Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.
In vacuum, going at 2.99×10^8 m/s.
The molecules which evaporate presumably take heat away from the liquid. So, I'd disagree with the classmate. Whether the amount of cooling would differ from the usual case wherein the molecules have different speeds is another question.
I guess the argument goes something along the lines of that the faster moving and therefore most kinetically energetic molecues evaporate and take away most heat. But if there's no faster moving molecules, 'cos they all have the same speed well, then presumably they'd all take away the same amount of heat. So, maybe the cooling would be less. No cooling though ??? Hmmmm dunno .... i think not ....
1) In any collision the momentum is conserved
(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')
candel all the m factors (because they appear in all the terms on both sides of the equation)
2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')
2) Elastic collision => conservation of energy
=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2
cancel all the 1/2 and m factors =>
2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>
4(vo)^2 = 2(v1')^2 + (v2')^2
now replace (v2') = -2(v1')
=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>
(v1')^2 = [4/6] (vo)^2 =>
(v1')^2 = [2/3] (vo)^2 =>
(v1') = [√(2/3)]*(vo)
Answer: (v1') = [√(2/3)]*(vo)
