A) 50%
B) 100%
C) 200%
D) 300%
Correct Answer: B
Solution :
: We know that\[{{p}^{2}}=\]link where p denotes momentum and k denotes kinetic energy of the body. Let initial \[K.E=100A:\] \[\therefore \]Finial \[K.E=400A;\] \[\therefore \] \[p_{1}^{2}=2m(100k)\] \[p_{2}^{2}=2m(400k)\] Fractional change in momentum \[=\frac{{{p}_{2}}-{{p}_{1}}}{{{p}_{1}}}\] % change in momentum\[=\frac{({{p}_{2}}-{{p}_{1}})}{{{p}_{1}}}\times 100\] \[=\left( \frac{{{p}_{2}}}{{{p}_{1}}}-1 \right)\times 100\left( \sqrt{\frac{4}{1}}-1 \right)\times 100\] \[=(2-1)\times 100\] \[\therefore \]Increase in momentum = 100%
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