Home / AP Physics 2: 2.8 Thermodynamics and Elastic Collisions: Conservation of Momentum – Exam Style questions with Answer- MCQ

AP Physics 2: 2.8 Thermodynamics and Elastic Collisions: Conservation of Momentum – Exam Style questions with Answer- MCQ

Question

A sample of gas X initially has a higher temperature than a sample of gas Y. The molecules of gas X have more mass than the molecules of gas Y. The samples are mixed together in an insulated container. Which of the following claims best describes the collisions between gas X and gas Y molecules?

A The gas X molecules will always lose momentum and kinetic energy, and the gas Y molecules will always gain momentum and kinetic energy.

B The gas X molecules will always gain momentum and kinetic energy, and the gas Y molecules will always lose momentum and kinetic energy.

C Neither gas molecules will have a change in momentum, but the gas X molecules will always lose kinetic energy, and the gas Y molecules will always gain kinetic energy.

D Molecules of both gases will sometimes gain and sometimes lose momentum, and sometimes gain and sometimes lose kinetic energy.

▶️Answer/Explanation

Ans:D

While the average speed of the gas X molecules is greater than the average speed of the gas Y molecules, the molecules of each gas have a variety of speeds. So the gas X molecule could have a greater speed or a smaller speed than the gas Y molecule. What happens in a collision depends on the relative speeds of the molecules.

Question

                    

An ideal gas is in a sealed cylinder with a moveable piston, as shown. In collisions between any two molecules of the gas, the linear momentum and the kinetic energy of the two-molecule system are the same before and after the collision. The gas is then heated until it reaches a certain temperature, and the heat source is removed. Which of the following claims correctly describes the linear momentum and kinetic energy in a collision between two gas molecules of the hotter gas?

A Both the linear momentum and the kinetic energy have the same values before and after the collision.

B The linear momentum has the same value before and after the collision, but the kinetic energy has a higher value after the collision than before the collision.

C The linear momentum has a higher value after the collision than before the collision, but the kinetic energy has the same value before and after the collision.

D Both the linear momentum and the kinetic energy have higher values after the collision than before the collision.

▶️Answer/Explanation

Ans:A

The fact that the gas is now hotter does not affect the application of conservation laws to the individual collisions.

Question

A rigid container holds an ideal gas. A gas molecule of mass m moving at speed v collides with a container wall. The molecule’s velocity is perpendicular to the wall before the collision. What type of collision occurs when the molecule strikes the wall, and what is the change in momentum of the molecule?

▶️Answer/Explanation

Ans:B

 Energy and momentum must be conserved in these types of collisions or the temperature (internal energy) of the gas will change; this means that the molecules have an elastic collision with the walls of the container. Since the molecules strike the stationary walls with a momentum of mv, they will rebound back at a speed of mv since momentum is conserved. As momentum is a vector quantity, this means that the change in momentum is 2mv.

Question

Oxygen molecules are in a closed container. One such molecule is traveling at 400m/s when it collides perpendicularly with the wall of the container. The mass of an oxygen molecule is \(5.3×10^{−26}\)  kg. The molecule can be treated as if it were a spherical particle, and the gas can be treated as ideal. What are the magnitudes of the change in the momentum and the change in the kinetic energy of the oxygen molecule due to the collision?

▶️Answer/Explanation

Ans:D

 The collision with the wall is elastic according to the model of ideal gases. Therefore, there is no change in kinetic energy due to the collision. The molecule bounces off the wall and reverses direction: \(Δp=mv_f−mv_o=(5.3×10^{−26} kg)\times (−400−400)m/s=4.2×10^{−23}kg.m/s\).

Question

                                     

A gas contains two types of charged particles. Negatively charged particle \(X^−\) has mass m and velocity \(+v_0\) , as shown in the figure. It collides head-on with positively charged particle \(Y^+\) that has mass 8m and velocity \(−v_0\)

. Electrostatic force then holds the particles together. What is the final velocity of the two-particle system?

A \(+\frac{7}{9}v_0\)

B \(-\frac{7}{9}v_0\)

C \(+v_0\)

D \(-v_0\)

▶️Answer/Explanation

Ans:B

 The system’s momentum before the collision is the same as the momentum after the collision: \(mv_0−8mv_0=(m+8m)v_f\)
, so \(v_f=-\frac{7}{9}v_0\) .

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