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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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thushananth01Best ResponseYou've already chosen the best response.0
I need your help :) @Callisto @VincentLyon.Fr @ganeshie8
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
dw:1354449116877:dw
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
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?
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
yes i got my answer as 0.5886 which is equal to PE after impact
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
(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?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
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
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
wht is the energy loss?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
dw:1354449520295:dw
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
how is KE at position3 = 0?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Velocity at the max point is 0, so KE=0 :
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
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)
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Oh... Is 1.2m the max. height of the rebound?
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
tht wsnt given in the question?
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
@Shadowys can u work it out using equations?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
eh? Since it isn't a perfect rebound, loss of kinetic energy = loss of potential energy =\(mg(hh_o)\) This is the potential energy is turned completely to kinetic energy during the drop, and vice versa during the rebound after hitting the ground.
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
@Shadowys what ws your answer? can u calculate and let me know
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
@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 0.9810.3924= 0.5886
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
why would the Kinetic energy at lowest point = Loss in potential energy?
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy  energy loss due to impact
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
mechanical energy is not really conserved since you've loss of rebound height, right?
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.
 one year ago

thushananth01Best ResponseYou've already chosen the best response.0
one more question K.E at position3 = 0?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
Final mechanical energy = Initial mechanical energy  energy loss. Hence Eloss = mg(h1h2) = 0.40 J with g = 10 m/s²
 one year ago
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