1. The problem statement, all variables and given/known dataA ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m.* What impulse was given to the ball by the floor? 2. Relevant equationsimpact speed is given by (1/2)mv^2 = mgh1. rebound speed is given by (1/2)mv^2 = mgh2. The impulse of the floor is the change in momentum. mv2(up) - mv1(down) = m(v1 + v2)up 3. The attempt at a solutionusing the speed impact speed equations, I have solved for v in both cases.* The ball hits the ground with a speed of 4.94 m/s.* Using mgh2 from its rebound height, one can find the potential energy and then solve for the velocity that the ball had when leaving the floor, which, algebraically, comes to 4.33 m/s. So then, momentum just before hitting floor is mv = 0.15kg*4.94m/s = 0.741 kg-m/s. momentum just after hitting floor = mv = 0.15kg*4.33m/s = 0.6495 kg-m/s. My textbook describes "impulse" as the change in momentum of an object. So I took the difference, which is .741 - .6495 = 0.0915. The correct answer, according to my book is 1.36 kg-m/s.Why did they add the momentums? My understanding is that impulse, is a difference, not a sum of momenta.I realize that, if one uses Impulse = momentum(final) - momentum(initial), you get the answer (because both - signs make the operation an addition)...but then the whole conecpt isn't very clear to me.It seems to me that the impact of the ball on the floor, is reducing the ball's momentum and velocity, thus, conceptually it's as if the floor is "taking away" energy from the ball. But mathematically, in this problem, it seems we are required to "add' this energy to the initial momentum of the ball....it doesn't make sense to me, I was wondering if someone could clarify the issue.