Two brothers, Dustin and Parker, have a combined mass of 168 kg. Atan ice skatin

Question

Two brothers, Dustin and Parker, have a combined mass of 168 kg. Atan ice skating rink, they stand close together on skates, at restand facing each other, with a compressed spring between them. thespring is kept from pushing them apart because they are holdingeach other. When they release their arms, Dustin moves off in onedirection at a speed of 0.90 m/s, while Parker moves off in theopposite direction at a speed of 1.2 m/s. Assuming that friction isnegligible, find Dustin’s mass.
a. 72 kg
b. 80 kg
c. 96 kg
d. 77 kg
e. 84 kg

Explanation / Answer

   Let Dustin's and Parker's masses be'm1' and 'm2' respectively.    According to law of conservation of linearmoemntum    total initialmomentum   =   total finalmomentum    (m1 + m2) *u   =   m1 *v1 + m2 *v2    initialvelocity   =   0    Dustin's finalvelocity   v1   =   0.90   m/s    Parker's finalvelocity   v2   =   -1.2   m/s                                 (-ve sign to account for opposite direction)    and   m1   +   m2   =   168   kg    =>   m2   =   168- m1    Substituting above values    (m1   +   m2) *0   =   m1 *0.9   +   (168 - m1) * ( -1.2)    =>   0.9 *m1   =   201.6 - 1.2 * m1    Dustin'smass    m1  =   201.6/ ( 1.2 + 0.9)                                  =   96.0 kg    Substituting above values    (m1   +   m2) *0   =   m1 *0.9   +   (168 - m1) * ( -1.2)    =>   0.9 *m1   =   201.6 - 1.2 * m1    Dustin'smass    m1  =   201.6/ ( 1.2 + 0.9)                                  =   96.0 kg

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