A metal baII of mass 2 kg moving with a velocity of 36 km/h has an head-on collision with a stationary ball of 3 kg. If after the collision, the two balls move together, the loss in K.E due to collision is

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40 J

60 J

100 J

140 J

answer is B.

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Detailed Solution

Applying the law of conservation of momentum, we have2×10=(2+3)∨ (∵36km/h=10m/s) ∴ V=4m/s Loss of K.E. =12×2×(10)2−12×5×(4)2 = 60 J

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