Figure
On a level track, a 1.0 kg1.0\,\mathrm{kg} cart moves at 4.0 m/s4.0\,\mathrm{m/s} to the right toward a 3.0 kg3.0\,\mathrm{kg} cart moving at 2.0 m/s2.0\,\mathrm{m/s} to the left, as shown.

The total momentum of the two-cart system is most nearly
A2.0 kg⋅m/s2.0\,\mathrm{kg\cdot m/s} to the left
B10 kg⋅m/s10\,\mathrm{kg\cdot m/s} to the left
C2.0 kg⋅m/s2.0\,\mathrm{kg\cdot m/s} to the right
D10 kg⋅m/s10\,\mathrm{kg\cdot m/s} to the right

What You’re Being Tested On:

Explore the learning objectives taken directly from the College Board’s AP® Physics 1 Curriculum. Ensure you’re prepared for the exact topics covered on the AP® exam, in-class tests, and quizzes, and gain confidence in your mastery of the material.

Topic 4.3: Conservation of Linear Momentum

Boundary Statements official exam scope notes

  • 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.

Learning Objective: 4.3.A

Describe the behavior of a system using conservation of linear momentum.

Essential Knowledge: 4.3.A.1

A collection of objects with individual momenta can be described as one system with one center-of-mass velocity.

Essential Knowledge: 4.3.A.1.i

For a collection of objects, the velocity of a system’s center of mass can be calculated using the equation v⃗cm=∑p⃗i∑mi=∑miv⃗i∑mi\vec{v}_{\mathrm{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}.

Essential Knowledge: 4.3.A.1.ii

The velocity of a system’s center of mass is constant in the absence of a net external force.

Essential Knowledge: 4.3.A.2

The total momentum of a system is the sum of the momenta of the system’s constituent parts.

Essential Knowledge: 4.3.A.3

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.

Essential Knowledge: 4.3.A.3.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.

Essential Knowledge: 4.3.A.3.ii

A system may be selected so that the total momentum of that system is constant.

Essential Knowledge: 4.3.A.3.iii

If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system. Relevant equation: J⃗=Δp⃗\vec{J} = \Delta \vec{p}

Essential Knowledge: 4.3.A.4

Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions.

Learning Objective: 4.3.B

Describe how the selection of a system determines whether the momentum of that system changes.

Essential Knowledge: 4.3.B.1

Momentum is conserved in all interactions.

Essential Knowledge: 4.3.B.2

If the net external force on the selected system is zero, the total momentum of the system is constant.

Essential Knowledge: 4.3.B.3

If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment.