anonymous
  • anonymous
A ball of mass 50g falls from a height of 2.0m and rebounds to a height of 1.2m. How much kinetic energy is lost on impact?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I need your help :) @Callisto @Vincent-Lyon.Fr @ganeshie8
Callisto
  • Callisto
|dw:1354449116877:dw|
anonymous
  • anonymous
yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Callisto
  • Callisto
At position 1: Total energy = PE = \(mgh_1\) At position 2: Total energy = (KE + energy loss) = total energy at position 1 So far so good?
anonymous
  • anonymous
yes i got my answer as 0.5886 which is equal to PE after impact
anonymous
  • anonymous
(Ep+Ek) at 1 = (Ep + Ek) at 2 mgh = 1/2m(V,)^2 Kinetic energy at the beginning = mgh (Ep+Ek) at 2 = (Ep + Ek) at 3 mgh = mgh1 + 1/2m(v,,)^2 so Loss in kinetic energy = mgh1?
Callisto
  • Callisto
Hmmm.. At position 3: Total energy = PE = \(mgh_2\) = KE at position 2 So, KE at position 2 = 0.05 x 9.81x 1.2 And from (KE + energy loss) = total energy at position 1 0.05 x 9.81x 1.2 + Energy loss = 0.05 x 9.81 x 2.0 So, you can get the energy loss
anonymous
  • anonymous
wht is the energy loss?
Callisto
  • Callisto
|dw:1354449520295:dw|
Callisto
  • Callisto
What do you think?
anonymous
  • anonymous
how is KE at position3 = 0?
Callisto
  • Callisto
Velocity at the max point is 0, so KE=0 :|
anonymous
  • anonymous
note: in perfect bounce, the height of rebound is equal to the original height of drop. So, the loss in potential energy =loss in kinetic energy.(For quick answer)
Callisto
  • Callisto
Oh... Is 1.2m the max. height of the rebound?
anonymous
  • anonymous
tht wsnt given in the question?
anonymous
  • anonymous
@Shadowys can u work it out using equations?
anonymous
  • anonymous
eh? Since it isn't a perfect rebound, loss of kinetic energy = loss of potential energy =\(mg(h-h_o)\) This is the potential energy is turned completely to kinetic energy during the drop, and vice versa during the rebound after hitting the ground.
anonymous
  • anonymous
@Shadowys what ws your answer? can u calculate and let me know
anonymous
  • anonymous
@Callisto Kinetic energy at lowest point = Loss in potential energy 1/2mv^2 = mgh Ek = 0.05*9.81*2=0.981 Ep + Ek at 2 = Ep + Ek at 3 0.981 = 0.05*9.81*1.2 + Ek Ek at 3 = 0.3924J there loss in kinetic energy = K.E before impact - K.E after impact 0.981-0.3924= 0.5886
anonymous
  • anonymous
.392J
anonymous
  • anonymous
why would the Kinetic energy at lowest point = Loss in potential energy?
Callisto
  • Callisto
Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy - energy loss due to impact
anonymous
  • anonymous
Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek
anonymous
  • anonymous
mechanical energy is not really conserved since you've loss of rebound height, right?
anonymous
  • anonymous
ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?
anonymous
  • anonymous
yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.
anonymous
  • anonymous
one more question K.E at position3 = 0?
anonymous
  • anonymous
yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.
anonymous
  • anonymous
thanks:)
anonymous
  • anonymous
you're welcome :)
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Final mechanical energy = Initial mechanical energy - energy loss. Hence Eloss = mg(h1-h2) = 0.40 J with g = 10 m/s²
anonymous
  • anonymous
correct!

Looking for something else?

Not the answer you are looking for? Search for more explanations.