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An object of mass 0.4 kg moving with a velocity of 4ms-1 collides with another object of mass 0.6 kg moving in the same direction with a velocity of 2ms-1. If the collision is perfectly inelastic, what is the loss in K E due to impact?

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m1 = 0.4 kg, u1 = 4 ms-1 m2 = 0.6kg, u2 = 2 ms-1

∴ Total K E of system before collision

ki = \(\frac{1}{2}\)m1u12 + \(\frac{1}{2}\)m2u22

\(\frac{1}{2}\)× 0.4 × 42 + × 0.6 × 22

= 3.2 + 1.2 = 4.4J

∴ As collision is perfectly inelastic, the common velocity after collision v is given by

v = 

= 2.8 ms-1

∴ Total K E of system after collision

kf = \(\frac{1}{2}\)(m1 + m2) v2 = \(\frac{1}{2}\)(1) × (2.8)2

= 3.92 J.

∴ Loss in K E = Δ K E = Δ K = Ki – Kf

= 4.4 – 3.92 J

= 0.48 J.

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