Two persons on ice skates stand face to face and then pusheach other away. Their

Question

Two persons on ice skates stand face to face and then pusheach other away. Their masses are 50kg and 60kg. Findthe ratio of their speeds immediately afterward. Which personhad the higher speed? I need to have this solution worked out from start tofinish. Thank you so much Two persons on ice skates stand face to face and then pusheach other away. Their masses are 50kg and 60kg. Findthe ratio of their speeds immediately afterward. Which personhad the higher speed? I need to have this solution worked out from start tofinish. Thank you so much

Explanation / Answer

    Mass of the first person, M1 = 50 kg     Mass of the second person, M2 = 60 kg     Speed of the first person just before push, U1 =0 m/s     Speed of the second person just before push, U2= 0 m/s     Speed of the first person just after push =V1     Speed of the second person just after push =V2     From the law of conservation of momentum,     Total momentum just after push = Total momentumjust before push     M1 V1 + M2 V2 = M1 U1 + M2 U2     M1 V1 + M2 V2 = 0     M1 V1 = - M2 V2     Momenta of both persons are same in magnitudeand opposite in direction just after the push.     Ratio of V1 and V2 ,    V1 / V2 =M2 / M1                                                      = 60 / 50                                                     = 6 / 5     As the velocity is inversely proportional to themass, the person with less mass possess more speed     i.e., Speed of the person of mass 50 kg is morethan the other person.

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