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NEET 2019
Exam
NEET
Year
2019
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NEET 2019 · Physics · Center of Mass, Momentum and Collision

An object flying in air with velocity (20 i +25 j -12 k ) suddenly...

An object flying in air with velocity (20i^+25j^−12k^)(20 \hat{i}+25 \hat{j}-12 \hat{k}) suddenly breaks in two pieces whose masses are in the ratio 1:51: 5. The smaller mass flies off with a velocity (100i^+35j^+8k^)(100 \hat{i}+35 \hat{j}+8 \hat{k}). The velocity of the larger piece will beAn object flying in air with velocity (20 i +25 j -12 k ) suddenly breaks in two pieces whose masses are in the ratio

Options

  1. A4i^+23j^−8k^4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}
  2. B−100i^−35j^−8k^-100 \hat{\mathrm{i}}-35 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}
  3. C20i^+15j^−80k^20 \hat{i}+15 \hat{j}-80 \hat{k}
  4. D−20i^−15j^−80k^-20 \hat{i}-15 \hat{j}-80 \hat{k}

Correct answer

A. 4i^+23j^−8k^4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}

Step-by-step solution

Let mm be the mass of an object flying with velocity vv in air. When it split into two pieces of masses in ratio 1:51: 5, the mass of smaller piece is m/6\mathrm{m} / 6 and of bigger piece is 5 m6\frac{5 \mathrm{~m}}{6}. This situation can be interpreted diagrammatically as below An object flying in air with velocity (20 i +25 j -12 k ) suddenly breaks in two pieces whose masses are in the ratio
As the object breaks in two pieces, so the momentum of the system will remains conserved i.e. the total momentum (before breaking) = total momentum (after breaking) \begin{aligned} & \mathrm{mv}=\frac{\mathrm{m}}{6} \mathrm{v}_1+\frac{5 \mathrm{~m}}{6} \mathrm{v}_2 \Rightarrow \mathrm{v}=\frac{\mathrm{v}_1}{6}+\frac{5 \mathrm{v}_2}{6} \\ & \text{ Here, } \quad \begin{aligned} \mathrm{v} & =20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} \\ \mathrm{v}_1 & =100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \end{aligned}\end{aligned}\begin{array}{rlrl} \Rightarrow & 20 \hat{\mathrm{i}}+25 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} & =\frac{(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})}{6}+\frac{5 \mathrm{v}_2}{6} \\ \Rightarrow & (120 \hat{\mathrm{i}}+150 \hat{\mathrm{j}}-72 \hat{\mathrm{k}}) & =(100 \hat{\mathrm{i}}+35 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})+5 \mathrm{v}_2 \\ \Rightarrow & \mathrm{v}_2 & =\frac{1}{5}(20 \hat{\mathrm{i}}+115 \hat{\mathrm{j}}-80 \hat{\mathrm{k}}) \\ & =4 \hat{\mathrm{i}}+23 \hat{\mathrm{j}}-16 \hat{\mathrm{k}} \end{array}
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