Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Due to friction from sliding against the ground, the player decelerates in 1 direction. Thus his momentum decreases.
Answer : The mass of a sample of water is, 888.89 grams
Explanation :
Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.
Formula used :
where,
q = heat = 2000 kJ = 
(1 kJ = 1000 J)
L = latent heat of vaporization of water = 
m = mass of sample of water = ?
Now put all the given values in the above formula, we get:
(1 kg = 1000 g)
Therefore, the mass of a sample of water is, 888.89 grams
