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Question
Hi
A ball is dropped vertically from restat a height of 12m. After striking the ground, it bounces back to a height of 9m. What fraction of kinetic energy does it lose on striking the ground?
Hi
A ball is dropped vertically from restat a height of 12m. After striking the ground, it bounces back to a height of 9m. What fraction of kinetic energy does it lose on striking the ground?
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Solution
Let, v1 be its velocity while hitting the ground.
Using,
v2 = u2 + 2as
=> (v1)2 = 0 + 2 × 9.8 × 12 = 235.2 m2/s2
Let, v2 e the velocity while leaving the ground.
Therefore,
(v2)2 = 0 + 2 × 9.8 × 9
=> (v2)2 = 176.4
Loss in KE = ½ m(v1)2 – ½ m(v2)2
Fraction of loss in KE = [½ m(v1)2 – ½ m(v2)2]/[ ½ m(v1)2] = [(v1)2 – (v2)2]/[(v1)2]
= (235.2 – 176.4)/235.2
= 0.25
Let, v1 be its velocity while hitting the ground.
Using,
v2 = u2 + 2as
=> (v1)2 = 0 + 2 × 9.8 × 12 = 235.2 m2/s2
Let, v2 e the velocity while leaving the ground.
Therefore,
(v2)2 = 0 + 2 × 9.8 × 9
=> (v2)2 = 176.4
Loss in KE = ½ m(v1)2 – ½ m(v2)2
Fraction of loss in KE = [½ m(v1)2 – ½ m(v2)2]/[ ½ m(v1)2] = [(v1)2 – (v2)2]/[(v1)2]
= (235.2 – 176.4)/235.2
= 0.25
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