A cube moving on a smooth horizontal surface collides inelastically with a small obstacle which is fixed on the ground. The percentage loss in kinetic energy in this collision is
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Given:
The cube collides inelastically with a small obstacle.
To find:
Percentage loss in kinetic energy.
Solution:
Let the mass of cube = m1
Mass of obstacle = m2
Velocity of cube before collision = v1
Velocity of system after collision = v2
From conservation of momentum we get:
m1v1 = (m1+m2)v2
v2 = m1v1/ (m1+m2)
Initial kinetic energy K = 1/2*m1*v1²
Final kinetic energy K' = 1/2*(m1v1/(m1+m2))²
Change in kinetic energy = 1/2 (m1v1/(m1+m2))² - 1/2*m1*v1²
Percentage change = {{1/2 [(m1v1/(m1+m2))² - m1*v1²]} / 1/2m1v1²} *100
= {{ [(m1v1/(m1+m2))² - m1*v1²]} / m1v1²} *100
Therefore the percentage loss in kinetic energy due to collision will be {{ [(m1v1/(m1+m2))² - m1*v1²]} / m1v1²} *100.
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