A) zero
B) 0.500 v
C) 0.765 v
D) 0.125 v
Correct Answer: C
Solution :
Angle traversed by the particle in \[\left( \frac{1}{8} \right)\] of the circumference of the circle \[=\left( \frac{1}{8} \right)\frac{2\pi r}{r}\] \[=\frac{\pi }{4}={{45}^{0}}\] So, change in velocity, ___ \[\Delta v=\sqrt{{{v}^{2}}+{{v}^{2}}-2{{v}^{2}}\,\cos ({{45}^{o}})}\] \[=\sqrt{2{{v}^{2}}-2{{v}^{2}}\times \frac{1}{\sqrt{2}}}\] \[=v\sqrt{2-\sqrt{2}}\] \[=0.765\,v\]
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