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An explosion blows a rock into three parts. Two pieces go off at right angles to each other, 1.0 kg piece with velocity of 12 m//s and other , 1.0 kg piece with a velocity of12 m//s and other 2.0 kg piece with a velocity of 8 m//s. If the third piece flies off with a velocity of 40 m//s compute the mass of the third piece. |
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Answer» Solution :Let `m_(1),m_(2)` and `m_(3)` be the MASSES of the three pieces. `m_(1)=1.0 kg, m_2=2.0 kg` Let `v_(1)=12 m//s, v_(2)=8 m//s, v_(3)=40 m//s`. Let `v_(1)` and `v_(2)` be directed along `x`- and `y`-axis respectively, and `v_(3)` be directed as shown. By the principle of coservation of momentum, INITIAL momentum is zero. Hence, along `x`-axis, `0=m_(1)v_(1)-m_(3)v_(3)costheta` along `y`-axis `0=m_(2)v_(2)-m_(3)v_(3) sin theta` `implies m_(1)v_(1)=m_(3)v_(3) costheta` and `m_(2)v_(2)=m_(3)v_(3) sintheta` By squaring and adding we get `m_(1)^(2)v_(1)^(2)+m_(2)^(2)v_(2)^(2)=m_(3)^(2)v_(3)^(2)` `impliesm_(3)^(2)=(1^(2)(12)^(2)+(2)^(2)(8)^(2))/((40)^(2))impliesm_(3)=0.5kg` |
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