If a skater of weight 3 kg has initial speed 32 m/s and second one of weight 4 kg has 5 m/s. After collision, they have speed (couple) 5 m/s. Then the loss in K.E. is

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48 J

96 J

Zero

None of these

answer is D.

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Detailed Solution

Loss in K.E. = (initial K.E. – Final K.E.) of system 12m1u12+12m2u22−12(m1+m2)V2 =123×(32)2+12×4×(5)2−12×(3+4)×(5)2= 1498.5 J

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