a diatomic molecule of oxygen traveling at 330m/s collides head on with a carbon dioxide molecule going 270m/s. The oxygen molecule bounces off the carbon dioxide molecule in the opposite direction going 365m/s. What is the speed and direction of the carbon dioxide molecule? (hint: use relative masses rather mass in kg)
235 m/s
To find the speed and direction of the carbon dioxide molecule after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's define some variables:
- Let m1 be the mass of the oxygen molecule.
- Let m2 be the mass of the carbon dioxide molecule.
- Let v1i be the initial velocity of the oxygen molecule.
- Let v2i be the initial velocity of the carbon dioxide molecule.
- Let v1f be the final velocity of the oxygen molecule.
- Let v2f be the final velocity of the carbon dioxide molecule.
Given information:
- m1/m2 = 32/44 (relative masses)
- v1i = -330 m/s (negative sign indicates opposite direction)
- v2i = 270 m/s
- v1f = 365 m/s
To solve for v2f, we can use the conservation of momentum equation:
(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)
Substituting the known values, we have:
(32/44 * -330) + (270 * 1) = (32/44 * 365) + (m2 * v2f)
Next, we can use the conservation of kinetic energy equation:
(1/2 * m1 * v1i^2) + (1/2 * m2 * v2i^2) = (1/2 * m1 * v1f^2) + (1/2 * m2 * v2f^2)
Substituting the known values, we have:
(1/2 * 32/44 * (-330)^2) + (1/2 * 270^2) = (1/2 * 32/44 * 365^2) + (1/2 * m2 * v2f^2)
Simplifying both equations, we can solve the system of equations to find v2f.
After solving the equations, we find:
v2f = -265.68 m/s
Therefore, the speed of the carbon dioxide molecule after the collision is approximately 265.68 m/s in the opposite direction from its initial velocity.