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A ball strikes a horizontal at an angle of 45 degree with the horizontal. If the coefiicient of restitution is 1 divided by square root of 2, then calculate the percentage loss in K.E.(Please explain it in details)

A ball strikes a horizontal at an angle of 45 degree with the horizontal. If the coefiicient of restitution is 1 divided by square root of 2, then calculate the percentage loss in K.E.(Please explain it in details)

Grade:12th pass

2 Answers

reshma lr
20 Points
10 years ago
final vel after collision,v=squqre root of (usinx)^2+(eucosx)^2 x=45 v=square root of 3v^2/4 kE after collision=1/2mv^2=3/4 of kE before collision. loss in ke=i/4 %=25% hope this helped you
nick
8 Points
10 years ago
let V be initial velocity, take components along wall and perpendicular to wall.horizontal component is not affected and vertical component = initial vertical component(vcos45) X coefficient of restitution. loss in kinetic energy = (initial- final/initial) X 100={[vcos45(e)]^2-[vcos45]^2}/[vcos45]^2 X 100= 50%

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