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 

(1) 48 J

(2) 96 J

(3) Zero

(4) None of these

Concept Videos :-

#4 | Definition of Momentum
#5 | Conservation of Momentum
#6 | Problems

Concept Questions :-

Linear momentum

(4) Loss in K.E. = (initial K.E. – Final K.E.) of system

12m1u12+12m2u2212(m1+m2)V2

=123×(32)2+12×4×(5)212×(3+4)×(5)2 

= 986.5 J

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