Question

A mixture of two ideal gases, A and B, is in a container. Molecules of gas A have half the mass of molecules of gas B. One molecule of gas A is moving with velocity +v when it collides head-on with a molecule of gas B moving with velocity −v. After the collision, the molecule of gas B moves with velocity +v/3. What is the velocity of the molecule of gas A after the collision?

Answer 1

In an elastic collision between molecule A with half the mass of molecule B and both initially moving at velocity ±v, molecule A will have a velocity of -5v/3 after the collision, according to conservation of momentum and energy.

Step-by-step explanation:

The student is asking about the final velocity of an ideal gas molecule A after it collides head-on with another molecule B, where molecule A has half the mass of molecule B. The scenario is a one-dimensional elastic collision, which conserves both momentum and kinetic energy.

In such a collision, the conservation of momentum and conservation of kinetic energy must both be satisfied. To find the final velocity of molecule A, let mA represent the mass of molecule A and mB represent the mass of molecule B. Given that mA = 0.5 × mB, and using the conservation of momentum:

mA×v + mB×(-v) = mA×v'A + mB×v/3

From conservation of kinetic energy:

0.5×mA×v^2 + 0.5×mB×v^2 = 0.5×mA×v'^2A + 0.5×mB×(v/3)^2

Using algebra, we solve for v'A and find it to be:

v'A = -5v/3

Thus, after the collision with molecule B, the velocity of molecule A will be -5v/3 in the opposite direction to its initial velocity.